solutions to ch.1-4 of rings of continuous functions by l. gillman and m. jerison
TRANSCRIPT
Solutions forRings of Continuous Functionsby L. Gillman and M. Jerison
version 1.0
Igor Khavkine
June 11, 2011
Preface to v1.0
Gillman and Jerison’s Rings of Continuous Functions is considered one ofthe most important texts in its field. A large part of the material containedin that book is in the form of problems, to be worked out by the reader.This challenge, though very pedagogical, can be daunting to someone whois just being introduced to the topics covered in the problems (e.g. advancedundergraduate or beginning graduate students).
Until now, there has been no comprehensive and readily available collec-tion of solutions for problems given in that book. Producing such a collectionwas the goal of this project. Unfortunately, time constraints have allowedonly the first four chapters to be completed for the current version.
The intended audience for this work is an advanced undergraduate orbeginning graduate student in mathematics or a related discipline, with anintroductory background in topology, abstract algebra and set theory. In thewords of the authors themselves:
This book is addressed to those who know the meaning of eachword in the title: none is defined in the text. [GJ, Preface]
This work attempts to be slightly more gentle to the reader. It aims tobe fairly self contained, when taken together with Gillman and Jerison’sprimary text. Whenever new terminology or a new concept is introduced, itis clearly defined or concisely explained, giving reference to the primary textif necessary.
Some results that might be useful for more than one problem are formu-lated as Lemmas and referenced in solutions whenever necessary. The solu-tions also heavily rely on solved problems preceding them (usually referencedas 1A.1), preceding questions from the same problem (usually referenced as(1.)), as well as results from the primary text (usually referenced as 1.11).Sometimes different questions in the same problem use similar reasoning,which is explained once and implicitly referred to in the questions that fol-low. For this reason, it is helpful to read the solution to each problem in itsentirety, even if the reader is only interested in a single question from thatproblem.
This document was typeset in LATEX with some custom formatting op-tions. The appearance of this document was meant to follow that of theprimary text as closely as possible. There are subtle differences in notation,but those, in the opinion of the author, are self explanatory. The electronic
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versions of the document was designed with hyperlinking capability in mind(for example when viewed in DVI or PDF formats). This feature can greatlyfacilitate navigation, especially when looking up referenced Lemmas or pre-vious problems. Those who intend to add to this work are encouraged tofollow the style of presentation that appears in this version.
I would like to thank my supervisor for this project, Dr. Robert Raphael,1
for his patience and guidance. I would like to thank the people at the McGillSociety of Undergraduate Math Students (SUMS) for providing a pleasantwork environment. I am also grateful to the people from the comp.text.tex
newsgroup for helpful tips and suggestions when LATEX was not cooperating.Last but not least, this work has been made possible by financial support fromthe Natural Sciences and Engineering Research Council of Canada (NSERC).
Distribution and Modification
The primary text contains 16 chapters, of which only four are considered inthis document. Conceivably, this work could evolve into a complete solutionsmanual for Rings of Continuous Functions. However this feat may require alot of time and contributions of many people. It is the intent of the authorto make it possible to contribute to this work as easily as possible.
Another wish of the author is to make these solutions available to any-one who is interested in the material for free distribution and use. Thusthese solutions can be used for a course without seeking explicit permissionof the author. Similar freedom is given to those who intend to extend, refor-mat, or republish this document, with the exception that this Preface staysunmodified.
For all of the above reasons, this work is placed under the GNU FreeDocumentation License (FDL).
Copyright c© 2002 Igor Khavkine.Permission is granted to copy, distribute and/or modify this doc-ument under the terms of the GNU Free Documentation License,Version 1.1 or any later version published by the Free SoftwareFoundation; with no Invariant Sections, with no Front-CoverTexts, and with no Back-Cover Texts. A copy of the licenseis included in the section entitled “GNU Free Documentation Li-cense” (Appendix A).
1Department of Mathematics and Statistics, Concordia University, Montreal, Canada.
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Contents
1 Functions on a topological space 61A p.20 continuity on subsets . . . . . . . . . . . . . . . . . 61B p.20 components of X . . . . . . . . . . . . . . . . . . . . . 71C p.21 C and C∗ for various subspaces of R . . . . . . . . 111D p.21 divisors of functions . . . . . . . . . . . . . . . . . . 161E p.21 units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181F p.22 C-embedding . . . . . . . . . . . . . . . . . . . . . . . . 201G p.22 pseudocompact spaces . . . . . . . . . . . . . . . . . 211H p.22 basically and extremally disconnected spaces . 24
2 Ideals and z-filters 292B p.30 prime ideals . . . . . . . . . . . . . . . . . . . . . . . . 292C p.31 functions congruent to constants . . . . . . . . . 302D p.31 z-ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . 302E p.31 z-ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . 312F p.31 finite spaces . . . . . . . . . . . . . . . . . . . . . . . . 322G p.31 prime vs. z-ideals in C(R) . . . . . . . . . . . . . . . 342H p.32 the identity function i in C(R) . . . . . . . . . . . 352I p.32 C(Q) and C∗(Q) . . . . . . . . . . . . . . . . . . . . . . 372J p.32 ideal chains in C(R), C(Q) and C(N) . . . . . . . . . 372K p.32 z-filters and C∗ . . . . . . . . . . . . . . . . . . . . . 382N p.35 the m-topology on C . . . . . . . . . . . . . . . . . . 38
3 Completely Regular Spaces 443B p.48 countable sets . . . . . . . . . . . . . . . . . . . . . . 443C p.48 Gδ-points of a completely regular space . . . . . 453D p.48 normal spaces . . . . . . . . . . . . . . . . . . . . . . . 463E p.49 nonnormal space . . . . . . . . . . . . . . . . . . . . . 483K p.49 the completely regular, nonnormal space Γ . . 493L p.51 extension of functions from a discrete set . . . 523M p.51 suprema in C(R) . . . . . . . . . . . . . . . . . . . . . 553N p.51 the lattice C(X) . . . . . . . . . . . . . . . . . . . . . 553O p.52 totally ordered spaces . . . . . . . . . . . . . . . . 583P p.53 convergence of z-filters . . . . . . . . . . . . . . . 62
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4 Fixed ideals. Compact space 644A p.60 maximal ideals; z-ideals . . . . . . . . . . . . . . . . 644B p.60 principal maximal ideals . . . . . . . . . . . . . . . . 664C p.61 finitely generated ideals . . . . . . . . . . . . . . . 674D p.61 functions with compact support . . . . . . . . . . 684E p.61 free ideals . . . . . . . . . . . . . . . . . . . . . . . . . 704F p.61 z-ultrafilters on R that contain no small sets . 724G p.61 base for a free ultrafilter . . . . . . . . . . . . . 734H p.62 the mapping τ# . . . . . . . . . . . . . . . . . . . . . . 744I p.62 the ideals Op . . . . . . . . . . . . . . . . . . . . . . . 754J p.62 P -spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 784K p.63 further properties of P -spaces . . . . . . . . . . . 814L p.63 P -points . . . . . . . . . . . . . . . . . . . . . . . . . . 844M p.64 the space Σ . . . . . . . . . . . . . . . . . . . . . . . . 854N p.64 a nondiscrete P -space . . . . . . . . . . . . . . . . . 90
References 91
A GNU Free Documentation License 92
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1 Functions on a topological space
1A p.20 continuity on subsets
Let f ∈ RX .Recall the following equivalent definitions of continuity. Given two topo-
logical spaces X and Y and a function f :X → Y , f is said to be continuous iffor every open set A ⊆ Y , f�[A] ⊆ X is open. Equivalently, f is continuousif for every closed set A ⊆ Y , f�[A] ⊆ X is closed. For future reference, theset of all continuous functions from X to R is denoted by C(X). It is alsoworth noting that below, when appropriate, subsets of X are endowed withthe subspace topology.
1. Let X =⋃ni=1 Ai where each Ai is closed in X. Suppose that for each
i, f |Ai is continuous on Ai. Show that f is continuous on X.
Proof. Let B′ be a closed subset of R, B = f�[B′] and Bi = (f |Ai)�[B′] foreach i. By continuity Bi is closed in Ai, and since Ai is closed in X, Bi isalso closed in X. Note that for each i, Bi = B ∩ Ai, and since Ai cover X,B =
⋃ni=1 Bi. As a finite union of closed sets B must also be closed. Hence
f is continuous on X.
2. Let X =⋃i∈I Ai where each Ai is open in X. Suppose that for each
i, f |Ai is continuous on Ai. Show that f is continuous on X.
Proof. Let B′ be an open subset of R, B = f�[B′] and Bi = (f |Ai)�[B′]for each i. By continuity Bi is open in Ai, and since Ai is open in X, Bi
is also open in X. Note that for each i, Bi = B ∩ Ai, and since Ai coverX, B =
⋃i∈I Bi. As a union of open sets B must also be open. Hence f is
continuous on X.
3. Let T be a family of closed subsets of X with X =⋃T and such that
every x ∈ X has an open neighborhood Nx that meets only finitely manymembers of T . (T is said to be locally finite.) Given that for each T ∈ T ,f |T is continuous on T , show that f is continuous on X.
Proof. For a given x ∈ X, let Nx be an open neighborhood of x which meetsonly the {Ti}ni=1 elements of T . Now for each i, let Si = Nx ∩ Ti. Since foreach i, Ti is closed in X, Si is also closed in Nx. Also, since for each i, f |Ti is
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continuous on Ti, then f |Si is continuous on Si. Now since T forms a coverfor X, whichever elements of T that Nx meets must form a cover for Nx. Sowe can write Nx ⊆
⋃ni=1 Ti and hence Nx =
⋃ni=1 Si. Now we can apply (1.)
to get that f |Nx is continuous on Nx. Obviously X =⋃x∈X Nx, and finally
we use (2.) to get that f is continuous on X.
1B p.20 components of X
First recall a few definitions. Given a topological space X, it is said tobe disconnected if it is possible to write X = A ∪ B where A and B arenon-empty disjoint open sets. Equivalently, X is said to be disconnected ifX = A ∪ B where A and B are non-empty disjoint closed sets. A set issaid to be connected if it is not disconnected. It follows that any topologicalspace can be written uniquely as X =
⋃i∈I Xi where the Xi are non-empty
disjoint maximal (under the ⊆ partial order) connected subsets, such subsetsare called connected components.
For the following problem we will find the following Lemmas useful.
Lemma 1.1. Let X and Y be topological spaces, and f :X → Y a continuousfunction. If X is connected then then so is f [X].
Proof. Suppose that X is connected but f [X] is not. Then we can writef [X] = A ∪ B, where non-empty A and B are open in f [X] and disjoint.Since f :X → Y is continuous, then so is f :X → f [X]. By continuity wehave non-empty f�[A] and f�[B] open in X and also disjoint since A and Bare. But then
X = f�[f [X]] = f�[A ∪B] = f�[A] ∪ f�[B]
But that implies that X is disconnected. This contradiction completes theproof.
Lemma 1.2 (Intermediate Value Theorem). Let X be a connected topologicalspace and f ∈ C(X). Then given x1, x2 ∈ X such that f(x1) = a1 andf(x2) = a2, where WLOG (without loss of generality) a1 ≤ a2, then
∀(a ∈ [a1, a2]) ∃(x ∈ X) (f(x) = a).
Proof. Suppose ∃(a ∈ [a1, a2]) such that ∀(x ∈ X) (f(x) 6= a). Then clearlya 6∈ f [X], and f [X] ⊆ ]−∞, a[ ∪ ]a,+∞[. We know that a1 ∈ ]−∞, a[ since
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a1 ≤ a and a2 ∈ ]a,+∞[ since a ≤ a1, then both of D1 = f [X]∩ ]−∞, a[ andD2 = f [X] ∩ ]a,+∞[ are non-empty, disjoint and open in f [X]. As well asf [X] = D1 ∪ D2, so f [X] is disconnected. Hence, by the above Lemma, Xcannot be connected. This contradiction completes the proof.
Lemma 1.3. Let X be a connected topological space, and u ∈ C(X) bepositive. It is given that there exists a square root s ∈ C(X) such thats2 = u,
∀(x ∈ X)(s(x) =
√u(x) ∨ s(x) = −
√u(x)
).
Where√u(x) denotes the unique positive root of the real number u(x). Then
∀(x ∈ X)(s(x) < 0) ∨ ∀(x ∈ X)(s(x) > 0).
Proof. First it is clear that if s(x) satisfies the given condition at every x ∈ Xthen s2 = u. Now suppose that there are x1, x2 ∈ X such that s(x1) < 0and s(x2) > 0. By Lemma 1.2, s�(0) is non-empty, but for all x ∈ X,|s(x)| =
√u(x) > 0. This contradiction completes the proof.
Lemma 1.4. Let X =⋃i∈I Xi, where the {Xi} are disjoint and and open.
Given fi ∈ C(Xi) for each i ∈ I, there exists a unique function f ∈ C(X)such that f |Xi = fi for each i ∈ I.
Proof. Uniqueness Given two such functions f, g ∈ C(X), by hypothesis,they agree on each Xi. Since {Xi} is a cover for X, f and g agree on X, orsimply f = g.
Existence Let f ∈ RX be defined such that f |Xi = fi for each i ∈ I, sincethe {Xi} are disjoint and cover X, f is well defined. Note that by continuity,each Ai is open, hence continuity of f follows from 1A.2.
1. Show that in C(X), all positive units have the same number of squareroots.
Proof. Let u, v ∈ C(X) be any two positive units. We want to establish abijection between the square roots of u and v. Let w = uv, w is also a positiveunit. Hence there must exist a unique positive square root r ∈ C(X), suchthat r2 = w. Let s ∈ C(X) be any square root of u. We can map s to a uniqueelement t ∈ C(X), s 7→ t = r/s. Note that t2 = r2/s2 = w/u = uv/u = v,hence t is a square root of v. This mapping is injective since the inverse existsand is obviously t 7→ r/t = s. It is also surjective because for any t2 = v,
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s = r/t is such that s2 = u and s 7→ t. Hence the mapping is a bijection, andthis establishes that the number of square roots for any pair of, and henceall, positive units in C(X) is the same.
2. Show that X is connected iff 1 has exactly two square roots.
Proof. Necessity Suppose s2 = 1, then ∀(x ∈ X)(s(x) = 1∨s(x) = −1). If Xis connected, then according to Lemma 1.3, there are only two possibilities.Either s(x) = 1 for all x ∈ X, or s(x) = −1 for all x ∈ X, both of which arein fact square roots of 1. Hence 1 has exactly two square roots.
Sufficiency Suppose that 1 has exactly two square roots, but X is disjoint.Then we can write X = A ∪ B where A and B are non-empty, disjoint andopen in X. But then, by Lemma 1.4, we can define at least four functions hon X of the form h|A = ±1 and h|B = ±1, each of which is a square root of1 on X. Hence 1 cannot have exactly two square roots. This contradictioncompletes the proof.
3. Show that for finite m, X has m connected components iff 1 has exactly2m roots. Also, show that this statement is false if m is infinite.
Proof. Necessity Assume that X has m connected components. Since m isfinite each component is open. If we have s ∈ C(X) such that s2 = 1, thens(x) = ±1 for each x ∈ X. By Lemma 1.3, s must be either positive ornegative on each component of X, so we can have at most 2m choices for s.By Lemma 1.4, we have that each one of those choices is in fact in C(X),hence 1 has exactly 2m roots.
Sufficiency Assume that 1 has exactly 2m roots. If X has n connectedcomponents where n is finite. Then from the above, 1 has exactly 2n roots.From the hypothesis 2n = 2m, which implies that n = m.
Conterexample. Now suppose that m is infinite, we take for example
X = {1, 1/2, . . . , 1/n, . . . , 0}
where m = ℵ0. Suppose s ∈ X and s2 = 1. Then there are two choicess(0) = 1 or s(0) = −1. For each choice of s(0), by continuity there mustexist an N ∈ N such that for each n > N , s(1/n) = s(0). If we fix s(0)and n, then there are 2n−1 possible choices for s, and all of the possibilities
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for n′ < n are already counted in that number. Hence the total number ofsquare roots of 1 is
2 · supn>N
2n−1 = 2 · ℵ0 = ℵ0.
But ℵ0 < 2ℵ0 = 2m (by Cantor’s theorem), this completes the proof.
4. Show that X is connected iff 0 and 1 are the only idempotents inC(X). (u ∈ C(X) is idempotent if u2 = u).
Proof. Necessity If u ∈ C(X) is idempotent, then for each x ∈ X, u(x) = 0or u(x) = 1. Assume that X is connected. Since u is idempotent it is its ownsquare root, so by Lemma 1.3, if for some x ∈ X, u(x) = 1, then u = 1. Onthe other hand if u is idempotent, but for all x ∈ X, u(x) 6= 1, then u = 0.Hence the only idempotents in C(X) are 0 and 1.
Sufficiency Assume that 0 and 1 are the only idempotents in C(X), butX is disconnected. Then we can write X = A ∪ B, where A and B are dis-joint, non-empty and open. Then we can construct at least four idempotentfunctions of the form u|A = 0 or 1 and u|B = 0 or 1, all of which are inC(X) by Lemma 1.4. This contradiction completes the proof.
5. If X is connected, then C(X) cannot be written as C(X) ∼= R ⊕ S,were R and S are non-trivial rings.
Proof. Suppose that C(X) ∼= R ⊕ S where R and S are non-trivial rings.Then we can write 1 = 1R ⊕ 1S. But now, we can construct at least foursquare roots of 1 of the form ±1R ⊕ ±1S. Hence, by 1B.2, X cannot beconnected. This contradiction completes the proof.
6. If we can write X = A ∪ B, where A and B are disjoint, non-emptyand open, then C(X) ∼= C(A)⊕ C(B).
Proof. Any f ∈ C(X) can be written as f = fA + fB, where fA|A = f |Aand fB|B = f |B, while fA|B = 0|B and fB|A = 0|A (fA, fB ∈ C(X) byLemma 1.4). Hence there is a surjective map C(A) ⊕ C(B) → C(X) ofthe form fA|A ⊕ fB|B 7→ fA + fB. On the other hand this map has aninverse given by C(X) → C(A) ⊕ C(B) of the form f 7→ f |A ⊕ f |B, so itis also injective and hence bijective. It is trivial to check that this map is ahomomorphism, and since it is bijective, it is also an isomorphism. HenceC(X) ∼= C(A)⊕ C(B).
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1C p.21 C and C∗ for various subspaces of RConsider the subspaces R, Q, N and N∗ = 1, 1/2, . . . , 1/n, . . . , 0 of R, and therings C and C∗ for each of these spaces. Each of these rings is of cardinalityc = 2ℵ0 .
For the following problem, we will find the following Lemmas useful.
Lemma 1.5. Let X and Y be topological spaces. If X is homeomorphic toY , then C(X) ∼= C(Y ).
Proof. Let f ∈ C(X), we construct a homomorphism C(X)→ C(Y ), of theform f 7→ f ◦ φ where φ:Y → X is a homeomorphism. This homomorphismhas an inverse of the form f ◦ φ 7→ (f ◦ φ) ◦ φ� = f , hence it is injective.Also, for any g ∈ C(Y ), we can write g = (g ◦ φ�) ◦ φ, where g ◦ φ� ∈ C(X),hence the homomorphism is surjective. So the homomorphism is bijective,and hence it is an isomorphism or C(X) ∼= C(Y ).
1. For each m ≤ ℵ0, each ring on R, N or N∗ contains a function havingexactly 2m square roots. If a member of C(Q) has more than one square root,it has exactly c of them.
Proof. Case R: Consider the function f ∈ C∗(R) defined by
f(x) =
{| sin(πx)| if x ∈ [0,m]0 if x 6∈ [0,m]
If m = ℵ0, then we replace [0,m] by [0,∞[. Let Ai = ](i− 1), i[ fori ∈ N and i ≤ m, and A =
⋃i≤mAi.
Case N: First recall that all functions from N to R are continuous. Nowconsider f ∈ C∗(N) defined by
f(x) =
{1 if 1 ≤ x ≤ m0 if x > m
Let Ai = {i} for i ∈ N and i ≤ m, and A =⋃i≤mAi.
Case N∗: Consider the function f ∈ C∗(N∗) defined by
f(1/n) =
{1/n if 1 ≤ n ≤ m0 if n > m or x = 0
f(0) = 0
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Let Ai = {1/i} for i ∈ N and i ≤ m, and A =⋃i≤mAi.
For all three cases, consider f |A, by Lemmas 1.3 and 1.4, f |A has exactly2m continuous square roots since the {Ai} are disjoint and open (both forfinite and infinite m). For the first case, the square root of f |A can becontinuously extended to the whole of R since the square root of f mustidentically vanish on R \A. Since C∗(R) ⊆ C(R), the same f works in C(R)as well.
Case Q: Consider any non-negative f ∈ C(Q). If f = 0, then its squareroot is unique and is itself. If f 6= 0, then consider the set A = pos f . A isopen, and it is a basic topological property of Q (and R) is that any open setcan be written as A =
⋃∞i=1Ai, where the {Ai} are disjoint open (bounded)
intervals.Pick a nonempty Ai = ]a, b[, where a and b are real numbers. Now we
construct monotone decreasing sequence {cn} contained in ]a, b[. Let c1 = b,then we can always find an irrational number c2 such that a < c2 < c1. Thesequence {cn} is constructed continuing by induction.
Now we define the disjoint open sets {Cn}, where Cn = ]cn+1, cn[, andC0 = ]a, infn cn[. Note that if C =
⋃∞n=0Cn, then C ∩ Q = Ai ∩ Q, so C
and Ai denote the same open set in Q. By Lemmas 1.3 and 1.4, there areexactly 2ℵ0 = c possible square roots for f |C. Each of those square rootscan be continuously extended to Q\C, e.g. with the unique non-negativeroot of f |(Q\C). Hence the number of square roots of any non-negative non-zero f ∈ C(Q) is bounded below by c, it is also bounded above by c (thecardinality of C(Q)), hence it is equal to c. So any non-negative f ∈ C(Q)has either one or c square roots.
2. C(R) has just two idempotents, C(N∗) has exactly ℵ0, and C(Q) andC(N) have exactly c.
Proof. Case C(R): R is connected, hence by 1B.4 its only idempotents are0 and 1.
Case C(N∗): If s ∈ C(X) is idempotent, then for each x ∈ X, s(x) = 0 ors(x) = 1. To show that C(N∗) has exactly ℵ0 idempotents repeat theargument in the second part of 1B.3, except replace −1, 1 by 0, 1 aspossible values of s(x).
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Case C(N) and C(Q): We can write both N and Q as a union of ℵ0 disjointopen sets. In particular N =
⋃n∈N{n}, and
Q = Q∩
[]−√
2,√
2[ ∪⋃n∈N
(]−(n+ 1)
√2,−n
√2[ ∪ ]n
√2, (n+ 1)
√2[)].
Hence by setting s to either 0 or 1 on each of these open sets, s isidempotent and s ∈ C(N) (resp. s ∈ C(Q)), by Lemma 1.4. Hencethe number of idempotents in each of the rings in question is at least2ℵ0 = c. And because of the cardinality of the rings, this number isalso bounded by c from above. Hence the total number of idempotentsin C(N) and C(Q) is c.
3. Every non-zero idempotent in C(Q) is a sum of two non-zero idem-potents. In C(N), and in C(N∗), some, but not all idempotents, have thisproperty.
Proof. Case C(Q): Let u ∈ C(Q) be a non-zero idempotent, hence posu isopen and non-empty. Hence posu can be written as a countable unionof disjoint non-empty open intervals. Let A = ]a, b[ be one of theseintervals, where a and b are real numbers. Note that we can alwaysfind an irrational number c such that a < c < b. Then we defineA1 = ]a, c[ and A2 = ]c, b[, so in the topology on Q, the sets A andA1 ∪ A2 are the same open set. Now we define u1, u2 ∈ RQ such thatu1|]−∞, c[ = u|]−∞, c[ and u1|]c,∞[ = 0, and u2|]c,∞[ = u|]c,∞[ andu2|]−∞, c[ = 0. Clearly u = u1 + u2, both u1 and u2 are non-zero,idempotent and u1, u2 ∈ C(Q), by Lemma 1.4.
Case C(N) and C(N∗): Consider the following functions u ∈ C(N) andv ∈ C(N∗)
u(x) =
{1 if x = 10 if x > 1
v(x) =
{1 if x = 10 if x < 1
Clearly, both u and v are idempotent and cannot be written as a sumof two non-zero idempotents.
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Now consider the following functions u ∈ C(N) and v ∈ C(N∗)
u(x) =
{1 if x = 1 or x = 20 if x > 2
v(x) =
{1 if x = 1 or x = 1/20 if x < 1/2
On the other hand, these idempotents can be written very simply as asum of two non-zero idempotents.
4. Except for the obvious identity C(N∗) = C∗(N∗), no two of the ringsin question are isomorphic.
Proof. C∗ is not isomorphic to C: First note that no C∗ is not isomor-phic to the corresponding C (except the afore mentioned case), sinceby 1.7 bounded functions are always taken to bounded functions byhomomorphisms.
C(R) � C(Q,N, N∗) (resp. C∗): Also, note that neither of C(R) or C∗(R)is isomorphic to any of the other rings in question since f [R] is con-nected for any f ∈ C(R), because R is connected, but any one of theother rings has a function with a disconnected image.
C, (N∗) � C(Q,N) (resp. C∗): Recall from 1C.2 that C and C∗ on N∗ haveexactly ℵ0 idempotents each, while C and C∗ on N and Q have exactlyc idempotents each. So since the number of idempotents is preservedunder isomorphism, the respective rings cannot be isomorphic.
C(N) � C(Q) (resp. C∗): Now recall from 1C.3 that every non-zero idem-potent in C of C∗ on Q can be written as a sum of two non-zeroidempotents from the same ring, while the same is not true for eitherof C or C∗ on N. This property is also preserved under isomorphism,hence the respective rings cannot be isomorphic. This completes theproof.
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5. Each of C(Q) and C(N) is isomorphic with a direct sum of two copiesof itself. C(N∗) is isomorphic with a direct sum of two subrings, just one ofwhich is isomorphic with C(N∗).
Proof. Case C(N): Let Neven = 2N and Nodd = −1 + 2N. N is naturallyhomeomorphic to both Neven and Nodd, hence, by Lemma 1.5, the re-spective rings are isomorphic. Finally by 1B.6, C(N) ∼= C(Neven) ⊕C(Nodd).
Case C(Q): Let Q− = ]−∞,√
2[∩Q and Q+ = ]√
2,∞[∩Q. We constructhomeomorphisms h±:Q→ Q±. They are defined as follows
h−: q 7→ q if q ≤ 0
h+: q 7→ q + 2 if q ≥ 0
Now suppose {an} is a monotone increasing sequence of rationals con-verging to
√2 with a1 = 0, and {bn} is a monotone decreasing sequence
converging to√
2 with b1 = 2. Then for each n ∈ N let
h−: q 7→ (q − (n− 1))(an+1 − an) + an if n− 1 < q ≤ n
h+: q 7→ ((n− 1)− q)(bn+1 − bn) + bn if −n ≤ q < −n+ 1
Since h± are homeomorphisms, then, by Lemma 1.5, the respectiverings are isomorphic. Finally by 1B.6, C(Q) ∼= C(Q−)⊕ C(Q+).
Case C(N∗): We can write N∗ = A ∪ B, where A = {1, 1/2, . . . , 1/n} andB = {1/(n + 1), . . . , 0} are open, non-empty and disjoint. There isa natural homeomorphism h:N∗ → B, given by h:x 7→ 1/(x−1 + n),hence, by Lemma 1.5, the rings C(N∗) and C(B) are isomorphic. Fi-nally by 1B.6, C(N∗) ∼= C(A)⊕ C(B).
6. The ring C(R) is isomorphic with a proper subring. But C(R) has noproper summand.
Proof. Consider the subring R ⊆ C(R) consisting of all functions constanton the interval [0, 1] (the fact that R is actually a subring is easy to see). Wecan construct a homomorphism h:C(R)→ R defined by h: f 7→ g, where
g(x) =
f(x) if x < 0f(0) if 0 ≤ x ≤ 1f(x− 1) if 1 < x
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This homomorphism is injective since it has an inverse given by h�: g 7→ f ,where
f(x) =
{g(x) if x ≤ 0g(x+ 1) if 0 < x
Since the inverse exists for every g ∈ R, then h is also surjective. So his bijective, and hence an isomorphism. The fact that C(R) has no propersummand follows directly from the fact that R is connected and 1B.5.
1D p.21 divisors of functions
1. If Z (f) is a neighborhood of Z (g), then f is a multiple of g—that is,f = hg for some h ∈ C. Furthermore, if X\ intZ (f) is compact, then h canbe chosen to be bounded.
Proof. Let Z ′ = X\Z (f), and h ∈ RX , defined by
h(x) =
{0 if x ∈ Z (f)f(x)/g(x) if x ∈ Z ′
Note that f, g are both continuous on clZ ′, and |g| > 0 on clZ ′ (since Z (g) ⊆intZ (f) = X\ clZ ′), hence h is continuous on clZ ′. It is also continuouson Z (f) (because it is constant). Now we can write X = Z (f) ∪ clZ ′ as aunion of two closed sets, on each of which h is continuous. Hence, by 1A.1,h ∈ C(X) and f = hg. Note that X\ intZ (f) = clZ ′, if it is compact thenh has to be bounded on clZ ′, and hence bounded everywhere on X.
2. Construct an example in which Z (g) ⊆ Z (g), but f is not a multipleof g. Consider X = R and f, g ∈ C(R) defined by
Example.
f(x) =
x if x < 00 if 0 ≤ x ≤ 1x− 1 if 1 < x
g(x) =
x2 if x < 00 if 0 ≤ x ≤ 1(x− 1)2 if 1 < x
Then Z (f) = Z (g) = [0, 1]. Suppose there is an h ∈ C(R) such that f = hg.Then h is uniquely defined on A = R\[0, 1]. However h is then unbounded on
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every neighborhood of [0, 1] in A, and hence cannot be continuously extendedto all of R. Therefore f is not a multiple of g.
3. If |f | ≤ |g|r for some real r > 1, then f is a multiple of g. Hence if|f | ≤ |g|, then f r is a multiple of g for every r > 1 for which f r is defined.
For this problem it is worth recalling the definition of continuity at apoint. Let X and Y be a topological spaces. Let f ∈ Y X . The function fis said to be continuous at x ∈ X if for any neighborhood U ⊆ Y such thatf(x) ∈ U , there is an neighborhood V ⊆ X such that x ∈ V and f [V ] ⊆ U .
We also introduce a piece of notation. Let S be a subset of a topologicalspace X, then the boundary of S is clA ∩ cl(X\S). For convenience theboundary of S is denoted by ∂S. Equivalently ∂S = (clS)\ intS. It is alsoworth noting that clS = S ∪ ∂S.
Proof. Let Z ′ = X\Z (g) (from the hypothesis we also have Z (g) ⊆ Z (f)),and h ∈ RX , defined by
h(x) =
{0 if x ∈ Z (g)f(x)/g(x) if x ∈ Z ′
Clearly f = gh. Since f and g are continuous and |g| > 0 on Z ′, we knowthat h is continuous on Z ′. It is also clear that h is continuous on Z (g). Sosince X = Z (g) ∪ clZ ′, it is sufficient to show that h is continuous on clZ ′
and use 1A.1.Suppose that x ∈ ∂Z ′, then h(x) = 0, f(x) = 0 and g(x) = 0. Let ε > 0,
U = ]−ε, ε[ and V = ]−ε1/(r−1), ε1/(r−1)[. Now since g is continuous, we canfind an open subset W ⊆ X such that x ∈ W and g[W ] ⊆ V . But, by thehypothesis, for any y ∈ W ∩ Z ′ we have
|h(y)| = |f(y)/g(y)| ≤ |g(y)|r−1 < ε,
and h(y) = 0 for each y ∈ W ∩ ∂Z ′. Since W is open, the set W ∩ clZ ′
is open in clZ ′ and by the above h[W ∩ clZ ′] ⊆ U , which implies that h iscontinuous in clZ ′ at every x ∈ ∂Z ′. Finally, this last fact implies that h iscontinuous on clZ ′ and completes the proof.
If |f | ≤ |g|, the for each r > 1 we have |f |r ≤ |g|r or |f r| ≤ |g|r (providedthat f r is defined). By the above argument f r must be a multiple of g.
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1E p.21 units
1. Let f ∈ C. There exists a positive unit u of C such that
(−1 ∨ f) ∧ 1 = uf.
Proof. Let A = {x ∈ X | f(x) ≤ −1} and B = {x ∈ X | f(x) ≥ 1}. Letu ∈ RX , defined by
u(x) =
{|f(x)|−1 if x ∈ A ∪B1 if x 6∈ A ∪B .
Note that u satisfies the hypothesis. For u to be continuous, it must becontinuous on each of A,B, cl(X\(A∪B)) (since they form a closed cover ofX). It is clear that u is continuous on both A and B, it is also continuouson cl(X\(A ∪ B)) since f [∂(A ∪ B)] = {1}, which matches with the valueof u on the boundary of cl(X\(A ∪ B)). Hence u is continuous on X, by1A.1.
2. TFAE (the following are equivalent)
(1) For every f ∈ C, there exists a unit u of C such that f = u|f |.
(2) For every g ∈ C∗, there exists a unit v of C∗ such that g = v|g|.
Proof. First, assume (1). Note that for f ∈ C(X) and unit u of C(X),we have f = u|f | iff |u(x)| = 1 for each x such that f(x) 6= 0 (in factu|(pos f) = 1 and u|(neg f) = −1), and u(x) can take on any value (as longas u is continuous) for each x where f(x) = 0. If u = 1 or u = −1 then u isalready a unit of C∗. If u takes on both values 1 and −1, and since u cannottake on the value 0 because u is a unit, we can write X = posu ∪ neg u andposu and neg u are disconnected, by the contrapositive of the IntermediateValue Theorem (Lemma 1.2). Therefore we can define a unit v of C∗ byv|(posu) = 1 and v|(neg u) = −1, and we still have f = v|f |. Hence, (1)implies (2).
Now, assume (2). Let f ∈ C, then there is g ∈ C∗ such that pos g = pos gand neg g = neg f (for example g = (−1∨f)∧1). Then there is a unit v ∈ C∗such that g = v|g|. Then the same unit v works for f as well, since f and gshare the same pos and neg sets, that is f = v|f |. Hence (2) implies (1).
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3. Describe the functions f in C(N) for which there exists a unit u ofC(N) satisfying f = u|f |. Do the same for C(Q) and C(R).
Description. Case C(N): Let f ∈ C(N) and u ∈ RN. Let u be defined as
u(x) =
{1 if x ∈ pos f ∪ Z (f)−1 if x ∈ neg f
.
Then u(x) satisfies the hypothesis, and since RN = C(N), we are done.
Case C(Q): Let f ∈ C(Q) and u ∈ RQ, defined by
u(x) =
{1 if x ∈ pos f−1 if x ∈ neg f
.
As long as u can be continuously extended to the rest of Q and stillremain a unit, we are done. By 1.15, u can be continuously extendedto all of Q iff posu = pos f and neg u = neg f are completely separated(have disjoint zero-set neighborhoods). Since Q is a metric space, by1.10, every closed set is a zero set, the above condition is equivalent topos f and neg f having disjoint closures (in fact this is also sufficientfor u to be a unit, since we can always make u assume the value 0 onlyat irrational numbers).
Case C(R): Let f ∈ C(R) and u ∈ C(R), where u satisfies the hypothesis.If there are x, y ∈ R such that u(x) = 1 and u(y) = −1, by theIntermediate Value Theorem (Lemma 1.2), for some x < z < y wehave u(z) = 0, then u cannot be a unit. Hence the condition on f iseither f ≤ 0 or f ≥ 0.
4. Do the same for the equation f = k|f |, where k belongs to C but isnot necessarily a unit.
Description. Case C(N): The argument from (3.) is still applicable.
Case C(Q): The argument from (3.) is still applicable.
Case C(R): In this case k may assume the value 0, so since R is a metricspace, the same conditions as for Q apply to R as well.
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1F p.22 C-embedding
1. Every C∗-embedded zero-set is C-embedded.
Proof. Let Z ∈ Z (X) be a C∗-embedded zero-set in X and f ∈ C∗(Z), thenthere is a function g ∈ C∗(X) such that g|Z = f . By 1.18, Z is C-embeddediff Z is completely separated from every zero-set disjoint from it. But, by1.15 two disjoint zero-sets are completely separated.
2. Let S ⊆ X; if every zero-set in S is a zero-set in X, then S is C∗-embedded in X.
Proof. Consider any two completely separated sets in S. They have disjointzero-set neighborhoods in S, which are also zero-sets in X. Hence these twosets are completely separated in X as well. So by Urysohn’s Theorem (1.17),the set S is C∗-embedded in X.
3. A discrete zero-set is C∗-embedded iff all of its subsets are zero-sets.
Proof. Necessity Suppose that a zero-set Z ∈ Z (X) is discrete and C∗-embedded. Now consider a subset Y ⊆ Z. Since Z is discrete, Y is alsodiscrete. So we can define a continuous function f ∈ C∗(Z) as
f(x) =
{0 if x ∈ Y1 if x 6∈ Y
Since Z is C∗ embedded, there is a function f ′ ∈ C∗(X) such that f ′|Z = f .Also, since Z is a zero set, there is a function g ∈ C∗(X) such that Z = Z (g).Then we can define another function h ∈ C∗(X) as h = (g2 + f ′2). Note thatZ (h) = Z ∩ Z (f ′) = Y . Hence every subset of Z is also a zero set in X.
Sufficiency Suppose that every subset Y ⊆ Z is also a zero-set in X. Anysubset of a discrete space is a zero-set. So any two disjoint subsets of Z arecompletely separated (they are their own zero-set neighborhoods). Also anytwo disjoint subsets of Z are zero sets in X. But that just means that theyare completely separated in X. Hence, by Urysohn’s Theorem (1.17), Z isC∗-embedded.
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4. A subset S of R is C-embedded [resp. C∗-embedded] iff it is closed.
Proof. Necessity Assume that the set S ⊆ R is C-embedded in R. Nowsuppose that S is not closed, then clS\S is non-empty. Take x ∈ clS\S,then there is a sequence {xn} ⊆ S such that limn→∞ xn = x (WLOG assumethat xn is monotone and increasing, a similar argument can be repeated fora monotone decreasing sequence). We define a function f ∈ C(R\{x}) asfollows
f(y) =
−1 if y ≤ x1
(−1)n + 2(y − xn)
(xn+1 − xn)if xn ≤ y < xn+1
0 if y > x
.
Since f is continuous and S ⊆ R\{x}, then f |S is also continuous. However,the function f cannot be continuously extended to all of R since no value off at x will make f continuous on R. For example limn→∞ f(x2n−1) = −1and limn→∞ f(x2n) = 1. This contradiction implies that S must be closed.
Sufficiency Suppose that S is closed in R. Then i|S is a homeomorphismwhich carries S to a closed set in R (that is S itself). So, by 1.19, S isC-embedded in R.
Since f is bounded, the exact same arguments apply to C∗.
5. If a (non-empty) subset S of X is C-embedded in X, then C(S) is ahomomorphic image of C(X). The corresponding result holds for C∗.
Proof. Consider the map φ:C(X) → C(S) defined by φ: f 7→ f |S. Clearly,the mapping φ is a homomorphism. Since S is C-embedded in X, then foreach function f ∈ C(S) there is a function g ∈ C(X) such that g|S = f .Hence φ is onto. This completes the proof.
1G p.22 pseudocompact spaces
First we recall a couple of definition. Let X be a topological space. X issaid to be Hausdorff if any pair of distinct points x, y ∈ X there is a pairof disjoint open sets one containing only x and the other only y. Also, atopological space X is said to be pseudocompact if C(X) = C∗(X).
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1. Any continuous image of a pseudocompact space is pseudocompact.
Proof. Let X and Y be topological spaces, of which X is pseudocompact.Let the function f ∈ Y X be continuous. Now consider W = f [X] and letg ∈ C(W ). Suppose that g is unbounded on W , then h = g ◦ f is alsounbounded. But then h ∈ C(X) but h 6∈ C∗(X), hence C(X) 6= C∗(X) andX cannot be pseudocompact. This contradiction completes the proof.
2. X is pseudocompact iff f [X] is compact for every f in C∗(X).
Proof. Necessity Suppose that X is pseudocompact. By the Heine-Borel the-orem, the set f [X] ⊆ R is compact iff it is closed and bounded. Boundednessof f [X] implies that f ∈ C∗(X). Now suppose that f [X] is not closed. Thenit is possible to construct an unbounded function g: f [X] → R (see argu-ment for 1.F4). But then g ◦ f ∈ C(X) and is unbounded, so X cannot bepseudocompact. This contradiction completes the proof.
Sufficiency Suppose that for each f ∈ C∗(X), the set f [X] ⊆ R is com-pact. Now suppose that X is not pseudocompact, then there is an unboundedfunction g ∈ C(X). But then tan−1 ◦g ∈ C∗(X), and f [X] is not closed (andhence not compact). This contradiction completes the proof.
3. Let X be Hausdorff space. If, of any two disjoint closed sets, at leastone is compact, or even countably compact, then X is countably compact.
For this problem, we need the following topological fact: a Hausdorffspace is countably compact iff every infinite every set has a limit point.
Proof. Suppose that X is not countably compact. Then there is an infiniteset A with no limit point. Take any point x ∈ X outside of A, then since Ahas no limit point, there is an open neighborhoodN of x such thatN∩A = ∅.Hence x 6∈ clA, which implies that A = clA or that A is closed. Since Ais infinite, it must have a countable subset {xn}. Consider the followingsubsets {yk = x2k} and {zk = x2k−1}, both of which are closed by the sameargument as above. Hence {yk} and {zk} are disjoint closed sets, so by thehypothesis, at least one of them must be compact. But since neither {yk} nor{zk} have limit points, each of their points must have an open neighborhooddisjoint from all other points in the same set, hence they must be discrete.But discrete infinite sets are not compact. This contradiction completes theproof.
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4. If, of any two disjoint zero-sets in X, at least one is compact, or evenpseudocompact, then X is pseudocompact.
It is fruitful to recall the following definition and topological fact. LetX be a topological space and (Y, d) a metric space. Given a sequence offunctions {fn} ⊆ Y X and a function f ∈ Y X , the sequence is said to convergeuniformly to f if for each compact subset K ⊆ X and for each ε > 0 thereis an N ∈ N such that d(f(x), fn(x)) < ε for each n > N and x ∈ X.Also recall that if {fn} are continuous functions from X to Y with f theirpointwise limit, then f is continuous iff {fn} converges uniformly to f .
Proof. Suppose that X is not pseudocompact, then there exists an unbound-ed continuous function f on X (WLOG suppose that f is non-negative). Weconstruct a sequence of real numbers {an} by induction:
a1 = inf ([1,∞[ ∩ f [X])
an+1 = inf ([2an + 1,∞[ ∩ f [X]) for n > 1.
Note that this sequence was chosen so that the subsequences {bk = a1−a2 +· · · − a2(k−1) + a2k−1} and {ck = a1 − a2 + · · · + a2k−1 − a2k} are monotone,respectively increasing and decreasing, and unbounded. Since f is unboundedand non-negative, the sequence {an} is guaranteed to exist and tend to ∞as n → ∞. Now we construct a sequence of continuous functions {gn} byinduction:
g0 = f
g1 = (a1 ∧ f)− (a1 ∨ f − a1)
g2k = (−a2k ∨ g2k−1)− (−a2k ∧ g2k−1 + a2k)
g2k+1 = (a2k+1 ∧ g2k)− (a2k+1 ∨ g2k − a2k+1)
[Try to perform this procedure for f = 1/x on ]0,∞[ to see what it does].Suppose K ⊆ X is compact, then for each n, the restricted function f |K
is continuous and hence must be bounded. Since f |K is bounded, then forN ∈ N such that aN > sup f |K, the tail sequence {gn|K}n>N is constant.Which implies that {gn|K} converges uniformly on each compact K ⊆ X,since each gn is continuous. The last statement implies that the functiong(x) = limn→∞ gn(x) is also continuous.
Finally we take two disjoint zero-sets A = {x ∈ X | g(x) ≥ 1} andB = {x ∈ X | g(x) ≤ −1}, which are non-empty by construction. By the
23
hypothesis, at least one of A or B must be pseudocompact, but both ofthem have unbounded functions defined on them, namely g|A and g|B. Thiscontradiction completes the proof.
1H p.22 basically and extremally disconnectedspaces
A space X is said to be extremally disconnected if every open set has an openclosure; X is basically disconnected if every cozero-set has an open closure.
1. X is extremally disconnected iff every pair of disjoint open sets havedisjoint closures. What is the analogous condition for basically disconnectedspaces?
Proof. Necessity Suppose that X is extremally disconnected. Let A and Bbe open and disjoint subsets of X. Then X\A is closed and contains B, sothat we have clB ⊆ X\A. But since X is extremally disconnected, clB isopen. Then X\ clB is closed and contains A, so that we have clA ⊆ X\ clB,hence A and B have disjoint closures.
Sufficiency Suppose that any two disjoint open subsets of X have disjointclosures. Take A ⊆ X to be any open set. Then A and int(X\A) must havedisjoint closures. There are two possibilities, either int(X\A) is empty ornot. In the latter case, we have clA ∩ cl(int(X\A)) = ∂A, hence ∂A = ∅,which implies that clA = A ∪ ∂A = A is open. In the former case, we haveclA = X, which is open. Therefore X is extremally disconnected.
The analogous condition for a basically disconnected space is as follows:the space X is basically disconnected iff for every pair of disjoint subsets Uand V , where U is open and V is cozero, they have disjoint closures.
Necessity Suppose that X is basically disconnected. Let U ⊆ X beopen and V ⊆ X be cozero. Then X\U is closed and contains V , henceclV ⊆ X\U . But since X is basically disconnected, the closure clV is openand hence X\ clV is closed and contains U . Hence clU ⊆ X\ clV , whichmeans that U and V have disjoint closures.
Sufficiency Suppose that, for any open U ⊆ X and cozero V ⊆ X,these sets have disjoint closures. Take V ⊆ X to be a cozero-set and U =int(X\V ). There are two possibilities, either int(X\V ) is empty or not. Inthe latter case, by hypothesis, the sets U and V have disjoint closures, hence
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∂V = (clV )∩U = ∅. Then clV = V ∪ ∂V = V is open. In the former case,we have clV = X, which is open. Therefore X is basically disconnected.
2. In an extremally disconnected space, any two disjoint open sets arecompletely separated. In a basically disconnected space, any two disjointcozero-sets are completely separated; equivalently, for every f ∈ C, pos fand neg f are completely separated.
Proof. Suppose that X is an extremally disconnected space. Let A and B,subsets of X, be disjoint open sets. These sets are completely separated ifthey have disjoint zero-set neighborhoods. Note that, by (1.), the sets A andB have disjoint open closures. So we can write
X = clA ∪ clB ∪ (X\(clA ∪ clB)),
where the terms in the union are disjoint and open, hence each of themis a zero-set. Therefore any two disjoint open sets have disjoint zero-setneighborhoods (their closures), and so are completely separated.
Suppose that X is a basically disconnected space. Let A,B ⊆ X becozero-sets. Then, by (1.), they have disjoint open closures, which are alsozero-set neighborhoods, by the same argument as above. So any two disjointcozero-sets are completely separated.
3. If X is basically disconnected, then for every f ∈ C, there exists aunit u of C such that f = u|f |.
Proof. The sets pos f and neg f are disjoint cozero-sets, and hence havedisjoint open closures. In fact from an intermediate result from Sufficiencyof the second part of (1.), pos f and neg f are their own closures. Then wedefine the function u ∈ RX as
u(x) =
{1 if x ∈ cl(pos f)−1 if x 6∈ cl(pos f)
Clearly f = u|f |. Also u ∈ C(X), by Lemma 1.4, and u is a unit in C(X).
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4. Every dense subspace X of an extremally disconnected space T isextremally disconnected. In fact, disjoint open sets in X have disjoint openclosures in T .
Before we begin, let us start off with a couple of Lemmas.
Lemma 1.6. Let T be a topological space and X ⊆ T a dense subset. Thefor any open A ⊆ T we have clA = cl(A ∩X).
Proof. First, consider p ∈ clA, this implies that any open neighborhood Nof p contains at least one point of A. Since A is open, the set A∩N is also aneighborhood of p. Note that since X is dense in T , every neighborhood of p,namely A∩N must contain points of X, hence A∩N∩X is non-empty. SinceN was arbitrary, we must conclude that p ∈ cl(A ∩X) or clA ⊆ cl(A ∩X).
Note that A∩X ⊆ A, which implies that cl(A∩X) ⊆ clA. So finally wehave clA = cl(A ∩X).
Lemma 1.7. Let T be a topological space and X ⊆ T a dense subset. Forany A ⊆ X, we have clX A = X ∩ clT A.
Proof. Note that clT A is closed in T , hence A ∩ clT A is closed in X, henceclX A ⊆ A ∩ clT A ⊆ clT A and clX A ⊆ X, hence clX A ⊆ X ∩ clT A.Now let x ∈ X ∩ clT A. For any open neighborhood N of x in X thereis an open neighborhood N ′ of x in T such that N ′ ∩ X = N . But sincex ∈ X clT A, we have that x ∈ X and that N ′ contains at least one point ofA. Hence N = N ′∩X also contains at least one point of A, which implies thatx ∈ clX A. Finally we have X ∩ clT A ⊆ clX A, and clX A = X ∩ clT A.
And now, we proceed to the proof.
Proof. Let A be an open subset of X, there there is an open subset A′ of Tsuch that A = X ∩ A′. Then we have clT A = clT A
′, by Lemma 1.6. Alsowe have clX A = X ∩ clT A
′, by Lemma 1.7. But clT A′ is open in T because
T is extremally disconnected, hence clX A is open in X. Therefore X is alsoextremally disconnected.
Now consider A and B disjoint open subsets of X, then we have A′ andB′ open subsets of T such that A = X∩A′ and B = X∩B′. Let C ′ = A′∩B′,the set C ′ is open, hence it must contain at least one point x ∈ X. But thenx ∈ X ∩ A′ = A and x ∈ X ∩ B′ = B which contradicts the fact that Aand B are disjoint. Hence A′ and B′ must be disjoint open subsets of T . Sosince T is extremally disconnected, they must have disjoint open closures in
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T . But clA′ = clA and clB′ = clB, by Lemma 1.6, hence A and B havedisjoint open closures in T .
5. Every open subspace of an extremally disconnected space is extremallydisconnected. (A closed subspace, however, need not even be basically dis-connected.)
Once again, we start off with a Lemma.
Lemma 1.8. Let X be a topological space and A an open subset. Then forany subset B open in A, we have clAB = A ∩ clX B.
Proof. Clearly clAB ⊆ clX B and clAB ⊆ A, hence clAB ⊆ A∩ clX B. Nowlet x ∈ A ∩ clX B, then for any open neighborhood N of x in A, the set Nis open in X since A is open and since x ∈ clX B, the neighborhood N mustcontain at least one point of B. Which implies that x ∈ clAB. So finally wehave A ∩ clX B ⊆ clAB and clAB = A ∩ clX B.
And now, we proceed to the proof.
Proof. Let X be an extremally disconnected topological space and A an opensubset. Then any subset B ⊆ A open in A is also open in X since A is open.Therefore clX B is open. But then clAB = A ∩ clX B, by Lemma 1.8, andsince clX B is open in X, the closure clAB is open in A. Hence A is alsoextremally disconnected.
6. X is extremally disconnected iff every open subspace is C∗-embeddedin X.
Proof. Necessity Suppose thatX is extremally disconnected, but that there isan open subset S ⊆ X which is not C∗-embedded in X. Then, by Urysohn’sExtension Theorem (1.17), there are disjoint subsets A,B ⊆ S in S that arecompletely separated in S but not inX. If at least one of A and B is not open,we can replace it WLOG by the interior of its zero-set neighborhood, hencewe can assume that A and B are open. Since S is open, the subsets A andB are also open and disjoint in X. So, by (2.), A and B must be completelyseparated in X. This contradiction implies that every open S ⊆ X must beC∗-embedded in X.
Sufficiency Suppose that every open S ⊆ X is C∗-embedded in X. Takeany two disjoint open subsets A,B ⊆ X. The set A ∪ B is open, and hence
27
C∗-embedded. We define f ∈ C(A∪B) such that f |A = 0 and f |B = 1, thefunction f is continuous on A∪B by Lemma 1.4. Since A∪B is C∗-embedded,the function f can be extended to a bounded continuous function f on all ofX. But this simply means that A and B are completely separated. Whichimplies that they have disjoint zero-set neighborhoods. Which implies thatthey have disjoint closures. Hence, by (1.), the space X must be extremallydisconnected.
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2 Ideals and z-filters
2B p.30 prime ideals
Recall that given a ring R, and ideal P is prime if ab ∈ P implies that atleast one of a or b is in P , where a and b are some elements of R.
Also recall that if R is a ring, and P and Q are ideals in R, we denoteby PQ the smallest (under set inclusion) ideal containing all products fg,where f ∈ I and g ∈ J .
1. An ideal P in C is prime iff P ∩ C∗ is a prime ideal in C∗.
Proof. Necessity Assume that P is prime. Then, clearly, P ∩ C∗ is an idealin C∗. Suppose that a, b ∈ C∗ such that ab ∈ P , then, since P is prime, atleast one of a or b is in P and hence in P ∩C∗. Hence P ∩C∗ is prime in C∗.
Sufficiency Assume that P is an ideal such that P ∩ C∗ is prime in C∗.Suppose that a, b ∈ C such that ab ∈ P . Then, by 1E.1, there are unitsu, v ∈ C such that ua, vb ∈ C∗. And since P is an ideal, we have (ua)(vb) ∈P ∩ C∗. Then, since P ∩ C∗ is prime, we have ua ∈ P ∩ C∗ or vb ∈ P ∩ C∗.But since P is an ideal, we have u−1ua = a ∈ P or v−1vb ∈ P . Hence P isprime in C.
2. If P and Q are prime ideals in C, or in C∗, then PQ = P ∩ Q. Inparticular, P 2 = P . Hence M2 = M for every maximal ideal M in C or C∗.
Proof. Take any a ∈ P and b ∈ Q, then ab ∈ P and ab ∈ Q, since both Pand Q are ideals. Note that since P ∩ Q is an ideal, we have PQ ⊆ P ∩ Q.Now take any f ∈ P ∩ Q. We can write f = (f 1/3)3, hence f 1/3 ∈ Pand f 1/3 ∈ Q, since both P and Q are prime. Therefore (f 1/3)2 ∈ P , and(f 1/3)2f 1/3 = f ∈ PQ. Hence P ∩Q ⊆ PQ, and PQ = P ∩Q.
So if P is a prime ideal, we have the identity P 2 = P ∩P = P . The sameis true for any maximal ideal M in C or C∗, since M is prime.
3. An ideal I in a commutative ring is an intersection of prime ideals iffa2 ∈ I implies a ∈ I.
Proof. Necessity Assume that I is an intersection of prime ideals. Supposethat a2 ∈ I, then a2 is contained in each of the prime ideals. So a is containedin each of the prime ideals, hence a ∈ I.
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Sufficiency Assume that I is an ideal in a commutative ring such thata2 ∈ I implies that a ∈ I. We know that I must be contained in some primeideal (for example a maximal ideal), hence I is contained in the intersectionof all prime ideals containing it. From 0.18 we know that the intersectionof all prime ideals containing I is the set of all elements of the ring of whichsome power belongs to I. Now take an element of the ring a, such that an ∈ Ifor some n ∈ N. Then if n is even, by the hypothesis, we have an/2 ∈ I. Andif n is odd, we have an+1 ∈ I, and hence a(n+1)/2 ∈ I. Iterating this way,we find that a ∈ I, hence I is equal to the intersection of all prime idealscontaining it.
2C p.31 functions congruent to constants
1. Let I be an ideal in C; if f ≡ r (mod I), then r ∈ f [X].
Proof. Since f ≡ r (mod I), we have f = r + i for some i ∈ I. Since I isan ideal, it contains no units. Hence for every i ∈ I, there is an x ∈ X suchthat i(x) = 0. Hence f(x) = r + i(x) = r. Therefore r ∈ f [X].
2. Let I be an ideal in C∗; if f ≡ r (mod I), then r ∈ clR f [X].
Proof. Since f ≡ r (mod I), we have f = r + i for some i ∈ I. Since I isan ideal, it contains no units, that is for every i ∈ I, there is no ε > 0 suchthat |i| ≥ ε. Hence for each ε > 0, the set i[X]∩ ]−ε, ε[ is non-empty. Hence0 ∈ clR i[X], and therefore r ∈ clR f [X].
2D p.31 z-ideals
1. Let I be a z-ideal in C, and suppose that f ≡ r (mod I). If g(x) = rwherever f(x) = r, then g ≡ r (mod I).
Proof. First note that, by the hypothesis, the function f − r is in I andZ (f − r) ⊆ Z (g − r). So since Z [I] is a z-filter, we have Z (g − r) ∈ Z [I].And since I is a z-ideal, we must have g−r ∈ I, hence g ≡ r (mod I).
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2. If f 2 + g2 belongs to a z-ideal I, then f ∈ I and g ∈ I.
Proof. First we use the facts that Z [I] is a z-filter and I is a z-ideal to get
Z (f 2 + g2) = Z (f) ∩ Z (g) ⊆ Z (f) ⇒ Z (f) ∈ Z [I]
⇒ f ∈ I.
Similarly, we have g ∈ I.
3. If I and J are z-ideals, then IJ = I ∩ J . Compare to 2B.2.
Proof. Suppose that a ∈ I and b ∈ J , so since I and J are ideals, we haveab ∈ I and ab ∈ J , and hence ab ∈ I ∩ J . Suppose that f ∈ I ∩ J . Notethat Z (f 1/3) = Z (f), so since I and J are z-ideals, we have g = f 1/3 ∈ Iand g ∈ J . Then g2 ∈ I and f = g2g ∈ IJ , hence IJ = I ∩ J .
The proof above is very similar to the one given for 2B.2, except that weuse the fact that I and J are z-ideals, instead of prime ideals, to show thatf 1/3 is in the same ideal as f .
4. Z [(I, J)] is the set of all Z1 ∩ Z2 where Z1 ∈ Z [I] and Z2 ∈ Z [J ].
Proof. Since (I, J) is an ideal, then Z [(I, J)] is a z-filter. So for any Z1 ∈ Z [I]and Z2 ∈ Z [J ] we have Z1, Z2 ∈ Z [(I, J)] and hence Z1 ∩ Z2 ∈ Z [(I, J)].Now suppose that Z ∈ Z [(I, J)], then there must exist i ∈ I and j ∈ J suchthat Z = Z (i + j). But Z (i) ∩ Z (j) ⊆ Z, hence Z ∈ Z [I] and Z ∈ Z [J ].Hence we can write Z = Z ∩ Z. This completes the proof.
2E p.31 z-ideals
First recall a z-filter on X is said to be prime if for any A,B ∈ Z (X) suchthat A ∪B ∈ F implies that at least one of A or B is in F .
For a z-filter F on X, TFAE:
(1) F is prime.
(2) Whenever A,B ∈ Z (X) such that A ∪ B = X, at least one of A or Bis in F .
(3) Given Z1, Z2 ∈ Z (X), there exists Z ∈ F such that one of Z ∩ Z1 orZ ∩ Z2 contains the other.
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Proof. This theorem is the analog of Theorem 2.9 for z-filters.(1)⇒(2). Assume that F is prime. By hypothesis, we have A ∪ B = X,
and X ∈ F since F is a z-filter. Therefore, since F is prime, at least one ofA or B must be in F .
(2)⇒(3). Consider the z-ideal P = Z�[F ]. Assuming (2), if for someg, h ∈ C we have gh = 0 (that is Z (g) ∪ Z (h) = X), then at least one ofZ (g) or Z (g) is in F , or at least one of g or h is in P , since P is a z-ideal.This is equivalent to condition (3) of Theorem 2.9, which implies condition(4) of 2.9 For every f ∈ C, there is a zero-set in Z[P ] = F on which f doesnot change sign.
Now take any Z1, Z2 ∈ Z (X), say Z1 = Z (g) and Z2 = Z (h) for g, h ∈ C.Consider the function |g| − |h|, by the condition (4) above, there must be aZ ∈ F such that |g| − |h| does not change sign, WLOG let us assume that itis non-negative on Z. Then on Z, wherever g vanishes h must also vanish.In other words Z ∩ Z1 ⊆ Z ∩ Z2, which completes the implication (2)⇒(3).
(3)⇒(1). Assume (3). Then suppose that for some Z1, Z2 ∈ Z (X) wehave Z1 ∪ Z2 ∈ F . Then we can find a Z ∈ F such that, WLOG, Z ∩ Z1 ⊆Z ∩ Z2. This implies that Z ∩ (Z1 ∪ Z2) = Z ∩ Z2, but Z ∩ (Z1 ∪ Z2) ∈ F ,since F is a filter. Therefore Z ∩ Z2 is in F , and hence Z2 ∈ F . Hence F isa prime z-filter.
2F p.31 finite spaces
Let X be a finite discrete space. In C(X):
1. f is a multiple of g iff Z (g) ⊆ Z (f).
Proof. Necessity Assume that f = gh for some h ∈ C(X). Then Z (f) =Z (g) ∪ Z (h), therefore Z (g) ⊆ Z (f).
Sufficiency Assume that Z (g) ⊆ Z (f), then we can construct h a follows
h(x) =
{0 if x ∈ Z (g)f(x)/g(x) if x 6∈ Z (g)
.
Clearly f = hg, and since X is discrete, we have h ∈ C(X).
2. Every ideal is a z-ideal.
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Proof. Let I be an ideal. Then take Z ∈ Z [I], then there is a function g ∈ Isuch that Z = Z (g). Now take any f ∈ C(X) such that Z (f) = Z. Then,by (1.), f must be a multiple of g, hence g ∈ I and I must be a z-ideal.
3. Every ideal is principal, and, in fact, is generated by an idempotent.
Proof. Let I be an ideal. Since X is finite, we have Z [I] ⊆ P(X) finite. SinceZ [I] is closed under finite intersections and ∅ 6∈ Z [I], then Z =
⋂Z [I] is
non-empty. Hence Z [I] must be principal and generated by Z, and sinceevery ideal is a z-ideal, I also must be principal.
In fact, we can define an idempotent i as
i(x) =
{0 if x ∈ Z1 if x 6∈ Z .
Then Z (i) = Z, and hence I = (i).
4. Every ideal is an intersection of maximal ideals. The intersection ofall maximal ideals in (0).
Proof. By (3.), every ideal is principal, hence every maximal ideal is principaland must be generated by a singleton. Take any ideal I, since it is principalwe can write I = (i), and let Z = Z (i). Then we take the collection ofmaximal ideals Mx = (mx) where Z (mx) = {x}, for each x ∈ X. Clearlythe ideal I is given by I =
⋂x∈ZMx.
If we take the intersection of all maximal ideals, we have J =⋂x∈XMx.
Clearly Z [J ] is generated by X, hence we must have J = (0).
5. Every prime ideal is maximal.
Proof. Let P be a prime ideal, then, since X is finite and discrete, anyZ ∈ Z [P ] can be expressed as a finite union of singletons. Note that sinceP is prime, the z-filter Z [P ] must also be prime, hence at least one of thesingletons must be in Z [P ]. Any (proper) z-filter cannot contain more thanone singleton, hence Z [P ] must contain a single singleton. This implies thatZ [P ] is principal and maximal, and so is P .
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2G p.31 prime vs. z-ideals in C(R)
1. Select a function l in C(R) such that l(0) = 0, while
limx→0
ln(x)/x =∞,
for all n ∈ N. Apply 0.17 to construct a prime ideal in C(R) that containsi but not l. This prime ideal is not a z-ideal (and hence is not maximal).
Proof. We define l ∈ C(X) as follows
l(x) =
{−1/ ln(x) if 0 < x ≤ e−1
1 if e−1 < x
l(0) = 0
l(−x) = −l(x).
It is easy to see that l satisfies the required properties. Now take the idealI = (i), it consists of all functions of the form f(x) = xg(x) for any g ∈C(R) (2.4). Note that l 6∈ I so the set of all powers of l is closed undermultiplication and disjoint from I. Hence by 0.17 we can construct a primeideal J such that I ⊆ J and it still does not contain any power of l. But sinceZ (l) = {0} and hence l ∈ Z�[Z [J ]], the ideal J cannot be a z-ideal.
2. Let O0 denote the ideal of all functions f in C(R) for which Z (f) isa neighborhood of 0. Define s ∈ C(R) as follows
s(x) =
{x sin(π/x) if x 6= 00 if x = 0
.
Then (O0, s) is not a z-ideal; and the smallest z-ideal containing (O0, s) isnot prime.
Proof. Suppose that (O0, s) is a z-ideal. Note that the zero-set of s is Z (s) ={−1,−1/2, . . . , 0, . . . , 1/2, 1}. Take t = s1/3, then Z (t) = Z (s) and hencet must be in (O0, s). Then we can write t = o + su for some o ∈ O0 andu ∈ C(R). Since Z (o) is a neighborhood of 0 we can find an ε > 0 such thato|[−ε, ε] = 0, we restrict our attention to this interval from now on. Nowwe can write t = su, hence wherever s(x) 6= 0 we have u(x) = t(x)/s(x) =(x sin(π/x))−2/3. But then limx→0 |u| =∞ and hence u cannot be continuous
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on R, which implies that t 6∈ (O0, s). This contradictions proves that (O0, s)is not a z-ideal.
Consider I = Z�[Z [(O0, s)]] the smallest z-ideal containing (O0, s). By2.12, the ideal I is prime iff the z-filter Z [(O0, s)] is prime. This z-filter isgenerated by the neighborhoods of 0 and the set
S = {−1,−1/2, . . . , 0, . . . , 1/2, 1}.We can write
S = {−1,−1/2, . . . , 0} ∪ {0, . . . , 1/2, 1},note that both sets in the union are zero-sets, hence if Z [(O0, s)] is prime itmust contain at least one of them. But neither of those sets is in this z-filterbecause they cannot be generated by finite intersections and supersets of thegenerating sets. Therefore J cannot be a prime ideal.
2H p.32 the identity function i in C(R)
1. The principal ideal (i) in C(R) consists precisely of all functions inC(R) that vanish at 0 and have a derivative at 0. Hence every non-negativefunction in (i) has a zero derivative at 0.
Proof. First note that the ideal (i) is the set of all f ∈ C(R) of the formf = ig. Hence, since i(0) = 0, for all f ∈ (i) we have f(0) = 0. Now takef ∈ (i) such that f = i · g, and consider its derivative at 0
f ′(0) = limh→0
f(h)− f(0)
h
= limh→0
hg(h)− 0
h= lim
h→0g(h)
= g(0).
Hence every function in (i) vanishes at 0 and has a derivative at 0. Conversely,consider a function f ∈ C(R) such that f(0) = 0 and f ′(0) exists. Then wecan define g = f/i and note that
limh→0
g(h) = limh→0
f(h)
h
= limh→0
f(h)− f(0)
h= f ′(0).
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Hence g is continuous and we can write f = i · g and f ∈ (i).
2. (i) is not a prime ideal.
Proof. Suppose that (i) is prime, then, by 2B.2, we must have (i)2 = (i).First note the identity (i)2 = (i2), in other words, the ideal (i) is the set ofall functions f ∈ C(R) such that f = i2 · g for some g ∈ C(R). Now take anyf ∈ (i)2, we can write f(x) = x2g(x) for some g ∈ C(R) and hence f(0) = 0.Consider its second derivative at 0,
f ′′(0) = limh→0
f(h)−f(0)h
− f(0)−f(−h)h
h
= limh→0
h2g(h)− 2f(0) + h2g(−h)
h2
= limh→0
g(h) + g(−h)
= 2g(0).
Hence every f ∈ (i)2 has a second derivative at 0. However, we have i·|i| ∈ (i)which does not have a second derivative at 0, hence i · |i| 6∈ (i)2. Thiscontradiction completes the proof.
3. The ideal (i, |i|) is not principal.
Proof. Suppose that this ideal is principal and (i, |i|) = (d) for some d ∈C(R). So, since i, |i| ∈ (d), there must exist g, h ∈ C(R) such that i = gdand |i| = hd. Note that for x > 0 we must have g(x) = h(x) and for x < 0we must have g(x) = −h(x). So, by continuity, we have g(0) = h(0) = 0.Also, since d ∈ (i, |i|), there must exist s, t ∈ C(R) such that si + t|i| = d.Now we can write s(gd) + t(hd) = d or (sg + th)d = d. Recall that wemust have Z (d) = {0}, and note that for every x where d(x) 6= 0 we have(sg + th)(x) = 1. Hence, by continuity, we must have sg + th = 1. But thenboth g and h cannot vanish at 0. This contradiction completes the proof.
4. Exhibit a principal ideal containing (i, |i|).
Proof. Note that both of i · |i|−1/2 and |i|1/2 = |i| · |i|−1/2 are continuousfunctions, hence (i, |i|) ⊆ (|i|1/2).
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2I p.32 C(Q) and C∗(Q)
The set of all f ∈ C(Q) for which limx→π f(x) = 0 is not an ideal in C(Q).But the bounded functions in this set do constitute an ideal in C∗(Q).
Proof. Let I ⊆ C(Q) be the set of all such functions, and let I∗ = I ∩C∗(Q).Consider the function f such that f(x) = x − π. Clearly f is in I, but fis a unit of C(Q) since f−1 is also continuous on Q. Hence I cannot be a(proper) ideal.
Now consider I∗, clearly it is closed under addition. Also, consider anyg ∈ C∗(Q), then |g| ≤ r for some r ∈ R. Then fg is also in I∗ sincelimx→π |f(x)|r = 0 and
0 ≤ limx→π|f(x)g(x)| ≤ lim
x→πf(x)r ⇒ lim
x→πf(x)g(x) = 0.
Therefore I∗ is an ideal in C∗(Q).
2J p.32 ideal chains in C(R), C(Q) and C(N)
1. Find a chain of z-ideals in C(R) (under set inclusion) that is in one-to-one, order preserving correspondence with R itself.
Proof. Consider the chain of z-filters Fr on Z (R), where for each r ∈ R wehave Fr is principal and generated by the interval [r,∞[. The correspondingz-ideals in C(R) satisfy the desired properties.
2. Find a chain of z-ideals in C(Q) (under set inclusion) that is in one-to-one, order preserving correspondence with Q itself.
Proof. The same construction as in (1.) works, except that the sets thatgenerate the z-filters have the form Q ∩ [r,∞[.
3. Find a chain of z-ideals in C(N) (under set inclusion) that is in one-to-one, order preserving correspondence with N itself.
Proof. Suppose that there is a bijection between two sets A and B, then thisbijection induces a natural bijection on the power sets P(A) and P(B) whichpreserves set inclusion. If F is a filter on A, then its image under the inducedbijection must also be a filter on B.
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Take the chain Fr of z-filters on Q from (2.), and consider the chain F ′rof corresponding filters on Q such that F ′r is the smallest filter containingFr (note that these retain the ordering of the generating z-filters). We knowthat there exists a bijection between Q and N, hence we can map the filtersF ′r to filters on N preserving the inclusion order. But since N is discrete,every filter on N is a z-filter. Hence we can construct a chain of z-idealsin C(N) by taking the z-ideals corresponding to the z-filters on N that weconstructed above.
2K p.32 z-filters and C∗
If M is a maximal ideal in C∗, and Z [M ] is a z-filter, then Z [M ] is a z-ultrafilter.
Proof. Suppose that I and J are ideals in C, such that I ⊂ J (properlycontains). Take any f ∈ J\I, it is either bounded or not. If it is notbounded, then there is a unit u of C such that uf is bounded (cf. 1E.1). Thefunction uf must be in J\I since if it was in I then so would f = u−1uf .Hence we have I ⊂ J implies that I ∩ C∗ ⊂ J ∩ C∗.
Now consider Z [M ], since it is a z-filter, then I = Z�[Z [M ]] is a z-ideal in I. Suppose that F is another z-filter such that Z [M ] ⊆ F . ThenJ = Z�[F ] is also an ideal in C, such that I ⊆ J . Then I∗ = I ∩ C∗ andJ∗ = J ∩ C∗ are ideals in C∗, so we have M ⊆ I∗ ⊆ J∗ and since M ismaximal we must have M = I∗ = J∗. Hence, by the argument above, wealso have I = J and hence F = Z [M ]. Therefore Z [M ] is a z-ultrafilter.
2N p.35 the m-topology on C
The m-topology is defined on C(X) by taking as a base for neighborhoodsystem at g all sets of the form
Mu = {f ∈ C | |g − f | ≤ u},
where u is a positive unit of C. The same topology results if it is requiredfurther that u be a bounded function.
We also need a few more definitions. The ring C is called a topologicalring if it is endowed with a topology and the operations (f, g) 7→ f + g and(f, g) 7→ fg are continuous. If C is regarded as a vector space (with scalarmultiplication given as (r, f) 7→ rf), it is called a topological vector space if it
38
is endowed with a topology and the operations (f, g) 7→ f+g and (r, f) 7→ rfare continuous.
The uniform norm topology on C∗ is defined by taking as a base for theneighborhood system at g all sets of the form
Uε = {f ∈ C∗ | |g − f | ≤ ε},
where ε > 0.
1. C is a topological ring.
Proof. Consider the addition operation on C, for any f, g ∈ C we have+: (f, g) 7→ f + g. Consider any f, g ∈ C, let h = f + g and take any Mu
neighborhood around h. Let v ∈ C be a positive unit and take an Mv ×Mv
neighborhood V around (f, g). Then for any (a, b) ∈ V , we have
|h− (a+ b)| = |(f − a) + (g − b)| ≤ |f − a|+ |g − b| ≤ 2v.
So, if we let v = u/3, then +[V ] ⊆ Mu. Hence addition is continuous at(f, g) (cf. 1D.3) and therefore on all of C × C.
Consider the multiplication operation on C, for any f, g ∈ C we have·: (f, g) 7→ fg. Consider any f, g ∈ C, let h = f + g and take any Mu
neighborhood around h. Let v ∈ C be a positive unit and take an Mv ×Mv
neighborhood V around (f, g). Then for any (a, b) ∈ V , we have
|h− ab| = |(f − a)g + (a− f)(g − b) + f(g − b)|≤ |f − a||g|+ |a− f ||g − b|+ |f ||g − b|≤ (|g|+ u+ |f |)v.
So, if we let v = u(u+|f |+|g|)−1/2, then ·[V ] ⊆Mu. Hence multiplicationis continuous at (f, g) and therefore on all C×C. Therefore C is a topologicalring.
2. The relative m-topology on C∗ contains the uniform norm topology,and the two coincide iff X is pseudocompact. In fact, when X is not pseudo-compact, the set of constant functions in C∗ is discrete (in the m-topology),so that C∗ is not even a topological vector space.
39
Proof. Let Tm denote the m-topology on C, and let Tu denote the uniformnorm topology on C∗. Let V ⊆ C∗ be an open set in Tu, then for everypoint in V , there is an ε > 0 such that there is a Uε neighborhood of thatpoint contained in V . But ε is also a positive unit of C, hence each Uε
neighborhood is also a Mε neighborhood. Therefore V is also open in Tm,hence Tu is contained in Tm.
Assume thatX is not pseudocompact, then C must contain an unboundedfunction f . Then there must exist an unbounded positive unit in C, e.g.|f | ∨ ε for any ε > 0. Let u be an unbounded positive unit of C, then u−1 isalso a positive unit such that 0 ∈ cl(u−1[X]). Consider anMu−1 neighborhoodU of 0, then intTm U is open in Tm and contains 0. However, note that no Uε
neighborhood of 0 is contained in U , since ε ≤ u−1 would imply that ε = 0.Therefore if X is not pseudocompact, then Tm is strictly bigger than Tu. Andif X is pseudocompact, then trivially Tm = Tu, since C = C∗.
3. The set of all units of C is open, and the mapping f 7→ f−1 is ahomeomorphism of this set onto itself.
Proof. Let U be the set of all units of C. Take any u ∈ U and let v = |u|/2.Since v is a positive unit, we can take an Mv neighborhood around u. Letf ∈Mv, then
|f | = |u− (u− f)| ≥ ||u| − |u− f || ≥ ||u| − v| ≥ |u|/2 > 0.
Hence each such neighborhood is contained in U , and therefore U is open.Now, consider the inverse operation inv: f 7→ f−1. Take any unit u ∈ U ,
and take a neighborhood Mv around u−1 and contained in U . Then takea neighborhood Mw around u and contained in U , letting f ∈ Mw (that is0 ≤ |u| − w ≤ |f | ≤ |u|+ w). Then we have
|f−1 − u−1| = |f − u||fu|
≤ w
(|u| − w)|u|≤ v,
if w ≤ u2v/(1 + uv). Therefore inv is continuous at u, and hence on allof U . The fact that inv is a bijection, with itself as the inverse, is trivial. Sosince it is continuous, inv must be a homeomorphism of U into itself.
40
4. The subring C∗ is closed.
Proof. Consider the set of all unbounded functions in C. For any unboundedf and any ε > 0, we can take an Mε neighborhood around f . Then everyfunction in this neighborhood is also unbounded. Hence the complement ofC∗ in C is open, and therefore C∗ is closed.
5. The closure of every ideal is a (proper) ideal. Hence every maximalideal is closed. Every maximal ideal in C∗ is closed.
Proof. The closure of an ideal is the set of all functions such that each neigh-borhood of each of those functions contains at least one element of the ideal.Now, let I be an ideal in C and J its closure in C.
Suppose that f, g ∈ J , then for any positive ideal u we can find a, b ∈ Isuch that |a− f | ≤ u/2 and |b− g| ≤ u/2. Recall that since I is an ideal, wehave (a+ b) ∈ I. Also,
|(f + g)− (a+ b)| = |(a− f) + (b− g)| ≤ u/2 + u/2 = u.
Therefore J is closed under addition.Suppose that f ∈ J , then for any positive ideal u we can find a ∈ I such
that |f − a| ≤ u. Recall that since I is an ideal, we have (−a) ∈ I. Also,we have |(−f) − (−a)| = |f − a| ≤ u. Therefore J is closed under takingadditive inverses.
Suppose that f ∈ J and g ∈ C, then for any positive unit u we can finda ∈ I such that |f − a| ≤ uh−1, where h = |g| ∨ 1 (note that h−1g ≤ 1).Since I is an ideal, we have ag ∈ I. Also, we have
|fg − ag| = |f − a||g| ≤ uh−1|g| ≤ u.
Therefore J is closed under multiplication from outside. Hence J mustbe an ideal (possibly an improper one).
Finally suppose that u ∈ J , where u is a unit of C. By (3.), the set ofall units of C is open, hence we can take a neighborhood U around u suchthat every element of the neighborhood is also a unit (that is not equal tozero anywhere). But an element of a proper ideal must vanish at least atone point. Therefore, since J is the closure of a proper ideal I, the ideal Jcannot contain any units, in other words J is also proper.
Let M be a maximal ideal in C and M ′ its closure. Then, by the above,M ′ is also an ideal and, by properties of closure, we have M ⊆M ′. But then,
41
by maximality of M , we have M ′ = M . In other words a maximal ideal Mmust be closed.
Let M be a maximal ideal in C∗ and M ′ its closure in C. Then M ′ ∩C∗is also an ideal in C∗, which contains M . But since M is maximal in C∗, wehave M ′ ∩ C∗ = M . Recall that, by (4.), C∗ is closed, and since M ′ is alsoclosed, then M must be closed as well.
6. Every closed ideal in C is z-ideal.
Proof. Consider a closed ideal I in C. Suppose g ∈ I and f ∈ C such thatZ (f) = Z (g). Now given a positive unit u of C, we define a function h asfollows:
h(x) =
f(x) + u(x)
g(x)if f(x) ≤ −u(x)
0 if |f(x)| ≤ u(x)f(x)− u(x)
g(x)if u(x) ≤ f(x)
.
First note that h is well defined, and that it is continuous, by construction,on {x ∈ X | |f(x)| ≤ u(x)} and {x ∈ X | |f(x)| ≥ u(x). Since these sets areclosed and form a closed cover or X, by 1A.1, h ∈ C.
Given x ∈ X such that |f(x)| ≤ u(x), we have
|f(x)− g(x)h(x)| = |f(x)− 0| ≤ u(x).
Given x ∈ X such that |f(x)| > u(x), we have
|f(x)− g(x)h(x)| = |f(x)− (f(x)± u(x))| = u(x) ≤ u(x).
Hence |f − gh| ≤ u, and since u was arbitrary, the function f must be inthe closure of I. But I is closed, so f must be in I. Therefore given anyZ ∈ Z [I], the ideal I must contain all functions f ∈ C such that Z (f) = Z.Therefore every closed ideal is also a z-ideal.
7. In the ring C(R), the z-ideal O0 of all functions that vanish on aneighborhood of 0, is not closed.
Proof. Consider the identity function i. Given any positive unit u of C(R)we define iu as follows:
iu(x) =
i(x) + u(x) if i(x) ≤ −u(x)0 if |i| ≤ u(x)i(x)− u(x) if u(x) ≤ i(x)
.
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Note that, since 0 ∈ Z (i) and since u must be bounded away from 0 bya positive constant in any bounded neighborhood of 0, we have that Z (iu)is a neighborhood of 0.
But then, |i − iu| ≤ u for any positive unit u. Hence i ∈ clO0. ButZ (i) = {0} which is not a neighborhood of 0, hence i 6∈ O0 and O0 is notclosed.
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3 Completely Regular Spaces
3B p.48 countable sets
Let X be a completely regular space.
1. A countable set disjoint from a closed set F is disjoint from somezero-set containing F .
Proof. Let S ⊆ X be a countable set and F ⊆ X be a closed set. Then sinceX is completely regular, for each s ∈ S there is a zero set Zs containing Fand disjoint from s. Then Z =
⋂s∈S Zs is also a zero set (intersection of
countably many zero sets) containing F and disjoint from S.
2. A C-embedded countable set S is completely separated from everydisjoint closed set.
Proof. Let F be a closed set disjoint from S. From (1.), we know that thereis a function f ∈ C(X) such that f [F ] = 0 and 0 6∈ f [S]. Therefore wecan define a continuous function g′ ∈ C(S) such that g′(s) = f−1(s), foreach s ∈ S. And since S is C-embedded, we can extend g′ to a continuousfunction g ∈ C(X). Note that h = fg satisfies the condition h[S] = {1} andh[F ] = {0}, hence S and F are completely separated.
3. Any C-embedded countable set is closed.
Proof. Let S be a C-embedded countable set. Note that, since X is Haus-dorff, points are closed sets. Then for any x 6∈ S, the singleton {x} is closed,and from (2.) we know that for every x 6∈ A there is a zero-set Zx contain-ing S and disjoint from x. Now we can write S =
⋂x 6∈S Zx, which is closed
(intersection of closed sets).
4. Any two countable sets, neither of which meets the closure of the other,are contained in disjoint cozero-sets.
Proof. Let A,B ⊆ X be countable subsets, neither of which meets the closureof the other. First, we can enumerate the elements of each set, that isA = {an}n∈N and B = {bn}n∈N. Recall that, since X is completely regular,we can find a function f ∈ C(X) such that f(a1) = 0 and f [clB] = {1}.
44
Then we can find a zero-set neighborhood ZA1 of a1 containing a cozero-
set neighborhood V A1 of a1, both disjoint from clB. Namely, we can have
ZA1 = {x | f(x) ≤ 1/3} and V A
1 = {x | f(x) < 1/3}. Similarly, we can find azero-set neighborhood ZB
1 of b1 containing a cozero-set neighborhood V B1 of
b1, both disjoint from clA∪ZA1 . Again, we can find a zero-set neighborhood
ZA2 of a2 containing a cozero-set neighborhood V A
2 of a2, such that ZA2 is
disjoint from clB ∪ ZB1 . Continuing by induction, we find that A and B are
contained in disjoint cozero-sets, respectively,⋃n∈N V
An and
⋃n∈N V
Bn .
5. A countable, completely regular space is normal.
Proof. Consider any two disjoint closed sets. Since the whole space is count-able, then so are they. Then, by (4.), we know that these sets are containedin disjoint cozero-sets. Since cozero-sets are open, any two disjoint closedsets are contained in disjoint neighborhoods, hence the space is normal.
3C p.48 Gδ-points of a completely regular space
Let p be a Gδ-point of a completely regular space X, and let S = X\{p}.
1. If g ∈ C∗(S), h ∈ C(X), and h(p) = 0, then g · (h|S) has a continuousextension to all of X.
Proof. We already know that g · (h|S) is continuous on S, so we define thefunction f ∈ RX as f |S = g · (h|S) and f(p) = 0. We need only prove thatf is continuous at p to show that f is continuous on all of X.
Since g is bounded, there must be an r ≥ 0 such that |g| ≤ r . Now takeany ε > 0, since h is continuous at p, there must be an open neighborhoodU of p such that h[U ] ⊆ ]−ε/r, ε/r[. Then f [U ] ⊆ ]−ε, ε[, therefore f iscontinuous at p and hence on all of X.
2. If Z is a zero-set in S, then clX Z is a zero-set in X.
Proof. Consider two zero-sets in X, namely Z = Z (f) and {p} = Z (h),recall that a Gδ point is a zero-set (3.11(b)), for f ∈ C(S) and h ∈ C(X).WLOG, assume that 0 ≤ f ≤ 1 and 0 ≤ h ≤ 1 and h[Z] = {1}. There aretwo possibilities, either Z = clX Z or Z ∪ {p} = clX Z. In the latter casep ∈ clX Z and clX Z = Z ∪ {p} = Z (fh). In the former case p 6∈ clX Z and
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clX Z = Z = Z ((1− f)h− 1), where we implicitly extend (1− f)h to all ofX as in (1.).
3D p.48 normal spaces
1. TFAE for any Hausdorff space X.
(1) X is normal.
(2) Any two disjoint closed sets are completely separated.
(3) Every closed set is C∗-embedded.
(4) Every closed set is C-embedded.
Proof. (1)⇒(2). Urysohn’s Lemma (3.13).(2)⇒(3). Urysohn’s Extension Theorem (1.17).(3)⇒(4). Take a closed set and any other closed set disjoint from it, since
X is normal, the two sets are completely separated. Finally use Theorem1.18.
(4)⇒(1). Assume (4). Given two disjoint closed sets A,B ⊆ X, the setA ∪ B is also closed. We define a continuous function f such that f |A = 0and f |B = 1. And since A∪B is C-embedded, we can extend f to all of X,hence any two closed sets are completely separated.
3. Every closed Gδ in a normal space X is a zero-set.
Proof. Let F =⋂n∈N Un be a closed Gδ set. Then, since X is normal, every
Un contains a zero-set neighborhood, say Zn, of F . Then
F ⊆⋂n∈N
Zn ⊆⋂n∈N
Un = F.
Hence F =⋂n∈N Zn is a zero-set, since it is a countable intersection of
countably many zero-sets.
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4. Every completely regular space with the Lindelof property (i.e. suchthat every open cover of X has a countable subcover) is normal.
Before we proceed, it is useful to consider a Lemma.
Lemma 3.1. A closed subspace of a Lindelof space is Lindelof.
Proof. Take X to be Lindelof and F ⊆ X a closed subspace. Then for anyopen cover {Uα}α∈A of F , we have an open cover {Uα}∪{X\F}. Then, sinceX is Lindelof, we can find a countable subcover {Vn}n∈N. If necessary weremove X\F , and we have the desired countable subcover of F . ThereforeF is also Lindelof.
And now, on to the proof.
Proof. Let X be completely regular and Lindelof. Take two disjoint closedsubsets A and B of X, by Lemma 3.1, they are also Lindelof. Since Xis completely regular, then for each x ∈ A we can find disjoint zero-setsneighborhoods Za
x and Zbx for x and B, respectively. Now we can find a
countable subcover {Zaxn}n∈N of A, and hence a zero-set Zb =
⋂n∈N Z
bxn
containing B and disjoint from A. Since Zb is still a closed set disjoint fromA, by a similar procedure we can find a zero-set Za containing A and disjointfrom Zb. Hence any two disjoint closed sets in X are completely separatedand X is normal.
5. Let X be a completely regular space. If X = S ∪K, where S is openand normal, and K is compact, then X is normal.
Proof. First, WLOG, assume that S and K are disjoint (or just replace Sand K by S and K\S).
Take A,B to be any two disjoint closed subsets of X. We can representthem as disjoint unions of A∩S, A∩K and B ∩S, B ∩K. Since any closedsubset of a compact set is also compact, we have that A∩K and B ∩K arecompact in X.
From 3.11(a) we know that in a completely regular space a compact setand a closed set disjoint from it are completely separated. Using this infor-mation we can construct the following neighborhoods (the neighborhoods aredenoted by Ni or Mi)
(a) N1 ⊇ A ∩K disjoint from M1 ⊇ B;
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(b) N2 ⊇ A disjoint from M2 ⊇ B ∩K;
(c) S ⊇ N3 ⊇ A ∩ S disjoint from S ⊇M3 ⊇ B ∩ S (S is normal).
Now we can form the following neighborhoods N = (N1 ∪ N3) ∩ N2 ⊇ Aand M = (M2 ∪M3) ∩M1 ⊇ B. If N and M are disjoint then we are done.Intuitively this seems true, but it is difficult to verify this claim by meansother than brute force. Namely
N ∩M = [(N1 ∪N3) ∩N2] ∩ [(M2 ∪M3) ∩M1]
= [(N1 ∩N2) ∪ (N3 ∩N2)] ∩ [(M2 ∩M1) ∪ (M3 ∩M1)]
= (N1 ∩N2 ∩M2 ∩M1) ∪ (N1 ∩N2 ∩M3 ∩M1)
∪ (N3 ∩N2 ∩M2 ∩M1) ∪ (N3 ∩N2 ∩M3 ∩M1)
= [
∅︷ ︸︸ ︷N1 ∩M1 ∩N2 ∩M2] ∪ [
∅︷ ︸︸ ︷N1 ∩M1 ∩N2 ∩M3]
∪ [N2 ∩M2︸ ︷︷ ︸∅
∩N3 ∩M1] ∪ [N3 ∩M3︸ ︷︷ ︸∅
∩N2 ∩M1]
= ∅
Hence any two disjoint closed subsets of X have disjoint neighborhoods andX is normal.
3E p.49 nonnormal space
Let X be a nonnormal, Hausdorff space.
1. X contains a closed set that is not a zero-set.
Proof. Suppose that every closed set in X is also a zero-set. Then any twodisjoint closed sets in X are contained in disjoint zero-sets, which impliesthat they are completely separated. But the latter statement implies that Xis normal. This contradiction completes the proof.
2. X has a subspace S with the following property: any two completelyseparated sets in S have disjoint closures in X, yet S is not C∗-embedded inX. Compare to Urysohn’s Extension Theorem
48
Proof. Consider a closed subset S ⊆ X. Then if two subsets A,B ⊆ Sare completely separated in S, there must be disjoint zero-sets of S, namelyZ1 ⊇ A and Z2 ⊇ B. Note that Z1 and Z2 are closed in S, and since S isclosed, they are also closed in X. Hence clX A ⊆ Z1 and clX B ⊆ Z2, hencethe closures of A and B are disjoint.
So any closed set S satisfies the desired property, now suppose that everyclosed set is C∗-embedded, from 3D.1(3), this implies that X is normal. Thiscontradiction completes the proof.
Urysohn’s Extension Theorem says that for a space S ⊆ X to be C∗-embedded any pair of sets that is completely separated in S are also com-pletely separated in X. This problem shows that requiring that they onlyhave disjoint closures is not enough.
3K p.49 the completely regular, nonnormal space
Γ
Let Γ denote the subset {(x, y) | y ≥ 0} of R×R, provided with the followingenlargement of the product topology: for r > 0, the sets
Vr(x, 0) = {(x, 0)} ∪ {(u, v) ∈ Γ | (u− x)2 + (v − r)2 < r2}
are also neighborhoods of the point (x, 0). This space Γ is called the Niemytzkiplane. Clearly, Γ satisfies the first countability axiom.
Also recall that the regular subspace topology is defined by the followingneighborhoods: for each ε > 0, the sets
Nε(x, y) = {(u, v) ∈ Γ | (u− x)2 + (v − y)2 < ε2}
are neighborhoods of the point (x, y).
1. The subspace D = {(x, 0) | x ∈ R} of Γ is discrete, and is a zero-setin Γ.
Proof. For any (x, 0) ∈ D, for any r > 0, the neighborhood Vr(x, 0) of (x, 0)does not contain any other elements of D. Hence D is discrete.
Consider the function f on Γ defined as f(x, y) = y for every (x, y) ∈ Γ.Clearly f is continuous and Z (f) = D.
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2. Γ is a completely regular space.
Proof. Consider a point p and a closed set F disjoint from it, then p musthave a neighborhood that does not meet F . Then p either has a Vr(p) orNε(p)neighborhood. In the latter case, we can define f(x, y) = (|(x, y)− p|/ε)∧1,which clearly satisfies f [F ] = {1} and f(p) = 0. In the former case, wecan define f as f(p) = 0 and f(q) = 1 for q ∈ ∂Vr(p), and require f to belinear on the segment between p and any such q, clearly f again satisfies theproperty f [F ] = {1} and f(p) = 0. To show that f is continuous from Γ to[0, 1], consider a subbasic closed set [a, b] ⊆ [0, 1], then f�[[a, b]] = clVb\Vawhich is a closed set in Γ. Hence f ∈ C(Γ) and therefore Γ is completelyregular.
3. The subspace ∆ = Γ ∩ (Q×Q)\D is dense in Γ. Hence |C(Γ)| = c.
Proof. Let p ∈ Γ, then it can have a Vr(p) or Nε(p) neighborhood. In thelatter case, since Q is dense in R there must be u, v ∈ Q such that |u− px| <ε/2 and 0 ≤ v− py < ε/2 so that (u, v) ∈ Nε(p). In the former case, we haveNr(p+(0, r)) ⊂ Vr(p), and we can apply the same argument as above. Hence∆ is dense in Γ.
Therefore every continuous function in C(Γ) is uniquely determined byits value on ∆. But since ∆ is countable, that is |∆| = ℵ0, we have
|C(Γ)| ≤ |R∆| = cℵ0 = c.
Also C(Γ) contains at least all the constant functions r for r ∈ R, hencec = |R| ≤ |C(Γ)|. Therefore |C(Γ)| = c.
4. The zero-set D is not C∗-embedded in Γ. Hence Γ is not normal.
Proof. In (1.) we have shown that the closed subspace D, with cardinality|D| = c, is discrete, hence any function on D is continuous. Therefore wehave |C∗(D)| ≥ {0, 1}D = 2c > c (by Cantor’s theorem) while |C∗(Γ)| =|C(Γ)| = c. So D cannot be C∗-embedded since not every function on D canbe continuously extended to all of Γ. But according to 3D.1(3), this impliesthat Γ is not normal.
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5. Every closed set in Γ is a Gδ.
Proof. Every open set V in Γ can be represented as V = U ∪ S, where U isan open set from the subspace topology and S ⊆ D. Then every closed setH in Γ can be represented as H = F ∩ (T ∪ (Γ\D)), where F is a closed setfrom the subspace topology and T ⊆ D. Note that it is known that Γ as asubspace of R × R is normal, hence every closed set F in this topology is azero-set and hence a Gδ (1.10). Also note that for any r > 0, we can writeT ∪ (Γ\D) = (Γ\D) ∪
⋃(t,0)∈T Vr(t, 0), which makes this set open. Hence H
can be written as the intersection of a Gδ and an open set, hence H is alsoa Gδ.
6. Γ contains a closed Gδ that is not a zero-set.
Proof. Since every closed set is a Gδ, by (5.), the statement of 3E.1 dictatesthe existence of such a set.
However, we can also exhibit such a set explicitly (almost). Consider thefollowing closed subsets of Γ, H = D\{(x, 0) | x ∈ Q} and K = D\{(x, 0) |x ∈ R\Q}. Since both of these sets are closed, they are also Gδ’s (5.). Now,suppose that both H and K are zero-sets, then they are completely separated(Theorem 1.15) and hence have disjoint open neighborhoods, say A and Brespectively. Clearly, each of these neighborhood must be a union of someVε(x, 0) neighborhoods.
Pick (x1, 0) ∈ H, then it is contained in a neighborhood Vε1(x1, 0) ⊆ Afor some ε1 > 0. Suppose that we have already picked 2n − 1 such pointsalternating between H and K, with corresponding ε1, . . . , ε2n−1, for somen ∈ N. Then we pick (x2n, 0) ∈ K such that x2n ∈ ]x2n−1 − ε2n−1, x2n−1[ (thisis possible since X\Q is dense in R), which is contained in Vε2n(x2n, 0) ⊆ Bfor some ε2n > 0. Since A and B are disjoint, we can bound ε2n form above,namely ε2n < ε2n−1/2 (this is not the tightest possible bound, but it willsuffice). Next we can pick (x2n+1, 0) ∈ H such that x2n+1 ∈ ]x2n, x2n + ε2n[(this is possible sinceQ is dense in R), which is contained in Vε2n+1(x2n+1, 0) ⊆A for some ε2n+1 > 0. Once again, since A and B have to be disjoint, we canbound ε2n+1 from above, namely ε2n+1 < ε2n/2. We continue this inductivelyand end up with two sequences {xn} and {εn}.
Clearly, {xn} is Cauchy, so it must converge to a limit, say x. Also, {εn}is bounded from above by {ε1/2n−1}, hence its limit is 0. Note that (x, 0) iseither in H or K, so it must be contained in an open neighborhood Vε(x, 0),
51
for some ε > 0, which is contained in only one of A or B. Finally, note thatε is bounded from above by each εn, in other words ε = 0. Which contra-dicts the initial assumption that H and K have disjoint open neighborhoods.Therefore at least one of H or K is not a zero-set.
3L p.51 extension of functions from a discrete set
Let X be a completely regular set.
1. Let {Vα} be a family of disjoint sets in X with nonempty interiors,and such that for each index α, the set
⋃σ 6=α Vσ is closed. Any set D formed
by selecting one element from the interior of each Vα is C-embedded in X.
Proof. Any such D is clearly discrete, hence any function on D is continuous.So let g ∈ C(D) be a continuous function on D. For each α there is a dα ∈ Dsuch that dα ∈ intVα. Since X is completely regular we can find fα ∈ C(X)such that fα(dα) = g(dα) and f [X\ intVα] = 0.
Now consider the following family of closed sets {Vα} ∪ {clX\⋃α Vα},
which forms a closed cover of X. We will show that this family is neighbor-hood finite (cf. 1A.3). Since for any σ we have
⋃α 6=σ Vα closed, then so is⋃
α Vα. Consider a point x ∈ X, since the Vα are disjoint, x can belong toat most one of them. Suppose that x is in neither of Vα’s, then there is aneighborhood around it contained in X\
⋃α Vα. If x is in Vσ, then
⋃α 6=σ Vα
is closed so there must be a neighborhood of x contained in the complementof that set, that is it meets at most two sets of the closed cover.
Finally we can define the function f ∈ RX such that f |Vα = fα|Vα, foreach α, and f | (X\
⋃α Vα) = 0. By 1A.3, the function f is continuous on all
of X and it extends g from D to X, hence D is C-embedded.
2. Let {xn}n∈N be a discrete set (not necessarily closed) in X, and let{rn}n∈N be any convergent sequence of real numbers. Then there existsf ∈ C∗(X) such that f(xn) = rn for all n ∈ N.
Before proceeding it is useful to introduce the generalized analogue of theWeierstrass M -test from analysis.
Lemma 3.2 (Weierstrass M -test). Let X be a topological space and {fn}be a sequence of continuous functions on X. Suppose that for each n ∈ N,we have |fn| ≤ Mn, where
∑∞n=1 Mn is a convergent series of reals. Then
f =∑∞
n=1 fn is continuous.
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Proof. Consider the sequence of partial sums sn =∑n
m=1 fm, each sn iscontinuous. Take any p, q ∈ N such that p ≤ q, then we have
|sq − sp| =
∣∣∣∣∣q∑
m=p+1
fm
∣∣∣∣∣≤
q∑m=p+1
|fm|
≤q∑
m=p+1
Mm
But since∑∞
n=1Mn converges, for any ε > 0 we can find an N ∈ N suchthat for any p, q > N we have |sq − sp| ≤
∑qm=p+1 Mm < ε. Hence the
sequence sn converges uniformly to∑∞
n=1 fn, and the function f =∑∞
n=1 fnis continuous.
And now, on to the proof.
Proof. Consider the sequence of real numbers {sn}n∈N where s1 = r1 andinductively sn+1 = rn+1 − rn. It is clear from the fact that {xn} is discretethat, for each n ∈ N, we can find open sets Vn such that Vn ∩ {xn}n∈N ={xm}nm=1. Also define V0 = ∅. From 3.11(a) we know that a compact set anda closed set disjoint from it are completely separated, noting that {xm}nm=1
is compact, since it is discrete and finite, this means that we can find acontinuous function fn from X to [−sn, sn] such that fn[X\Vn−1] = {sn}and fn[{xm}n−1
m=1] = {0}. Then, by the Weierstrass M -test (Lemma 3.2), thefunction f =
∑∞n=1 fn is continuous and satisfies the desired properties.
3. If X is infinite, then C∗(X) contains a function with infinite range.
Proof. According to 0.13, every infinite Hausdorff space contains a discretecountably infinite subset {xn}n∈N. By (2.), we can find a continuous functionf on X such that f(xn) = 1/n (this function can be bounded by (−1∨f)∧1if necessary). This function f clearly has infinite range.
4. Let D be a countable discrete set in X. TFAE and imply that D isclosed.
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(1) D and any disjoint closed set are completely separated.
(2) D and any disjoint closed set have disjoint neighborhoods.
(3) D is C-embedded in X.
Proof. (1)⇒(2). If two sets are completely separated, then, by 1.15, theyhave disjoint zero-set neighborhoods.
(2)⇒(3). Assume that D and any disjoint closed set have disjoint neigh-borhoods. First note that we can choose disjoint open neighborhoods Un, foreach n ∈ N, each one containing only one point from D, say dn ∈ D, whereD = {dn}n∈N. We do so inductively. Since D is discrete, we can find an openset containing only d1 and no other elements of D. Since X is completelyregular, we can take a zero-set neighborhood of d1 contained in that openset. Now suppose that for some n ∈ N, the points dm, where m < n, havedisjoint zero-set neighborhoods. Then we again find an open set containingonly dn from D, intersect it with the complement of the union of the obtainedzero-set neighborhoods, which gives us an open neighborhood of dn, insidewhich we can take the next zero-set neighborhood containing dn. Finally, wecan take the interiors of these zero-set neighborhoods and call them {Un}n∈N.
Now, let U =⋃n∈N Un, then X\U is a closed set disjoint from D, hence
there must exist disjoint open neighborhoods A and B, containing D andX\U , respectively. Then, for each n ∈ N, the set U ′n = Un ∩A is sill an openneighborhood of X, inside which we can take a zero-set neighborhood Zn ofdn.
Next we show that Wm =⋃n6=m Zn is closed. Consider x ∈ Wm, its is
contained in either exactly one of Un or X\U . If x ∈ Un and n = m, thenx ∈ Um ⊆ X\Wm. If x ∈ Un and n 6= m, then x ∈ Un ∩ (X\Zn) ⊆ X\Wm.If x ∈ X\U , then x ∈ B ⊆ X\Wm. Hence X\Wm is open, or Wm is closed,for any m ∈ N. Finally, since the elements of D are picked from the interiorsof Zn as in (1.), the set D is C-embedded.
(3)⇒(1). Assume that D is C-embedded. But then, directly from 3B.2,since D is also countable, we have that D is completely separated from anyclosed set disjoint from it.
Assume condition (2). Then take any point x ∈ X\D. Since X is com-pletely regular, it is also assumed to be Hausdorff, so {x} is a closed set.Hence, by hypothesis, D and x have disjoint neighborhoods, which impliesthat X\D is open or that D is closed.
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3M p.51 suprema in C(R)
1. Construct a sequence of functions fn in C(R), with fn ≤ 1, for whichsupn fn does not exist in C(R)—that is, whenever g ∈ C(R) satisfies g ≥ fnfor all n, then there is exists h ∈ C(R) such that h ≤ g, h 6= g and h ≥ fnfor each n.
Proof. Consider the sequence of fn defined by
fn(x) =
n(x+ 1) if − 1 ≤ x ≤ (−1 + 1/n)1 if |x| ≤ (1− 1/n)−n(x− 1) if (1− 1/n) ≤ x ≤ 10 if |x| ≥ 1
.
Now suppose that there exists g ∈ C(R) such that g ≥ fn for all n. WLOGwe can assume that g vanishes at some point and it is clear that g|[−1, 1] ≥ 1,so by the Intermediate Value Theorem (Lemma 1.2) we can find x ∈ R suchthat g(x) = 1/2. Then h = g((0∨g)∧1) satisfies the desired properties, andhence supn fn does not exist in C(R).
2. Construct a sequence of functions fn in C(R) for which supn fn exists inC(R), but is not the pointwise supremum—that is (supn fn)(x) 6= supn fn(x)for at least one x.
Proof. Consider the sequence fn = 1 ∧ (n |i|). Clearly supn fn = 1, yet
supnfn(x) =
{1 if x 6= 00 if x = 0
.
3N p.51 the lattice C(X)
Let X be a completely regular space.
1. Let f ≥ 0 in C(X) be given. If
g = supn∈N
(1 ∧ nf)
exists in C(X), then g is 1 on pos f and 0 on X\ cl pos f .
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Proof. First we define the functions fn = 1 ∧ nf . Now suppose that x ∈pos f , then there exists n ∈ N such that n > 1/f(x) > 0. Hence for eachm > n we have fm(x) = 1, hence g(x) ≥ 1. Clearly for each n ∈ N andx ∈ X\ cl pos f we have fn(x) = 0, hence g(x) ≥ 0.
Note that 1 is an upper bound for each fn, hence for x ∈ pos f wehave 1 ≤ g(x) ≤ 1, hence g(x) = 1. And for any x ∈ X\ cl pos f , sinceX is completely regular, there exists a continuous 0 ≤ h ≤ 1 such thath(x) = 0 and h[cl pos f ] = {1}. Again, h is an upper bound for each fn so0 ≤ g(x) ≤ h(x) = 0, hence g(x) = 0.
2. Let V be an open set, and let B denote the family of all functions ≤ 1in C that vanish on X\V . If f = supB exists in C, then f is 1 on V and 0on X\ clV .
Proof. Since X is completely regular, every open set is a union of cozero-sets (3.2). Hence for every x ∈ V , there is a cozero-set Vx ⊆ V containingx. Hence there exists a non-negative g ∈ C(X) such that Vx = pos g, andtherefore the function b = (g/g(x)) ∧ 1 is in B, so f(x) ≥ 1. But since 1is an upper bound for B, we have 1 ≤ f(x) ≤ 1, hence f(x) = 1. Also, foreach x ∈ X\ clV , since X is completely regular, we can find a continuous0 ≤ h ≤ 1 such that h(x) = 0 and h[clV ] = {1}. Again, h is an upper boundfor B, and for each b ∈ B we have b(x) = 0, hence 0 ≤ f(x) ≤ h(x) = 0 orf(x) = 0.
For the following questions, let {fα} be a family of functions in C, andfor r ∈ R, define
U r = cl⋃α
{x | fα(x) > r}.
3. If g ∈ C, and g ≥ fα for every α, then, for each x,
g(x) ≥ sup{r | x ∈ U r}.
Proof. Suppose that for some x ∈ X we have g(x) < sup{r | x ∈ U r}, thenwe can find a real s, such that g(x) < s < sup{r | x ∈ U r}. By continuity weknow that there is a neighborhood V of x such g[V ] ⊆ ]−∞, s[, which impliesthat for each α, fα[V ] ⊆ ]−∞, s[. Which in turn implies that x 6∈ U s, andtherefore s ≥ sup{r | x ∈ U r}. This contradiction completes the proof.
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4. If X is basically disconnected (cf. 1H), and if {fα} is a countablefamily, then each U r is open.
Proof. A countable union of cozero-sets is a cozero-set, hence⋃α{x | fα(x) >
r} is also a cozero-set. In a basically disconnected set, the closure of a cozero-set is open. So each U r is both open and closed.
5. X is basically disconnected iff every countable family with an upperbound in C has a supremum n C.
Proof. Necessity Assume that X is basically disconnected. Consider anycountable family of functions {fα} ⊆ C. First, recall Lemma 3.12 and notethat the U r, as defined above, satisfy the desired conditions (4.) if we reversethe order of inclusion and replace inf with sup in the Lemma. So let us definef as in Lemma 3.12, for each x,
f(x) = sup{r | x ∈ U r}.
Therefore f is continuous. Now we want to show that f is an upper boundfor {fα}. Suppose that for some x, we have f(x) < fα(x). Then we canwrite f(x) < s < fα(x) for some s ∈ R. Then x ∈ U s, which implies thats ≤ f(x) and produces a contradiction. Hence f is an upper bound of {fα},but from (3.) we know that it also must be the least upper bound. Thereforeany countable subfamily of C has a supremum in C.
Sufficiency Assume that every countable subfamily of C with an upperbound has a supremum in C. Take a cozero-set V , say such that there is anon-negative f ∈ C such that V = X\Z (f), and any open set U disjointfrom it. Now define the functions fn = 1 ∧ nf ; by assumption they musthave a supremum g = supn fn in C. From (1.) we know that g[V ] = {1} andg[U ] = {0}. Since U and V are completely separated, they must be containedin disjoint zero-sets and hence have disjoint closures. But according to 1H.1,we have just proved that X is basically disconnected.
6. X is extremally disconnected iff every family with an upper bound inC has a supremum in C.
Proof. Necessity Assume that X is extremally disconnected. Recall that, inan extremally disconnected space, open sets have open closures. So, similarly
57
to (4.), but this time for an arbitrary subfamily {fα} of C, the U r are opensets. The rest of the argument is exactly the same as in the first part of (5.).
Sufficiency Assume that every subfamily of C with an upper bound hasa supremum in C. Now take two disjoint open sets U and V . Let B be thefamily of all functions ≤ 1 that vanish on X\V , then, by the assumption, wehave g = supB in C. Now, according to (2.) we have that g[V ] = {1} andg[U ] = {0}. Since these two sets are completely separated, they are containedin disjoint zero-sets and hence must have disjoint closures. But according to1H.1, we have just proved the at X is extremally disconnected.
3O p.52 totally ordered spaces
Let X be a totally ordered set (of more than one element). We make X intoa topological space by taking as a subbase for the open sets the family of allrays {x | x > a}, denoted by ]−∞, a[, and {x | x < b}, denoted by ]b,∞[.Consistent with notation, basic open intervals are denoted as ]a, b[ and closedones as [a, b].
A nonempty subset S of X is called an interval of X if whenever anelement x of X lies between two elements of S, then x ∈ S. When aninterval is an open set, it is called an open interval. For example, the set ofall positive rationals less than
√2 is an open interval in the totally ordered
space Q.The topology on X is called the interval topology, because the open in-
tervals form a base.
1. Every open set is expressible in a unique way as a union of disjointmaximal open intervals.
Proof. Given an open set, we define an equivalence relation on it as follows:x ≡ y iff ]x, y[ ⊆ U or ]y, x[ ⊆ U . Taking equivalence classes, we see thateach of them is clearly an interval, open, and equivalence classes are alwaysdisjoint. Now suppose that we can write an open set as a union of disjointopen intervals such that no union of a subset of these intervals is itself asinterval (i.e. they are maximal). Then taking equivalence classes with respectto the same equivalence relation as above, it is clear that the equivalenceclasses are the same as the maximal open intervals.
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2. X is a Hausdorff space.
Proof. Let a < b be distinct points of X. Then there are two cases:
(a) There exists c ∈ ]a, b[. Then a and b have disjoint neighborhoods,]−∞, c[ and ]c,∞[, respectively.
(b) a and b are consecutive elements. Then a and b also have disjointneighborhoods, ]−∞, b[ and ]a,∞[, respectively.
3. For any nonempty subset A, if supA exists, then supA ∈ clA.
Proof. By definition, for any neighborhood ]u, v[ of supA, there must be anelement a ∈ A such that u < a ≤ supA. But these neighborhoods form theneighborhood base at supA. Hence every neighborhood of supA containspoints of A, in other words supA ∈ clA.
4. For A ⊆ X, the relative topology on A contains the interval topology,but the two need not be the same.
Proof. The subbasic sets of the interval topology on A are of the form {x ∈A | x < a} and {x ∈ A | x > b} for some a, b ∈ A. But since A ⊆ X, theseare also some of the subbasic sets of the subspace topology on A. Whichhave the form A ∩ ]−∞, a[ and A ∩ ]b,∞[ for some a, b ∈ X.
The two, however, need not be the same. Consider X = R\{x | |x| < 1}.In the relative topology on X, the point 0 is an isolated point. But in itsinterval topology X is homeomorphic to R.
5. If A is an interval of X, then the relative topology on A does coincidewith the interval topology.
Proof. Take a subbasic open set of the relative topology on A, say B = {x ∈A | x < b} (resp. x > b) for some b ∈ X. We know that either b ∈ A orb 6∈ A. In the latter case either B = ∅ or B = A, which are both open setsin the interval topology on A. In the former case B is a subbasic open setof the interval topology itself. So the interval topology on A contains therelative topology on A, and, by (4.), the reverse also holds. Therefore thetwo topologies coincide.
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6. X is connected iff X is Dedekind-complete (0.6) and has no consecu-tive elements.
Proof. Necessity Assume that X is connected. Suppose that a, b ∈ X areconsecutive elements, assuming that WLOG a < b. Then X is disconnected,since we can write it as a union of two disjoint open sets, namely X =]−∞, b[ ∪ ]a,∞[. Hence X cannot have consecutive elements. Now, supposethat X is not Dedekind-complete. Then, there must be a subset B ⊆ A suchthat B has an upper bound but no sup in A. First we construct the setL =
⋃b∈B ]−∞, b[, then we let U be the set of all upper bound of B (which
are all also upper bounds of L), which we can write as U =⋃u∈U ]u,∞[.
Clearly X = L ∪ U and by construction, both L and U are non-empty,disjoint and open. Which implies that X is disconnected. Hence X must beDedekind-complete.
Sufficiency Assume that X is Dedekind-complete and has no consecutiveelements. Suppose that X is disconnected, then we can write X = A′ ∪ B′,where A′ and B′ are non-empty, open and disjoint. Since A′ is open, from(1.) we know that we can write it uniquely as a union of disjoint maximalintervals. Let A be one of these intervals, and let B = B′ ∪ (A′\A), theset B is clearly still open. Every point of B bounds A either from belowor from above. WLOG assume that B contains points that bound A fromabove, in fact let U be the set of all upper bounds of A that are contained inA. Since A is Dedekind-complete, it must have a sup, and U must have aninf. Suppose supA 6= inf U , then supA and inf U are consecutive elements,hence we must have supA = inf U . Since A and U are disjoint, we must haveeither supA ∈ A or supA ∈ U , WLOG assume that a = supA ∈ A. Since Ais open, there must be a neighborhood of the form ]c, d[, for some c, d ∈ X,around a contained in A. But any such neighborhood will contain points ofU , hence A cannot be open as per the original assumption. Hence X cannotbe disconnected.
7. X is compact iff it is lattice-complete (0.5). Thus, X is compact iff itis Dedekind-complete and has both a first element and a last element.
Proof. Necessity Assume that X is compact. Consider the family of closedsets of the form [x,∞[, for each x ∈ X. This family has the finite inter-section property, hence the intersection
⋂x∈X [x,∞[ is non-empty, since X is
compact. Every point in that intersection satisfies the properties of supX,
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hence there is only one point in the intersection, and it is the last element ofX. Similarly, X has a first element.
Now, let S ⊆ X be any subset. Consider the family of closed intervals{[x,∞[ | x ∈ S} ∪ {]−∞, x[ | x is an upper bound of S}. Once again, thisfamily of closed sets has the finite intersection property, so its intersectionis non-empty, since X is compact. By arguments similar to those of theparagraph above, that intersection must contain a single point which is supS.So every subset of X has a sup and similarly an inf.
Sufficiency Assume that X is lattice-complete. Let T be a family ofclosed subsets of X with the finite intersection property. Let A be the setof all a ∈ X such that [a,∞[ meets the intersection of every finite subfamilyof T . Note that A is non-empty, since at least inf X ∈ A. Now, take F tobe the intersection of some finite subfamily of T . The set F is closed, byassumption, inf F must exist in X, and by (3.) supF ∈ F . Since [inf F,∞[contains F , it must also meet every other intersection of a finite subfamilyof T , hence inf F ∈ A. Also, note that supF ≥ b = supA, for if a > supF ,for some a ∈ A, we would have [a,∞[ ∩ F = ∅ which is contrary to theconstruction of A. So for every such F , we have b = supA ≥ inf F andb ≤ supF (also recall that, by (3.), supF ∈ F since it is closed), whichbasically means that b is in F . But this implies that b ∈
⋂T . Hence X is
compact.
8. X has a totally ordered compactification. Hence X is completely reg-ular.
Proof. We know that X has an essentially unique Dedekind completion (0.6),which can be augmented with inf X and supX to create an essentially uniquelattice completion X ′ of X. Since X ′ is lattice complete, it must be compact(7.). The space X, as a subspace of X ′, must be completely regular sinceit is a subset of a compact space (3.14). Now it suffices to show that theinterval topology on X is the same as its subspace topology from X ′.
Consider a subbasic open set of the subspace topology on X, say ]−∞, b[∩X (resp. ]b,∞[ ∩ X) for some b ∈ X ′. Then there must exist a subsetB ⊆ X such that b = supB (resp. b = inf B). But then, we can write B =⋃x∈B ]−∞, x[ (resp. B =
⋃x∈B ]x,∞[), hence the interval topology contains
the subspace topology on X, and since the subspace topology automaticallycontains the interval topology on X (4.), the two topologies must coincide.HenceX is completely regular and has a totally ordered compactification.
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9. X is normal.
Proof. Given the disjoint closed sets H,K ⊆ X, we want to construct afunction f ∈ C(X) such that f [H] = {0} and f [K] = {1}. First we definef(x) = 0 if x ∈ H and f(x) = 1 if x ∈ K. Note that the set X\(H ∪K) isopen, so it must be composed of disjoint maximal open intervals (1.), let Abe one of these intervals. Then there are several possibilities.
First, consider the trivial possibilities. Suppose that A is disconnectedfrom X\A, then we can set f |A to any continuous function since clA = A.Also, suppose that A is disconnected from the set of lower (or upper) boundsof A (or if the set of bounds is empty), but not disconnected from X\A.Then inf A ∈ H ∪ K must exist and clA = A ∪ {inf A}, and we can setf(x) = f(inf A) for all x ∈ A, since f is already defined for inf A (resp.supA).
Then, suppose that A is not disconnected from X\A, so both extremam1 = inf A,m2 = supA ∈ H∪K exist and we have clA = A∪{inf A, supA}.Then f is already defined for both m1 and m2 and there are two remainingcases. If f(m1) = f(m2), then we simply define f(x) = f(m1) for every x ∈A. And if f(m1) 6= f(m2), then, since X is completely regular, we can find acontinuous function g such that g(m1) = f(m1) and g[[m2,∞[] = {f(m2)}.Then we can define f |A = g|A.
Finally, f is well defined and is such that f |(H ∪ K) is continuous andf | cl(X\(H ∪K)) is also continuous. So, by 1A.3, f is continuous as well asf [H] = {0} and f [K] = {1}. Therefore X is normal.
3P p.53 convergence of z-filters
For this problem we need a few definitions for a completely regular space X.A point p ∈ X is said to be a cluster point of a z-filter F if every neighborhoodof p meets every member of F . The z-filter F is said to converge to the limitp ∈ X if every neighborhood of p contains a member of F . We also definethe following notation: the z-filter on a completely regular space X that iscomposed of all the zero-sets containing a given point p is denoted by Ap (infact Ap is a z-ultrafilter (3.18)).
Finally consider a small lemma:
Lemma 3.3. Let X be a topological space, and F any z-filer on X. Then Fis an intersection of prime z-filters.
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Proof. According to Theorem 2.8, every z-ideal in C(X) is an intersectionof prime ideals. Note that Z�[F ] is a z-ideal, hence we can write Z�[F ] =⋂α Pα, where each Pα is a prime ideal. But according to 2.12(a), each Z [Pα]
is a prime z-filter. Hence we can write F =⋂αZ [Pα].
Let F be a z-filter on a completely regular space X, and let p be a clusterpoint of F .
1. F converges to p iff F is contained in a unique z-ultrafilter.
Proof. Necessity If F is a z-filter converging to p, the Ap is the unique z-ultrafilter containing F (3.18(d)).
Sufficiency Assume that F is contained in a unique z-ultrafilter. Accord-ing to Lemma 3.3, F is an intersection of all the prime z-filters containingit. Suppose F could be written as the intersection of more than one primez-filter, then, since every prime z-filter is contained in a unique z-ultrafilter(2.13), F must be contained in more than one z-ultrafilter. But, by assump-tion, F is contained in a unique z-ultrafilter, hence F must be prime. Finally,since F is prime and p is its cluster point, by Theorem 3.17, F converges top.
2. If X is compact and p is the only cluster point of F , then F convergesto p.
Proof. Since X is compact, for each z-filter the intersection of all its elementsis non-empty (since a z-filter has the finite intersection property). Hence, by3.16, every z-filter on X has a cluster point. So every z-ultrafilter wouldhave a cluster point and converge to it (3.16(b)), then, according to 3.18,every z-ultrafilter has the form Ap. If F was contained in more than onez-ultrafilter, then it would have more than one cluster point (3.18(a)). SoF is contained in the unique z-ultrafilter Ap, and therefore it converges to p(1.).
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4 Fixed ideals. Compact space
in the sequel, all given spaces are assumed to be completelyregular.
4A p.60 maximal ideals; z-ideals
1. Maximal fixed ideal coincides with fixed maximal ideal, and maximalfree ideal with free maximal ideal.
Proof. A fixed maximal ideal is a maximal ideal which is fixed, and a maximalfixed ideal is a fixed ideal which is maximal with respect to the propertiesof being fixed and proper. Clearly a fixed maximal ideal is also a maximalfixed ideal. Suppose I is any maximal fixed ideal. Then
⋂Z [I] is non-empty,
hence Z [I] has at least one cluster point. So, by 3.18(a), I is contained inan ultrafilter converging to that point (hence a fixed ultrafilter). Taking thepreimage of that ultrafilter we end up with a fixed maximal ideal containingI. Hence a maximal fixed ideal is also a fixed maximal ideal.
Taking similar definitions for free maximal ideal and maximal free ideal.Again, it is clear that a free maximal ideal is a maximal free ideal. Now, takeI to be a maximal free ideal. It must be contained in some maximal ideal,which must also be free, since it contains I. Hence a maximal free ideal isalso a free maximal ideal.
2. C and C∗ are semi-simple (i.e. the intersection of all maximal idealsis (0).
Proof. Consider any f ∈ C such that f 6= 0, and its zero-set Z = Z (f).Take a point x ∈ X\Z, then, by complete regularity of X, there exists g ∈ Csuch that g(x) = 0 and g[Z] = {1}. Note that f 2 + g2 is a unit of C, so fcannot belong to the same ideal as g. So any maximal ideal that contains g,does not contain f . But 0 is in every maximal ideal, so the intersection ofall maximal ideals is (0) or C is semi-simple.
Consider f ∈ C∗, such that f 6= 0, and, for a fixed 0 < ε < sup |f |[X], theset E = {x ∈ X | |f(x)| ≤ ε}. Suppose that Z (f) = ∅, then f cannot be inany fixed maximal ideal, but we know that fixed maximal ideals always exist(4.4). If Z (f) 6= ∅, then we can pick a point x ∈ X\E, and, by completeregularity, find g ∈ C∗ such that g(x) = 0 and g[E] = {1}. Then f 2 + g2 is a
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unit of C∗, so f cannot belong to the same (maximal) ideal as g. But 0 is inevery maximal ideal, so the intersection of all maximal ideals is (0) and C∗
is semi-simple.
3. Prove directly that M p (where for p ∈ X, M p = {f ∈ C | f(p) = 0})is a maximal ideal in C and so is M ∗
p (M ∗p = {f ∈ C∗ | f(p) = 0}) in C∗.
Proof. Consider f, g ∈ M p, then f(p) = g(p) = 0 and so (f + g)(p) =f(p) + g(p) = 0, hence f + g ∈ M p. Consider f ∈ M p and g ∈ C, thenf(p) = 0 and so (fg)(p) = f(p)g(p) = 0, hence fg ∈ M p. Hence M p isan ideal in C. Now, consider f 6∈ M p and the ideal I = (M p, f), notethat p 6∈ Z (f). Since X is completely regular, we can find g ∈ C such thatg(p) = 0 and g[Z (f)] = {1}, note that g ∈ M p ⊂ I. But then f 2 + g2 is aunit in C, so I cannot be proper. Therefore M p is a maximal ideal in C.
By arguments similar to those of the previous paragraph, we can say thatM ∗
p is an ideal in C∗. Now, take f 6∈ M ∗p and the ideal I = (M ∗
p, f), notethat p 6∈ Z (f). Also, consider the set E = {x ∈ X | |f(x)| ≤ ε} for some0 < ε < |f(p)|. Since X is completely regular, we can find g ∈ C∗ such thatg(p) = 0 and g[E] = {1}, note that g ∈ M ∗
p. But then f 2 + g2 is a unit inC∗, so I cannot be proper. Therefore M ∗
p is a maximal ideal in C∗.
4. Either in C or in C∗, if f belongs to every maximal ideal that g belongsto, then Z (g) ⊆ Z (f).
Proof. Suppose that there exists x ∈ Z (g) such that x 6∈ Z (f). Then g ∈M x (resp. M ∗
x) but f 6∈M ∗x), so by the hypothesis, we must have x ∈ Z (f).
Therefore Z (g) ⊆ Z (f).
5. The following algebraic condition is necessary and sufficient that anideal I in C be a z-ideal: given f , if there exists g ∈ I such that f belongsto every maximal ideal containing g then f ∈ I.
Proof. Necessity Assume that I ⊂ C is a z-ideal. Let f ∈ C and supposethat there exists g ∈ I such that f belongs to every maximal ideal containingg. Then, by (4.), we have Z (g) ⊆ Z (f), which implies that Z (f) ∈ Z [I],which in turn implies that f ∈ I, since I is a z-ideal.
Sufficiency Assume the condition in question. Let Z ∈ Z [I], say Z =Z (g), for some g ∈ I. Now, consider any f ∈ C such that Z (f) = Z. Since
65
every maximal ideal is a z-ideal, if g belongs to one, then so does f . So, bythe hypothesis, we have f ∈ I, which implies that I is a z-ideal.
4B p.60 principal maximal ideals
1. A point p of X is isolated iff the ideal M p (resp. M ∗p) is principal.
Proof. Necessity Assume that p is isolated, then we can define the functionf ∈ C (resp. C∗) to be f(p) = 0 and f(x) = 1 for any x 6= p. Then, forany g ∈ C (resp. C∗) such that g(p) = 0, we can write g = fg. Hence themaximal ideal M p = (f) (reps. M ∗
p) is principal.Sufficiency Assume that M p (resp. M ∗
p) is principal and given by (f).First, note that Z (f) must be {p}. For if we could find q ∈ Z (f) such thatq 6= p, then (f) would also be contained in M q 6= M p (resp. M ∗
q) (3.18(c))and (f) would not be maximal. Since (f) is a maximal idea, it must bea z-ideal. So it must also contain f 1/3, since Z (f 1/3) = Z (f). But thenwe can write f 1/3 = gf , for some g ∈ C (resp. C∗). Now, suppose that pis not isolated, that is every neighborhood of p contains points other thanp. then the function g = f−2/3 restricted to X\{p} is unbounded in everyneighborhood of p, and hence g cannot be continuous (much less bounded)on X. This contradiction completes the proof.
2. X is finite iff every maximal ideal in C (resp. C∗) is principal.
Proof. Necessity Assume that Xis finite, then clearly C(X) = C∗(X). But,by 2F.3, every ideal in C(X) (and hence C∗(X)) must be principal.
Sufficiency Assume that every maximal ideal in C (resp. C∗) is principal.Then, by (1.), we have that X is discrete. It remains to be shown that Xmust also be compact. So, since X is discrete, it can be compact only if it isfinite.
Case C: Let M be a maximal ideal in C. Since it is maximal, it must alsobe principal, or M = (f) for some f ∈ C. But then
⋂Z [M ] = Z (f),
which must be non-empty. So every maximal ideal in C is fixed, whichimplies that X is compact (Theorem 4.11).
Case C∗: Let M be a free maximal ideal in C∗. Since it is maximal, it mustalso be principal, or M = (f) for some f ∈ C∗. But then
⋂Z [M ] =
Z (f) = ∅. Now, consider the function f 1/3, we can write f = (f 1/3)3,
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so f 1/3 ∈ M , since a maximal ideal is also prime (0.15). BecauseM = (f), we can write f 1/3 = gf , for some g ∈ C∗. But g mustbe f−2/3, which is continuous because f is nowhere zero, but clearlyunbounded. This contradicts the hypothesis, so every maximal ideal inC∗ is fixed. Again, by Theorem 4.11, this proves that X is compact.
4C p.61 finitely generated ideals
1. Every finitely generated ideal in C is fixed.
Proof. Consider the ideal I = (f1, . . . , fn) in C, then the family of zero-sets {Z (f1), . . . ,Z (fn)} forms a subbase for the z-filter Z [I]. Therefore⋂
Z [I] =⋂ni=1 Z (fi), which is non-empty because Z [I] is a z-filter and is
closed under finite intersections.
2. A necessary and sufficient condition that every finitely generated idealin C∗ be fixed is that X be pseudocompact.
Proof. Necessity Assume that every finitely generated ideal in C∗ is fixed,but X is not pseudocompact (that is C 6= C∗). Then C must contain anunbounded unit u that is bounded away from 0, say u = |f | ∨ 1 where f isan unbounded function in C. Note that u−1 ∈ C∗, but u 6∈ C∗, hence (u−1)is an ideal. This ideal is finitely generated and also free, since Z (u−1) = ∅,which contradicts the hypothesis. Hence X must be pseudocompact.
Sufficiency Assume that X is pseudocompact, that is C∗ = C. Then,directly from (1.), we know that every finitely generated ideal in C∗ is fixed.
3. If X is infinite, then both C and C∗ contain fixed ideals that are notfinitely generated.
Proof. Since X is infinite and Hausdorff, it must contain a discrete countablyinfinite subset (0.13), say N . Let p ∈ N and consider the family F of zero-sets such that, for each Z ∈ F , the zero-set Z contains p and Z∩N has finitecomplement in N . Clearly F is a z-filter on X. Note that F is different fromthe z-filter of all zero-sets that contain N . Suppose that F is a finite subsetof N , note that, since N is discrete, clF does not meet N\F and clN\F does
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not meet F . Then there exists a cozero-set containing F and disjoint fromN\F , by 3B.4, and hence there exists a zero-set Z such that Z ∩N = N\F .As a consequence, we have
⋂F ∩N = {p}.
If⋂F = Z, for some Z ∈ Z (X), then F is also different from the
principal z-filter generated by Z. Since N is countably infinite, we can writeN = L∪M , where L and M are disjoint and both countably infinite. Since Nis discrete, we can say that clL∩M = ∅ and clM ∩L = ∅, then again thereexists a zero-set Z ′ such that Z ′ ∩N = L (3B.4). Clearly Z ′′ = Z ∪ Z ′ is inthe principal z-filter generated by Z, but Z ′′ is not in F , since N\(N ∩Z ′′) =M\{p} is infinite.
Consider the z-ideal J = Z�[F ] in C (resp. C∗), clearly J is a fixed idealsince p ∈
⋂F . If J is finitely generated, then F must have a finite a finite
subbase. In other words F is principal, say⋂F = Z. But from the previous
paragaph, we know that this is impossible. This contradiction completes theproof.
4D p.61 functions with compact support
The support of f , denoted by S(f), is, by definition, the closure of X\Z (f).Let CK(X) denote the family of all functions in C having compact support.
1. If X is compact, then CK = C; otherwise, CK is both an ideal in Cand an ideal in C∗.
Proof. If X is compact, then, since the support of any function is a closedset and hence compact, every function in C is also in CK or C = CK .
Assume that X is not compact. Then take any f, g ∈ CK , the supportof f + g is a closed subset of S(f) ∪ S(g) (which is clearly compact). So thesupport of f + g is also compact. Take any f ∈ CK and g ∈ C, the supportof fg is a closed subset of S(f) ∩ S(g) (which is clearly compact). So thesupport of fg is also compact. The support of a unit in C is X, and hencecannot be compact. Therefore CK contains not unit of C and satisfies theproperties of an ideal in C. Since every function in CK is non-zero only on asubset of a compact set, it must be bounded. Hence CK ∩C∗ = CK , but anyideal in C intersected with C∗ is an ideal in C∗. Hence CK is also an idealin C∗.
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2. CK(Q) = (0).
Proof. Consider any non-zero f ∈ C(Q), then, for any x ∈ X\Z (f), the setS(f) is a closed neighborhood of x. Suppose that S(f) is compact, then x hasa compact neighborhood. But no point of Q has a compact neighborhood.This contradiction completes the proof.
To verify the claim of the previous paragraph, consider a point x of Q, andsuppose that it has a compact neighborhood K. Since K is a neighborhoodof X, it must contain an open interval ]a, b[ with a < x < b and a, b ∈ Q. Butthen we can always find an irrational i such that a < i < b, and construct afunction which is unbounded in every neighborhood of i, yet still continuouson Q and K (cf. 1C.5, Case C(Q)). But it is impossible to construct anunbounded function on a compact set. This contradiction completes theproof.
3. CK is a free ideal iff X is locally compact but not compact.First, recall that a space X is said to be locally compact if every point of
X has a compact neighborhood.
Proof. Necessity Assume that CK is a free ideal, then X cannot be compact(Theorem 4.11). Also, for each x ∈ X there must exist f ∈ CK such thatf(x) 6= 0. Therefore each x ∈ X has a compact neighborhood, namely S(f).So X is locally compact.
Sufficiency Assume that X is locally compact but not compact, so thatCK is a (proper) ideal in C. Consider any x ∈ X, by hypothesis, it musthave a compact neighborhood. Since X is completely regular, the cozero-setsform a base for the open sets of X (3.2), so there exists f ∈ C such thatX\Z (f) is a subset of the compact neighborhood of x and contains x. HenceS(f) must also be compact, since it is a closed subset of a compact set, andf ∈ CK . Therefore, for each x ∈ X, there exists f ∈ CK such that f(x) 6= 0,hence CK is free.
4. And ideal I in C of C∗ is free iff, for every compact set A, there existsf ∈ I having no zeros in A.
Proof. Necessity Assume that I is free, then for every x ∈ X, there existsfx ∈ I such that X\Z (fx) is a neighborhood of x. Let A be a compact set,then consider a cover of A consisting of sets of the form X\Z (fx) for each
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x ∈ A. Since A is compact, there must exists a finite subcover of A, say thecomplements of Z (fx1), . . . ,Z (fxn). Then the function g = f 2
x1+ · · ·+ f 2
xn isalso in I and has no zeros in A.
Sufficiency Assume that, for every compact A ⊆ X, there exists f ∈ Ihaving no zeros in A. Note that, since X is Hausdorff, the singleton {x} iscompact. So for every x ∈ X there is a function f ∈ I such that f(x) 6= 0.Therefore I is free.
5. CK is contained in every free ideal in C and in every free ideal in C∗.
Proof. Let I be any free ideal in C or C∗. let f ∈ CK , then, by (4.), thereexists g ∈ I such that g does not vanish on S(f). But this implies thatZ (g) ⊆ intZ (f). So, by 1D.1, we can find h ∈ C∗ (since S(f) is compact)such that f = gh. So f ∈ I or CK ⊆ I.
4E p.61 free ideals
1. Let f ∈ C∗. If f belongs to no free ideal in C∗, then Z (f) is compact.But the converse is false. Compare Lemma 4.10.
Proof. If f belongs to no free ideal, then Z (f) belongs to no free z-filter onX. So, according to Lemma 4.10, the zero-set Z (f) must be compact.
Now, consider the case X = N, I = (j) and f ∈ C∗ given by the char-acteristic function χN\{1} (the characteristic function χA of the set A is de-fined to be 1 for every point in A and 0 for every point outside it). ThenZ (f j) = Z (f) = {1}, which is compact, but f j ∈ I, which is free.
2. The intersection of all free maximal ideals in C coincides with the setof all f in C for which Z (f) meets every noncompact zero-set.
Proof. Since we are only dealing with z-ideals in this question, we can talkabout z-filters instead. All the results are equally valid for the correspondingz-ideals.
According to Theorem 4.10, a zero-set is noncompact iff it belongs toat least one free z-filter. Hence every element of a free z-ultrafilter mustbe noncompact, and every noncompact zero-set belongs to at least one freez-ultrafilter.
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Suppose that a zero-set intersects every noncompact zero-set, then, forany free ultrafilter, it must intersect each one of its elements. Hence, byTheorem 2.6(b), this zero-set must be in every free z-ultrafilter.
Suppose that a zero-set is in every free z-ultrafilter, then, for any freez-ultrafilter, it must intersect each of its elements. Hence it must intersectevery noncompact zero-set.
3. A z-filter is a base for the closed sets iff it is free.
Proof. Necessity Assume that a z-filter is a base for the closed sets. Since Xis completely regular, in order to have a free z-filter, X cannot be compact,which also implies that it must have have more than one point. So given twodistinct points of X, by complete regularity, they must have disjoint zero-set(closed) neighborhoods, say H and K. By hypothesis, both H and K can bewritten as intersections of elements from the z-filter. So since H ∩K = ∅,the z-filter must be free.
Sufficiency Assume that a z-filter is free. Then, according to 3.2, thefilter is a base for the closed sets if for every closed H ⊆ X, and everyx 6∈ H, there is a set F in the filter containing H but not x. Suppose thisis not the case, then there must exist a closed H ⊆ X and x 6∈ H such thatevery element of the filter does not contain H or contains x.
Recall that, since X is completely regular, the zero-sets are a base for theclosed sets. So for a given closed set H there exists a zero-set Z1 containingit. Also, since the z-filter is free, for a given x 6∈ H, it must contain anelement Z2 which does not contain x. Hence Z1 ∪ Z2 ⊇ Z2, which is also inthe z-filter, contains H and does not contain x. Hence, by contradiction, thez-filter must be a base for the closed sets in X.
4. Let S be a compact set X, I a free ideal in C(X), and J the set of allrestrictions f |S, for f ∈ I. Then J = C(S).
Proof. First, we show that J is an ideal in C(S). Take any a, b ∈ J , thenthere must exist f, g ∈ I such that a = f |S and b = g|S. Since I is an ideal,f + g ∈ I, hence (f + g)|S = a + b ∈ J . Take any a ∈ J and b ∈ C(S),then there exist f ∈ I and g ∈ C(X) such that a = f |S and b = g|S(recall that S is C-embedded, by 3.11(c)). Since f ∈ I, we have fg ∈ I, so(fg)|S = ab ∈ J .
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Note that, by 4D.4, since I is free and S is compact, there must existf ∈ I such that Z (f) ∩ S = ∅. Then f |S ∈ J , but f |S is a unit of C(S).Hence J must be improper or J = C(S).
4F p.61 z-ultrafilters on R that contain no smallsets
Let F denote the family of all closed subsets of R whose complements are offinite Lebesgue measure (the Lebesgue measure of a set A ⊆ R is denoted byν(A)).
1. F is a free z-filter.
Proof. Since R has infinite measure, any set whose complement has finitemeasure must have infinite measure. Let A,B ∈ F , then R\(A ∩ B) hasmeasure ≤ ν(R\A)+ν(R\B) which is still finite, hence A∩B ∈ F . ConsiderA ∈ F and closed B ⊆ R such that A ⊆ B. Then R\B ⊆ R\A, or ν(R\B) ≤ν(R\A), which is finite, hence B ∈ F . Lastly, ∅ 6∈ F , since ν(R\∅) =ν(R) =∞. Therefore F is a z-filter.
Take any x ∈ R, then ]x−1, x+1[ has finite measure, so R\]x− 1, x+ 1[ ∈F . Therefore
⋂F = ∅ or F is free.
2. F is not a z-ultrafilter.
Proof. Assume that F is a z-ultrafilter. Consider the set
Z =∞⋃
n=−∞
[2n, 2n+ 1].
It and its complement both have infinite measure. So if Z ′ is another zero-setthat is disjoint from Z, we have Z ⊆ R\Z ′, hence the complement of Z ′ hasinfinite measure and Z ′ 6∈ F . Therefore Z intersects every element of F ,hence it must be an element of F , by Theorem 2.6(b). This contradictioncompletes the proof.
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3. Any z-ultrafilter containing F contains only sets of infinite measure.
Proof. Suppose that U is a z-ultrafilter containing F and Z ∈ U such thatZ has finite measure. Then we can find an open set U of finite measurecontaining Z, so that R\U ∈ F ⊆ U . But (R\U) ∩ Z = ∅, hence U cannotcontain Z.
4G p.61 base for a free ultrafilter
1. A free ultrafilter cannot have a countable base.
Proof. Suppose that U is a free ultrafilter with a countable base {Un}n∈N.First note that no Un can be finite. For if Un, for some n, is finite, then foreach x ∈ Un we can find Umx such that x 6∈ Umx . Then V =
⋂x∈Un Umx
is a finite intersection of elements of U , hence it is also in U . But thenV ∩ Un = ∅ ∈ U , and that would imply that U is an improper filter.
Consider the set U1, since it is infinite, we can pick two distinct pointsa1 and b1. Now, suppose that we already have 2n distinct points An ={a1, . . . , an} and Bn = {b1, . . . , bn} such that am, bm ∈ Um, for 1 ≤ m ≤ n.Consider the set Un+1, since it is infinite, Un+1\(An ∪ Bn) is still infinite, sowe can pick two distinct points an+1, bn+1 ∈ Un+1\(An ∪ Bn). Continuingby induction, we end up with two disjoint sets A = {a1, a2, . . .} and B ={b1, b2, . . .}, both intersecting each Un.
Since A and B intersect every element of the base for U , they must alsointersect every element of U . From Theorem 2.6(b) (although this theoremis about z-ultrafilters, it can be adjusted to ultrafilters by considering theunderlying space X under the discrete topology), we must have both A andB as elements of U . But this would imply that A ∩ B = ∅ ∈ U , making itan improper filter. This contradiction completes the proof.
2. More generally, a free ultrafilter, each of whose members is of power≥ m, cannot have a base of power ≤ m.
Proof. Suppose U is a free ultrafilter which satisfies the specified conditions.And suppose that it has a base of power m, namely {Uα}α∈m (the argumentis exactly the same for any base of power less than m). Recall from (1.), thatno Uα can be finite, hence m is at least countably infinite.
Consider the set U0, since it is infinite, we can pick two distinct pointsa0, b0 ∈ U0. Now, consider α+ 1 < m (a successor ordinal) and suppose that
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we already have the distinct points Aα = {a0, . . . , aα} and Bα = {b0, . . . , bα},such that aβ, bβ ∈ Uβ, for 0 ≤ β ≤ α. Consider the set Uα+1, since it hascardinality ≥ m and Aα ∪Bα has cardinality < m, the set Uα+1\(Aα ∪Bα) isstill infinite. So we can pick two distinct points aα+1, bα+1 ∈ Uα+1\(Aα∪Bα).Next, consider γ < m (a limit ordinal) and suppose that for each α < γ wealready have distinct points Aα = {a0, . . . , aα} and Bα = {b0, . . . , bα} suchthat aβ, bβ ∈ Uβ, for 0 ≤ β ≤ α. Let A′γ =
⋃α<γ Aα and B′γ =
⋃α<γ Bα, then
consider the set Uγ. Since the cardinality of A′γ and B′γ is |γ| < m and thecardinality of Uγ is m, the set Uγ\(A′γ∪B′γ) is still infinite, so we can pick twodistinct points aγ, bγ ∈ Uγ\(A′γ ∪ B′γ). Continuing by transfinite induction,we end up with two disjoint sets A =
⋃α<mAα and B =
⋃α<mBα, both
intersecting each Uα.Since A and B intersect every element of the base for U , they must also
intersect every element of U . From Theorem 2.6(b), we must have both Aand B as elements of U . But this would imply that A∩B = ∅ ∈ U , makingit an improper filter. This contradiction completes the proof.
4H p.62 the mapping τ#
Let X and Y be spaces with the same underlying point set, such that theidentity mapping τ :X → Y is continuous.
Before proceeding we should provide another definition. Given two topo-logical spaces X and Y and a continuous map τ :X → Y , we define the mapτ# acting on subfamilies of Z (X) as follows: given a subfamily F of Z (X),we have
τ#F = {Z ∈ Z (Y ) | τ�[Z] ∈ F}.
1. If F is a z-filter on X, then τ#F = F ∩ Z (Y ).
Proof. First, note that, since τ is a continuous identity map from X to Y ,any open set in Y is also open in X. Hence the topology on X contains thetopology on Y , which implies that Z (Y ) ⊆ Z (X). Now, consider Z ∈ τ#F .We already know that Z ∈ Z (Y ). We also know that Z = τ�[Z] ∈ F , henceZ ∈ F ∩ Z (Y ). Finally, consider Z ∈ F ∩ Z (Y ). Then τ�[Z] = Z ∈ F ,hence Z ∈ τ#F . Therefore τ#F = F ∩ Z (Y ).
2. If U is a z-ultrafilter on X, τ#U need not be a z-ultrafilter on Y .
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Proof. Suppose that we are dealing with X = Y = [0, 1], where Y is endowedwith the usual interval topology and X is discrete. Consider a point p ∈ Yand the z-filter F = Z [Op] of all zero-set neighborhoods of p in Y . ThenF is a base for a filter F ′ on X. Note that F ′ is not maximal, since anultrafilter must contain either {p} or X\{p} (Lemma 4.32) and F ′ containsneither. Hence there must exist an ultrafilter U ′ on X which contains bothF ′ and X\{p}.
Recall that the zero-sets on Y are precisely the closed sets (1.10), hence,by (1.), we know that U = τ#U ′ is the family of all sets in U ′ that are closedin Y . But then U must contain Z [Op], hence the only z-ultrafilter thatcontains U is Z [M p] (4I.23), which is the principal z-filter generated by {p}.However, by construction, U cannot contain {p}, therefore U cannot be az-ultrafilter.
4I p.62 the ideals Op
For p ∈ X, let Op denote the set of all f in C for which Z (f) is a neighbor-hood of p.
1. Op is a z-ideal in C, Op ⊆M p, and⋂
Z [Op] = {p}.
Proof. Take f, g ∈ Op, then Z (f + g) ⊇ Z (f) ∩ Z (g) which is still aneighborhood of p, hence f + g ∈ Op. Take f ∈ Op and g ∈ C, thenZ (fg) = Z (f)∪Z (g) ⊇ Z (f) which is a neighborhood of p, hence fg ∈ Op.Therefore Op is an ideal in C.
Now, take f ∈ Op and consider any g ∈ C such that Z (g) = Z (f). ThenZ (g) is also a neighborhood of p, hence g ∈ Op. Therefore Op is a z-ideal.
If f ∈ Op, then f(p) = 0, hence f ∈M p. Therefore Op ⊆M p. Clearly,p ∈
⋂Z [Op]. Take x ∈ X different from p, then, by complete regularity,
x and p have disjoint zero-set neighborhoods, hence there exists Z ∈ Z [Op]such that p ∈ Z but x 6∈ Z. Therefore
⋂Z [Op] = {p}.
2. M p is the only maximal ideal, fixed or free, that contains Op.
Proof. First, we show that the z-filter Z [Op] converges to p. Let U be anyneighborhood of p, then, by complete regularity, it must contain a zero-set
2Apologies for the forward reference.3Once again, apologies for the forward reference.
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neighborhood Z of p. But we know that Z ∈ Z [Op], so Z [Op] converges top.
According to 3.18(d), since Z [Op] converges to p, the unique z-ultrafiltercontaining it is Ap = Z [M p] (cf. 3P). Therefore the unique maximal idealcontaining Z�[Z [Op]] = Op is M p.
3. If Op 6= M p, then Op is contained in a prime ideal that is not maximal.
Proof. According to Theorem 2.8, since Op is a z-ideal, it must be an in-tersection of prime ideals. So if Op 6= M p, because Op is contained in nomaximal ideal other than M p (2.), Op must be contained in prime ideal thatis not maximal.
4. If P is a prime ideal in C, and P ⊆M p, then Op ⊆ P .
Proof. Take f ∈ Op, then f vanishes on some open neighborhood U of p. Bycomplete regularity, we can find g ∈ C such that g(p) = 1 and g[X\U ] = {0}.Note that fg = 0 ∈ P , and g 6∈ M p ⊇ P , hence since P is prime we musthave f ∈ P . Therefore Op ⊆ P .
5. If f ∈M p\Op, then there exists a prime ideal, containing Op and f ,that is not a z-ideal (and hence not maximal).
Proof. Consider the function g ∈ C defined as:
g(x) =
−1/ ln |f(x)| if 0 < |f(x)| ≤ e−1
1 if |f(x)| ≥ e−1
0 if f(x) = 0.
Clearly, Z (g) = Z (f), so gn ∈ (Op, f) only if gn is a multiple of f . But(gn/f)|(X\Z (f)) is unbounded in every neighborhood of Z (f) (cf. 2G.1),hence it cannot be continuously extended to all of X. Since no power of gbelongs to (Op, f), by 0.17, there must exist a prime ideal containing (Op, f)but not g. This prime ideal cannot be a z-ideal since Z (g) = Z (f).
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6. If f ∈M p\Op, then there exists a prime ideal containing Op, but notf , that is not a z-ideal.
Proof. Consider the function g ∈ C defined as:
g(x) =
{exp(−1/|f(x)|) if f(x) 6= 00 if f(x) = 0
.
Note that, for each n ∈ N, we have limx→0 exp(−1/|x|)/xn = 0, in otherwords limf(x)→0 f
n(x)/ exp(−1/|f(x)|) = ±∞. So no power of f is a multipleof g (cf. 2G.1), and f 6∈ (Op, g). Therefore, by 0.17, there must exist a primeideal containing (Op, g) and not containing f . This prime ideal cannot be az-ideal since Z (f) = Z (g).
7. Op is a countably generated ideal iff p has a countable base of neigh-borhoods.
Proof. Necessity The neighborhoods of p form a z-filter, namely the z-filterZ [Op]. So if Op is countably generated, say by (fn)n∈N, then the zero-setsZn = Z (fn) form a countable subbase of Z [Op]. Taking all finite intersec-tions of elements of the subbase, we end up with a countable base for theneighborhoods of p.
Sufficiency Assume that p has a countable base {Zn}n∈N of neighbor-hoods. Then these zero-sets are a base for the z-filter Z [Op]. For eachn ∈ N, pick fn ∈ C such that Z (fn) = Zn. Then (fn)n∈N generate the z-idealZ�[Z [Op]] = Op.
8. There exists a countably generated ideal I containing Op iff p is aGδ-point.
Proof. Necessity Assume that there exists a countably generated ideal I =(fn)n∈N, where each fn ∈ C, which properly contains Op. By (2.), the idealI must be contained in the maximal ideal M p, hence, for every f ∈ I, wehave p ∈ Z (f). Since I is generated by the fn, the z-filter Z [I] must have{Z (fn)}n∈N as a subbase. Hence we must have F =
⋂Z [I] =
⋂n∈NZ (fn).
So since Z [Op] ⊂ Z [I], we must have F ⊆⋂Z [Op] = {p}, by (1.). But
we already know that p ∈ F , so F = {p}. Recall, that each Z (fn) is a Gδ
(1.10), hence F = {p} must be a Gδ as well, since it is an intersection ofcountably many Gδ’s. Therefore p is a Gδ-point.
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Sufficiency Assume that p is a Gδ point, hence it must be a zero-set, say{p} = Z (g) for some g ∈ C. According to 1D.1, any f ∈ C such that Z (f)is a neighborhood of p must be a multiple of g. Therefore Op ⊂ (g).
4J p.62 P -spaces
For any X, every maximal ideal in C(X) is prime. When, conversely, everyprime ideal in C(X) is maximal, we call X a P -space. (Recall our blanketassumption that X is completely regular).
This definition is stated in terms of an algebraic property of C(X); there-fore, if C(X) is isomorphic with C(Y ), and X is a P -space, then Y is aP -space.
Before proceeding it is useful to state another definition and a couple ofLemmas. A ring R is said to be a von Neumann regular ring (or simply aregular ring, if the meaning is clear from context) if, for every element a ∈ R,there exists b ∈ R such that aba = a.
Lemma 4.1. Let R be a ring and P ⊆ R a prime ideal. Then, R/P is anintegral domain (i.e. a ring with no zero divisors).
Proof. Take a′, b′ ∈ R/P , such that a′b′ = 0. In other words, if a, b ∈ R arerespective representatives for a′ and b′, then (a + P )(b + P ) = ab + P = P ,that is ab ∈ P . Since P is prime, at least one of a or b is in P . Hence atleast one of a′ or b′ is 0. Therefore R/P is an integral domain.
Lemma 4.2. The property of being (von Neumann) regular is preserved un-der homomorphisms.
Proof. Consider rings R and Q, and a surjective homomorphism φ:R → Q.Suppose that R is regular, then take any a ∈ Q which we can write as φ(a′),for some a′ ∈ R. Since R is regular, we can find b′ ∈ R such that a′b′a′ = a′.Let b = φ(b′), then we can write φ(a′b′a′) = φ(a′) or aba = a. Therefore Q isalso regular.
Since an integral domain allows the cancellation law (e.g. ab = ac⇒ b =c), an immediate corollary of the above Lemma is that a regular integraldomain is a field. And now, on to the problem.
The following assertions are equivalent. It will be obvious from several ofthem that every discrete space is a P -space.
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(1) X is a P -space.
(2) For all p ∈ X, M p = Op, i.e. every function in C is a constant on aneighborhood of p.
(3) Every zero-set is open.
(4) Every Gδ is open.
(5) Every ideal in C(X) is a z-ideal.
(6) Every ideal is an intersection of prime ideals.
(7) For every f, g ∈ C, the ideal (f, g) is the principal ideal (f 2 + g2).
(8) For every f ∈ C, there exists f0 ∈ C such that f 2f0 = f (i.e. C is aregular ring).
(9) Every ideal is an intersection of maximal ideals.
(10) Every cozero-set in X is C-embedded.
(11) Every principal ideal is generated by an idempotent.
Proof. Since the number of equivalent statements is quite large, the implica-tion graph is a bit complicated. It is illustrated below.
(11) (10)
((QQQQQQQQQQQQQQQQ (7)
��(2) // (3) //
��
OO 66mmmmmmmmmmmmmmmm(4) // (5) //
66mmmmmmmmmmmmmmmm(6) // (8)
rreeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
(1)
OO
// (9)
OO
(1)⇒(2). By 4I.3, if M p 6= Op, for a given p ∈ X, then there exists aprime ideal containing Op that is not maximal.
(2)⇒(3). Take any Z ∈ Z (X), say Z = Z (f), for some f ∈ C(X). From(2), we know that f is zero on some neighborhood Up of p, for every p ∈ Z.Therefore Z is open, since Up ⊆ Z for each p ∈ Z.
(3)⇒(4). Since X is Hausdorff, for every p ∈ X, the singleton {p} iscompact. Suppose that V is a Gδ, according to Theorem 3.11(b), this implies
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that, for each p ∈ V , there is a zero-set Zp ⊆ V containing {p}. So, sinceevery zero-set is open, V must also be open.
(4)⇒(5). Every zero-set is a Gδ (1.10), hence every zero-set is open.Let I ⊂ C be any ideal, f ∈ I, and g ∈ C such that Z (g) = Z (f). ThenZ (g) = intZ (g) ⊇ Z (f), hence g is a multiple of f , by 1D.1. Thereforeg ∈ I, and I is a z-ideal.
(5)⇒(6). By Theorem 2.8, every z-ideal is an intersection of prime ideals.Since every ideal is a z-ideal, every ideal is an intersection of prime ideals.
(6)⇒(8). Consider f ∈ C, by assumption, the ideal (f 2) must be anintersection of prime ideals. Suppose f 2 belongs to a prime ideal, then fmust belong to the same ideal. Therefore f ∈ (f 2) or there exists f0 ∈ Csuch that f 2f0 = f .
(8)⇒(1). From (8), we know that C is regular. Suppose that P is aprime ideal in C, then C/P is also regular, by Lemma 4.2. Hence, from thecorollary to Lemma 4.1, we know that C/P is a regular integral domain andhence a field. Therefore P is maximal, in other words X is a P -space.
(5)⇒(7). Since f, g ∈ (f, g), we also have f 2 + g2 ∈ (f, g), hence (f 2 +g2) ⊆ (f, g). Also, note that Z (f),Z (g) ⊇ Z (f) ∩ Z (g) = Z (f 2 + g2),hence Z (f),Z (g) ∈ Z [(f 2 + g2)]. Since (f 2 + g2) must be a z-ideal, we havef, g ∈ (f 2 + g2) and (f, g) ⊆ (f 2 + g2). Therefore (f, g) = (f 2 + g2).
(7)⇒(8). Let f ∈ C and g = 0, then, by assumption, we must have(f) = (f,0) = (f 2 + 02) = (f 2). Therefore there exists f0 ∈ C such thatf 2f0 = f .
(1)⇒(9). We already know that (1)⇒(5), hence every ideal in C is a z-ideal. Then, from Theorem 2.8, every ideal is an intersection of prime ideals.But, by assumption, all prime ideals are maximal.
(9)⇒(6). A maximal ideal is necessarily prime, hence, by assumption,every ideal is an intersection of prime ideals.
(3)⇒(10). Let Z be a zero-set and V = X\Z its cozero-set. Then,by assumption, Z is open, hence Z and V are disconnected. Suppose thatf ∈ C(V ), then we can define g ∈ C as g|V = f and g|Z = 0, here g iscontinuous by Lemma 1.4. Therefore V is C-embedded in X.
(10)⇒(8). Consider f ∈ C(X) and let Z = Z (f). By assumption, weknow that V = X\Z is C-embedded in X. Consider f ′0 ∈ C(V ) definedas f ′0 = (f |V )�, we extend this function to a continuous function f0 on X.Clearly f 2f0 = f .
(3)⇒(11). We already know that (3)⇒(5), hence every ideal is a z-ideal. So given a principal ideal (f), for some f ∈ C, the same ideal can
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be generated by any function with the same zero-set as f . Consider theidempotent i defined as i|Z (f) = 0 and i|X\Z (f) = 1. By assumption, Z isopen, hence Z and X\Z are disconnected, so i is continuous by Lemma 1.4.Therefore the idempotent i generates the principal ideal (f).
(11)⇒(3). Consider a zero-set Z and the ideal (f), for some f ∈ Cwith Z (f) = Z. If it is generated by an idempotent i, then we must haveZ (i) = Z (f). The function i is an idempotent iff i[X] ⊆ {0, 1}, thereforef [X\Z] = {1}. This implies that Z and X\Z are completely separated andthus have disjoint open neighborhoods. But X = Z ∪ X\Z, hence Z mustbe open.
Finally, if X is discrete, then every subset of X is both open and a zero-set. Therefore, by (3), X is a P -space.
4K p.63 further properties of P -spaces
Please keep in mind the blanket assumption that all spaces under considera-tion are completely regular. Sometimes this property will not be mentionedexplicitly, for example in the case of subspaces of a completely regular space(3.1).
1. Every countable subset of a P -space (4J) is closed and discrete. Henceevery countable P -space is discrete, and every countably compact P -space isfinite.
Proof. Consider a countable subset D = {xn}n∈N of X and the sequenceof reals {1/n}n∈N. Then, according to 3L.2, we can construct a continuousfunction f such that f(xn) = 1/n, for all n ∈ N. From 4J(2), we know thatfor every point p ∈ X there is neighborhood Up around it on which f isconstant. Therefore, for f , the neighborhoods {Uxn} are disjoint, each onecontaining only xn. Hence D is discrete.
Also, from 4J(4), we know that every Gδ is open. Hence the set X\D =⋂n∈NX\{xn} is also open. Therefore D is closed.
Suppose a P -space is countably compact and infinite. Then we can find aclosed discrete countable subset D = {xn}n∈N of X (in fact, every countablesubset is such a set). So we can construct a countable cover of X consistingof {Uxn}n∈N ∪ {X\D} (where the Uxn are the same as in the previous para-graphs). Clearly, this cover has no finite subcover. Therefore X has to befinite.
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2. Every countable set in a P -space is C-embedded. Hence every pseu-docompact P -space is finite.
Proof. Consider a countable subset D = {xn}x∈N of X and any closed set Fdisjoint from it. Since X is completely regular, we can find pairs of disjointzero-sets neighborhoods Zn, Z
′n such that xn ∈ Zn and F ⊆ Z ′n. Then D and
F have disjoint neighborhoods, since⋂n∈N Z
′n is a zero-set (which is open
(4J(3))) and contains F , and⋃n∈N intZn is an open set which contains D.
Therefore, according to 3L.4(3), the set D must be C-embedded.Suppose a pseudocompact P -space is infinite, then it must have a count-
able subset D = {xn}n∈N which is discrete (1.). Then we can define f ∈ C(D)such that f(xn) = n, for each n ∈ N. So, since every countable subset isC-embedded, we can find g ∈ C such that g|D = f . In other words C con-tains an unbounded function, therefore the a pseudocompact P -space mustbe finite.
3. If X is a P -space, and every function in C(X) is bounded on a subsetS, then S is finite.
Proof. Suppose S ⊆ X is infinite, but every continuous function is boundedon it. Then S must have a countable subset D = {xn}n∈N. From (2.), weknow that D must be C-embedded. Since D is discrete (1.), we can definef ∈ C(D) such that f(xn) = n for each n ∈ N. We can continuously extendf to g ∈ C(X), which is unbounded on S. Therefore S must be finite.
4. Every subspace of a P -space is a P -space.
Proof. Consider S ⊆ X and U ⊆ S, where U is a Gδ in S. We can writeU =
⋂n∈N Un, where each Un is open in S. But for each Un, there is an open
U ′n ⊆ X such that Un = S ∩ U ′n. By 4J(4), the set U ′ =⋂n∈N U
′n is open in
X, hence U = S ∩ U ′ is open in S. Therefore S is a P -space.
5. Every (completely regular) quotient space of a P -space is a P -space.First, a definition. Let (X, T ) be a topological space and R an equivalence
relation on X. Consider the induced projection map πR:X → X/R definedby πR:x 7→ Rx, where Rx denotes the equivalence class to which x belongs
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and X/R the set of such equivalence classes. We say that X/R is a quotientspace when it is endowed with the topology
TR = {U ⊆ X/R | π�R [U ] ∈ T }.
And now, on to the proof.
Proof. Consider an equivalence relation R on X and U ⊆ X/R, where U isa Gδ in X/R. We can write U =
⋂n∈N Un, where each Un is open in X/R.
But for each Un, we must have U ′n = π�R [Un] open in X. By 4J(4), the set
U ′ =⋂n∈N U
′n is open in X, hence U = πR[U ′] is open in X/R. Therefore
X/R is a P -space.
6. Finite products of P -spaces are P -spaces, but infinite products neednot be.
Proof. Consider the product space X =∏
α∈AXα, where A is finite. Thesubbase for the open sets of X is given by the sets of the form π�
α [U ], whereU is open in Xα and πα:X 7→ Xα is the corresponding projection map. Soevery open set in X can be expressed as a union of finite intersections of setsof the form π�
α [U ]. A Gδ is a countable intersection of open sets, distributingintersections over unions, it can be expressed as a union of countable inter-sections of sets of the form π�
α [U ]. Therefore every Gδ in X is open if everycountable intersection of sets of the form π�
α [U ] is open.Consider the families of sets {Uαβ}β∈Bα ⊆ P(Xα), where Bα is countable,
for each α ∈ A. Then consider the set
U =⋂α∈A
( ⋂β∈Bα
π�α [Uαβ]
)
=⋂α∈A
π�α
[ ⋂β∈Bα
Uαβ
]=
⋂α∈A
π�α [Uα].
Each Uα is a Gδ in Xα, and hence open, by 4J(4). Therefore each π�α [Uα] is
open in X, and U is open as well, since A is finite.Consider the product space
∏∞i=1[0, 1]. Then the set
∏∞i=1[1/4, 3/4] is a
Gδ, but clearly not open.
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7. Every P -space is basically disconnected (1H).
Proof. From 4J(4), we know that every zero-set is open, hence every cozero-set is closed. Therefore the closure of every cozero-set is itself, which is open.Hence the space is basically disconnected.
8. Every P -space—more generally, every basically disconnected space—has a base of open-and-closed (clopen) sets.
Proof. At first, this question might appear ambiguous, since it is not clearwhether it is referring to the base for closed or open sets. However, thisdistinction is immaterial. If there exists a base for the closed sets consistingof clopen sets, then the family of complements of these sets forms a base forthe open sets consisting of clopen sets.
Since X is completely regular, the zero-sets form a base for the closedsets. Since X is basically disconnected, every cozero-set is closed (1H.1),and hence every zero-set is open. Therefore the zero-sets form a clopen basefor the closed sets.
9. Neither C(R) nor C(Q) is a homomorphic image of C(N).
Proof. Since every discrete space is a P -space (4J), according to 4J(4), thering C(N) is regular. For C(R) (resp. C(Q)), for the function f(x) = x, itis not possible to find g ∈ C(R) (resp. C(Q)) such that f 2g = f . Then, byLemma 4.2, it is clear that neither C(R) nor C(Q) is a homomorphic imageof C(N), since they are not regular.
4L p.63 P -points
If M p = Op, then P is called a P -point of X. Thus, X is a P -space iff everypoint is a P -point (4J(2)).
1. p is a P -point iff every Gδ containing p is a neighborhood of p.
Proof. Necessity Assume that p is a P -point, then every continuous functionon X is constant on some neighborhood of p. Hence every zero-set containingp is a neighborhood of p. Since X is Hausdorff, the singleton {p} is compact.Suppose that V is a Gδ containing p, according to Theorem 3.11(b), thereis a zero-set Z ⊆ V containing {p}. Therefore V is also a neighborhood of p.
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Sufficiency Assume that every Gδ containing p is a neighborhood of p.Since every zero-set is a Gδ, every zero-set containing p is a neighborhood ofp. Therefore M p = Op or p is a P -point.
2. A P -point of X is a P -point in any subspace containing it. Hence theset of all P -points of X is a P -space.
Proof. Consider p ∈ S ⊆ X, where p is a P -point in X. Every Gδ in Sis the intersection of S and a Gδ in X. And any neighborhood of p in X,intersected with S, is a neighborhood of p in S. According to (1), every Gδ
in X containing p is a neighborhood of p, hence every Gδ in S containing pis also a neighborhood of p. Therefore p is a P -point in S.
3. Let f ∈ RX . If f is continuous at a P -point p, then f is constant ona neighborhood of p.
Proof. Let r = f(p). Since f is continuous at p, for every n ∈ N, we can findan open set Un ⊆ X such that f [Un] ⊆ ]r − 1/n, r + 1/n[. Then U =
⋂n∈N Un
is a Gδ containing p and hence a neighborhood of p (1.). But, by construction,f [U ] = {r}.
4. If p is a P -point and M p is countably generated, then p is isolatedand M p is principal.
Proof. If M p = Op is countably generated, then Op must be principal, by4I.8. But M p is principal only if p is isolated (4B.1).
4M p.64 the space Σ
Let U be a free ultrafilter on N, let Σ = N∪{σ} (where σ 6∈ N), and define atopology on Σ as follows: all points of N are isolated, and the neighborhoodsof σ are the sets U ∪ {σ} for U ∈ U .
1. N is a dense subspace of Σ . Every set containing σ is closed; henceevery subset of Σ is open or closed. Σ is a normal space (cf. 3B.5 and 3D.5);in fact, every closed set is a zero-set.
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Proof. Clearly every neighborhood of σ meets N, hence σ ∈ clN or clN = Σ .Consider any S ⊆ Σ such that σ ∈ S. Then X\S ⊆ N is open in Σ , since
X\S is a union of isolated points. Hence S is closed.We already have that every set containing σ is closed. If a set S ⊆ Σ
does not contain σ, then it is a subset of N and open, because S is a unionof isolated points. Hence every subset of Σ is open or closed.
First, we show that Σ is Hausdorff. Consider two distinct points x, y ∈ Σ .At least one of them, say x, cannot be σ. Then the singleton {x} must beboth closed and open, hence it is its own neighborhood. Also, Σ\{x} mustbe a neighborhood of y not containing x. Now, consider two disjoint closedsets H,K ⊆ Σ . Then at least one of them, say H, does not contain σ. SinceH is a union of isolated points, it must itself be open. But since H is closed,we have X\H a neighborhood of K. Therefore Σ is normal.
Consider any closed set F ⊆ Σ , then either F contains σ or it is of theform F = X\V , where V = U ∪ {σ} for some U ∈ U . In the latter case,because both F and X\F are open, we can define f [F ] = {0} and f [X\F ] ={1} with f ∈ C(Σ ), by Lemma 1.4. If F contains σ and X\F is finite, thenagain F and X\F are both open, and f [F ] = {0} and f [X\F ] = {1} withf ∈ C(Σ ). Lastly, if σ ∈ F and X\F = {xn}n∈N is infinite, then we candefine f [F ] = {0} and f(xn) = 1/n for each n ∈ N. Since the preimage,with respect to f , of any open interval ]a, b[ ⊆ R is either finite (and henceopen) or has a finite complement (and hence also open), the function f iscontinuous. Therefore any closed subset of Σ is a zero-set.
2. The point σ does not have a countable base of neighborhoods. HenceΣ is not metrizable.
Proof. If σ were to have a countable base of neighborhoods, then the freeultrafilter U on N would also have a countable base. But that is impossible,by 4G.1. If we could impose a metric d on Σ , then every point p ∈ Σwould have a countable base of neighborhoods, namely {B1/n}n∈N whereBε = {x ∈ Σ | d(p, x) < ε}. Hence Σ is not metrizable.
3. Σ is extremally disconnected (1H), and so is every subspace.For the following proof we will require a small Lemma concerning ultra-
filters.
Lemma 4.3. A filter F on a set X is an ultrafilter iff, for any S ⊆ X, thefilter F contains exactly one of S or X\S.
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Proof. Necessity Assume that F is an ultrafilter, then F is prime becauseit is the image, under Z , of some maximal (hence prime) ideal in C(X),where X is considered under the discrete topology (Theorems 2.5, 2.12).Note that S ∪ (X\S) = X ∈ F , hence at least one of S or X\S must be inF . Note that both S and X\S cannot be in F , since then we would haveS ∩ (X\S) = ∅ ∈ F , which would mean that F is not proper.
Sufficiency Suppose that F is a filter which contains exactly one of S orX\S, for each S ⊆ X. Take any S ⊆ X not in F and construct the filter F ′by adjoining S to F . But, by hypothesis, we must have X\S in F . Henceboth S and X\S are in X, so S ∩ (X\S) = ∅ ∈ F ′ and F ′ cannot be aproper filter. Therefore F must be an ultrafilter.
And now, on to the proof.
Proof. Suppose U is an open subset of Σ , then either σ ∈ U or σ 6∈ U . In theformer case, U is already closed, hence clU = U . In the latter case, eitherΣ\U is open or it is not. If Σ\U is open, again U is closed and clU = U .If Σ\U is not open, then it does not contain a neighborhood of σ, whichimplies that N\U is not an element of the ultrafilter U . Then, by Lemma4.3, we must have U ∈ U . Since U is not closed, clearly its closure mustbe clU = U ∪ {σ}. But since U ∈ U , we have clU open. Therefore Σ isextremally disconnected.
Every subspace of Σ that does not contain σ is discrete, therefore it isautomatically extremally disconnected. Let Y ⊆ X such that σ ∈ Y . Thenany open U ⊆ Y such that σ ∈ U is already closed, hence clU = U . IfU ⊆ Y is open but σ 6∈ U , then if it is not closed, its closure must beclU = Y ∩ (U ∪{σ}). Moreover, it must be of the form U = Y ∩V , for someV ∈ U (see previous paragraph for reasoning). But them U ∪ {σ} is open,and hence clU = Y ∩ (U ∪ {σ}) is also open. Therefore every subspace of Σis extremally disconnected.
4. The z-ideal Oσ is prime but not maximal. Hence Σ is not a P -space(4J).
Proof. Consider two closed sets (every closed set is a zero-set (1.)) H,K ⊆ Σ .Suppose that Z = H ∪K is a neighborhood of σ but neither of H or K is.There are two possibilities, either both H and K contain σ or only one ofthem does, say H. In the former case we must have H = (N\U) ∪ {σ} andK = (N\V ) ∪ {σ}, for some U, V ∈ U . In the latter case H = (N\U) ∪ {σ}
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and K = N\V , for some U, V ∈ U . But then U ∩V ∈ U , so (N\V )∪(N\U) =N\(U ∩V ) is not in U , hence Z = N\(U ∪V )∪{σ} cannot be a neighborhoodof σ. Therefore at least one of H or K must be a neighborhood of σ and Oσ,since it is a z-ideal, must be prime.
Note that {σ} is a closed but not open zero-set. Therefore Oσ is notmaximal, since M σ is strictly greater than Oσ. Therefore σ is not a P -pointand Σ is not a P -space.
5. The dense subspace N is C∗-embedded in Σ , but not C-embedded.Moreover, every subspace of Σ is C∗-embedded.
Proof. Consider two disjoint zero-sets in H,K ⊆ N (since N is discrete, anyset is a zero-set). They both cannot be in the ultrafilter U , since otherwiseit would not be a proper filter. Suppose only one, say H, is in U , then N\Kmust also be in U , by Lemma 4.3. Note that (N\K) ∪ {σ} is open, so Kis closed (and hence a zero-set) in Σ . Also, H ∪ {σ} is closed (and hence azero-set) in Σ disjoint from K. If, on the other hand, neither H nor K isin U , then both H and K are closed (and hence zero-sets) in Σ , by similarreasoning as in the previous case. Hence, by Urysohn’s Extension Theorem(1.17), N is C∗-embedded in Σ . However, N is not C-embedded since {σ} isa zero-set and the smallest zero-set containing N is Σ itself (Theorem 1.18).
Consider any subspace Y of Σ . Let H,K ⊆ Y be any two disjoint zero-sets. Suppose that neither H nor K contains σ, then they can be treated assubsets of N and are completely separated in Σ for the same reasons as inthe previous paragraph. Suppose only one zero-set, say H, contains σ, thenH is automatically closed in Σ (1.). If K 6∈ U , again K is already closed inΣ . If K ∈ U , then there must exist a set K ′ closed in Σ and not containing σsuch that K = Y ∩K ′. Note that H\{σ} is an open set, hence K ′\(H\{σ})is a closed set (hence zero-set) in Σ that contains K and is disjoint from H.Therefore, by Urysohn’s Extension Theorem (1.17), every subspace of Σ isC∗-embedded.
6. C∗(Σ ) is isomorphic with C∗(N). But C(Σ ) is not isomorphic withC(N).
Proof. First, let us show some properties of continuous functions on Σ . Con-sider f ∈ C(Σ ), then r = f(σ) must be an accumulation point of f [Σ ].
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Suppose that r is isolated from f [N], then there is an open interval U con-taining r such that f�[U ] = {σ}. But {σ} is not open, hence f cannot becontinuous.
Suppose that f ∈ C∗(N), then the z-filter f#U is prime on the compactspace [inf f, sup f ]. Since ever z-filter on a compact space must be fixed, theintersection A =
⋂f#U must be non-empty. Note that each a ∈ A is an
accumulation point of f [U ], for each U ∈ U . Suppose there are two distinctpoints a, b ∈ A (WLOG, let a < b), then we can find a point c ∈ ]a, b[ andconstruct the sets U = f�[] inf f, c]] and V = f�[]c, sup f ]]. Since U ∪V = N,exactly one of U or V must be in U (Lemma 4.3). If U is in U , then b is nota point of accumulation of f [U ], so b cannot be in A. If V is in U , then a isnot a point of accumulation of f [V ], so a cannot be in A. Therefore A hasexactly one point.
Since f#U is prime and⋂f#U = {r}, for some r ∈ R, then f#U must
converge to r, by Theorem 3.17. This implies that, for f to be continuouslyextended to all of Σ , its value at σ must be r.
We already know that f 7→ f |N is a surjective homomorphism from C∗(Σ )to C∗(N). The previous paragraph shows that its is bijective, so since it isclear that the inverse is also a homomorphism, the two rings C∗(Σ ) andC∗(N) are isomorphic. Finally, C(Σ ) cannot be isomorphic to C(N), since Nis discrete and hence a P -space (4J), but Σ is not (4.).
7. If 0 ≤ h ≤ k in C(Σ ), then h ∈ (k).
Proof. We have h ∈ (k) if h = kg for some g ∈ C(Σ ). Let V = Σ\Z (k)and consider the function g′ = (h|V )/(k|V ), clearly (h|V ) = g′(k|V ) andg′ ∈ C∗(V ), since by hypothesis h(x)/k(x) ≤ 1 for each x ∈ V . By (5.),the subspace V of Σ is C∗-embedded, so we can find g ∈ C∗(Σ ) such thatg|V = g′. But then h = kg, hence h ∈ (k).
8. The ideal (f, g) in C(Σ ) is the principal ideal (|f |+ |g|). Hence everyfinitely generated ideal is principal.
Proof. Note that, since Σ is extremally disconnected (3.), it is also basicallydisconnected. Hence, by 1H.3, there exists a unit u of C such that |f | = uf ,that is |f | ∈ (f, g) and similarly |g| ∈ (f, g). Which implies that |f | + |g| ∈(f, g) and (|f |+ |g|) ⊆ (f, g).
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Note that 0 ≤ |f |, |g| ≤ |f |+ |g|, hence we have |f |, |g| ∈ (|f |+ |g|) (7.).From the previous paragraph, we know that we can find h ∈ C∗(Σ ) such that|f | = hf , but then we also have f = h|f |. Hence f, g ∈ (|f |, |g|) ⊆ (|f |+ |g|).Therefore (f, g) ⊆ (|f |+ |g|) and (f, g) = (|f |+ |g|).
9. The only z-ideals containing Oσ are Oσ and M σ.
Proof. Suppose that Z ∈ Z [M σ]\Z [Oσ]. Then Z is closed but not open,hence Σ\Z ∈ U . Note that we must also have Z ′ = (Σ\Z) ∪ {σ} ∈ Z [Oσ].Let F be any z-filter that contains both Z [Oσ] and Z, then we must haveZ ∩ Z ′ = {σ} ∈ F . Which implies that F = M σ.
4N p.64 a nondiscrete P -space
Let S be an uncountable space in which all points are isolated except for adistinguished point s, a neighborhood of s being any set containing s whosecomplement is countable.
1. The closed set {s} is not a zero-set. S is a nondiscrete P -space (4J).
Proof. Suppose that f ∈ C(S) and f(s) = 0. Since f is continuous at s, foreach ε ∈ {1/n}n∈N, we can find a neighborhood Uε of s such that f [Uε] ⊆]−ε, ε[. Note that, for each x ∈ Z (f) =
⋂n∈N U1/n, we have f(x) = 0. But
since the complement of each U1/n is countable, the complement of Z (f) isalso countable. Therefore Z (f) itself must be uncountable and Z (f) 6= {s}.
Since each point p of S, except s, is isolated, we have Op = M p. From theprevious paragraph, we have that the zero-set of any function which vanishesat s has countable complement, hence each zero-set containing s is also aneighborhood of s. Therefore S must be a P -space (4J(2)).
2. No member of Z [M s] is countably compact.
Proof. First, a note about the topology on S. Every set that does not contains is open, since it is a union of isolated points. Every set that contains s isclosed, because its complement is open. A set that contains s is open iff itcontains a neighborhood of s, i.e. its complement is countable. A set thatdoes not contain s is closed iff its complement is open, i.e. it is countable.
Let Z ∈ Z [M s]. Since Z is uncountable (1.), we can choose a countablesubset {xn}n∈N. Then we can construct a countable open cover {Um}m∈N of
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Z, where Um = (Z\{xn}n∈N) ∪ {xm}. Each Um is open because its comple-ment is countable. Note that no subset of {Um}m∈N is a cover of Z. ThereforeZ cannot be countably compact.
3. S is basically disconnected but not extremally disconnected (1H).
Proof. Since S is a P -space (1.), every zero-set is open (4J(3)) and henceevery cozero-set is closed. Therefore every cozero-set has an open closure(itself) and hence S is basically disconnected.
Since S is uncountable, we can write S\{s} = U ∪V , where U and V aredisjoint uncountable sets. Clearly, both U and V are open but not closed,also clU = U ∪ {s} and clV = V ∪ {s}, so the closure of these sets are notdisjoint. But, according to 1H.1, this implies that S cannot be extremallydisconnected.
4. Let T be the topological sum of Σ (4M) and S (i.e. Σ and S aredisjoint open sets whose union is T ). Then T is basically disconnected, butit is neither extremally disconnected nor a P -space.
Proof. Both Σ and S are basically disconnected (4M.3 and (3.)). Since Tis the topological sum of Σ and S, clearly every cozero-set V in T can bewritten as V = U1 ∪ U2, where U1 is a cozero-set in Σ and U2 is a cozero-setin S. But then clT V = clT (U1∪U2) = clΣ U1∪clS U2. Since clΣ U1 and clS U2
are both open, the set clT V must be open as well. Therefore T is basicallydisconnected.
Recall that S is not extremally disconnected (3.), hence there must existopen U ⊆ S such that clS U is not open in S. But then U is open in T andclT U = clS U , which is also closed but not open in T . Hence T cannot beextremally disconnected.
References
[GJ] Leonard Gillman—Meyer Jerison. Rings of Continuous Functions. D.van Nostrand Company, Inc., 1960.
[Dug] James Dugundji. Topology. Allyn and Bacon, Inc., 1966.
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