[undergraduate texts in mathematics] real mathematical analysis || a taste of topology

2 A Taste of Topology

1 Metric Space Concepts

It may seem paradoxical at first, but a specific math problem can be harder to solve than some abstract generalization of it. For instance, if you want to know how many roots the equation

t5 - 4t4 + t 3

- t + I = 0

can have, then you could use calculus and figure it out. It would take a while. But thinking more abstractly, and with less work, you could show that any nth degree polynomial has at most n roots. In the same way many general results about functions of a real variable are more easily grasped at an abstract level - the level of metric spaces.

Metric space theory can be seen as a special case of general topology, and many books present it that way, explaining compactness primarily in terms of open coverings. In my opinion, however, the sequence/subsequence approach provides the easiest and simplest route to mastering the subject. Accordingly it gets top billing throughout this chapter.

A metric space is a set M, the elements of which are referred to as points of M, together with a metric d having the three properties that distance has in Euclidean space. The metric d = d (x, y) is a real number defined for all points x, y E M and d (x, y) is called the distance from the point x to the point y. The three distance properties are: for all x, y, Z EM,

C. C. Pugh, Real Mathematical Analysis© Springer Science+Business Media New York 2002

52 A Taste of Topology Chapter 2

(a) positive definiteness: d(x, y) ::: 0, and d(x, y) = 0 if and only if x =y.

(b) symmetry: d(x, y) = dey, x).

(c) triangle inequality: d(x, z) ::s d(x, y) + dey, z).

The function d is also called the distance function. Strictly speaking, it is the pair (M, d) which is a metric space, but we will follow the common practice of referring to "the metric space M ," and leave to you the job of inferring the correct metric.

The main examples of metric spaces are JR, JRm, and their subsets. The metric on JR is d(x, y) = Ix - yl where x, y E JR and Ix - yl is the magnitude of x - y. The metric on JRm is the Euclidean length of x - y where x, y are vectors in JRm. Namely,

d(x, y) = J(XI - YI)2 + ... + (xm - Ym)2

for x = (Xl, ... , xm) and y = (YI, ... , Ym). Since Euclidean length satisfies the three distance properties, d is a bona

fide metric and it makes JRm into a metric space. A subset M C JRm becomes a metric space when we declare the distance between points of M to be their Euclidean distance apart as points in IRm. We say that M inherits its metric from JRm and is a subspace of JRm. Figure 22 shows a few subsets of 1R2 to suggest some interesting metric spaces.

There is also one metric that is hard to picture but valuable as a source for counter-examples, the discrete metric. Given any set M define the distance between distinct points to be 1 and the distance between any point and itself to be O. This is a metric. If M consists of three points, say M = {a, b, c}, you can think of the vertices of the unit equilateral triangle as a model for M. See Figure 23. They have mutual distance 1 from each other. If M consists of one, two, or four points can you think of a model for the discrete metric on M? More challenging is to imagine the discrete metric on R All points, by definition of the discrete metric, lie at unit distance from each other.

Convergent sequences and subsequences

A sequence of points in a metric space M is a list PI, P2, ... where the points Pn belong to M. Repetition is allowed, and not all the points of M need to appear in the list. Good notation for a sequence is (Pn), or (Pn)nEN. The notation {Pn} is also used but it is too easily confused with the set of points making up the sequence. The difference between (Pn)nEN and {Pn : n E N} is that in the former case the sequence prescribes an ordering of the points, while in the latter the points get jumbled together.

Section 1 Metric Space Concepts 53

--+-

--+-

"T-

--+-

Figure 22 Five metric spaces: a closed outward spiral, a Hawaiian earring, a topologist's sine circle, an infinite television antenna, and Zeno's maze.

Figure 23 The vertices of the unit equilateral triangle form a discrete metric space.

For example, the sequences 1,2,3, ... and 1,2,1,3,2,1,4,3,2,1, ... are different sequences but give the same set of points, namely N.

Formally, a sequence is a function f : N -+ M. The nth term in the sequence is f (n) = Pn. Clearly, every sequence defines a function f :


N ~ M and conversely, every function f : N ~ M defines a sequence in M. The sequence (Pn) converges to the limit P in M if

'<IE > ° 3N EN such that

n E Nand n :::: N => d (Pn, p) < E.

Limits are unique in the sense that if (Pn) converges to P and if (Pn) also converges to p' then P = p'. The reason is this. Given any E > 0, there are integers N, N ' such that if n :::: N then d (Pn, p) < E, while if n :::: N ' then d(Pn, p') < E. Then for all n :::: max{N, N ' },

d(p, p') :s d(p, Pn) + d(Pn, p') < E + E = 2E.

But E is arbitrary and so d(p, p') = ° and P = p'. (This is the E-principle again.)

We write Pn ~ p, or Pn ~ P as n ~ 00, or

lim Pn = P n-HX)

to indicate convergence. For example, in JR the sequence Pn = 1 I n converges to ° as n ~ 00. In JR2 the sequence (lin, sin n) does not converge as n ~ 00. In the metric space Q (with the metric it inherits from JR) the sequence 1,1.4,1.414,1.4142, ... does not converge.

Just as a set can have a subset, a sequence can have a subsequence. For example, the sequence 2,4, 6, 8, ... is a subsequence of 1, 2, 3,4, . .. . The sequence 3, 5, 7,11, l3, 17, ... is a subsequence of 1, 3, 5, 7, 9, ... , which in tum is a subsequence of 1,2,3,4, .... In general, if (Pn)nEN and (qkhEN are sequences and if there is a sequence 1 :s nj < n2 < n3 < ... of integers such that for each kEN, qk = Pnk then (qk) is a subsequence of (Pn). Note that the terms in the subsequence occur in the same order as in the mother sequence.

1 Theorem Every subsequence of a convergent sequence converges and it converges to the same limit as does the mother sequence.

Proof Let (qk) be a subsequence of (Pn), qk = Pnk' where nj < n2 < .... Assume that (Pn) converges to P in M. Given E > 0, there is an N such that for all n :::: N, d(Pn, p) < E. Since nj, n2, ... are integers, k :s nk for all k. Thus, if k :::: N then nk :::: Nand d (qk. p) < E. Hence (qk) converges to p. D

A common way to state Theorem 1 is that limits are unaffected when we pass to a subsequence.


Continuity

In linear algebra the objects of interest are linear transformations. In real analysis the objects of interest are functions, especially continuous functions. A function f from the metric space M to the metric space N is just that; f : M ~ Nand f sends points p E M to points f pEN. A function is also called a transformation, a mapping, or a map. The function maps M

to N. The way you should think of functions - as devices, not formulas - is discussed in Section 4 of Chapter 1.

The most common type of function maps M to R It is a real-valued function of the variable p EM.

Definition A function f : M ~ N is continuous if it satisfies the f, ~

condition:

"IE > 0 and V p E M 38 > 0 such that

q E M and d(p, q) < 8 :::} d(fp, fq) < E.

Here and in what follows, the notation f p is used as convenient shorthand for f(p).

Consider the case of a real valued function of a real variable f : (a, b) ~

ffi. as in Chapter 1. There we defined f to be continuous when

"IE > 0 and "Ix E (a, b) 38 > 0 such that

y E (a, b) and Ix - yl < 8 :::} If x - fyl < E.

In case M = (a, b) and N = ffi., the two E, 8 definitions are identical except in the second we write out explicitly the distance in ffi. as Ix - y I instead of d (x, y). Thus, every continuous real function of areal variable is an example of a continuous mapping from the metric space (a, b) to the metric space ffi., and conversely, every continuous map from the metric space (a, b) to the metric space ffi. is a continuous function of a real variable.

How is continuity expressed using sequences? Nothing could be simpler or more natural.

2 Theorem f : M ~ N is continuous if and only if it sends each convergent sequence in M to a convergent sequence in N, limits being sent to limits.

Proof Suppose that f is continuous and (Pn) is a convergent sequence inM,

lim Pn = p. n--+oo


Then (f (Pn)) is a sequence in N. Continuity implies that given E > 0, there exists 0 > 0 such that d(x, p) < 0 implies d(fx, fp) < E. Convergence implies that there is an N such that for all n ::: N, d(Pn, p) < O. Then d(f(Pn), fp) < E and

lim f(Pn) = fp· n-HX)

We prove the converse in contrapositive form: if f is not continuous then it does not preserve convergence. f being not continuous means that for some P E M there is an E > 0 such that no matter how small we take 0, there will always be points x E M with d(x, p) < 0 but d(fx, fp) ::: E.

Take 01 = 1, 02 = 1/2, ... On = lin,

For each On there is a point Xn with d(xn, p) < On d(f(xn), fp) ::: E. Thus

lin and

lim Xn = P n-HX)

but f(xn) does not converge to fp.

See also Exercise 17.

3 Corollary The composite of continuous functions is continuous.

Proof Let f : M -+ Nand g : N -+ P be continuous and assume that

lim Pn = P n-HXl

D

in M. Since f is continuous, Theorem 2 implies that lim f(Pn) = fp· n--+oo

Since g is continuous, Theorem 2 implies that lim g (f (Pn)) = g (f p) and n--+oo

therefore g 0 f : M -+ P is continuous. See Figure 24. D

go!

Figure 24 The composite function g 0 f.


Moral The sequence condition is the easy way to tell at a glance whether a function is continuous.

Homeomorphism

Vector spaces are isomorphic if there is a linear bijection from one to the other. When are metric spaces isomorphic? They should "look the same." The letters Y and T look the same; and they look different from the letter O. If I : M --+ N is a bijection and I is continuous and the inverse bijection 1-1 : N --+ M is also continuous then f is a homeomorphismt (or a "homeo" for short) and M, N are homeomorphic. We write M ~ N to indicate that M and N are homeomorphic. ~ is an equivalence relation: M ~ M since the identity map is a homeomorphism M --+ M; M ~ N clearly implies that N ~ M; and the previous corollary shows that the composite of homeomorphisms is a homeomorphism.

Geometrically speaking, a homeomorphism is a bijection that can bend, twist, stretch, and wrinkle the space M to make it coincide with N, but it can not rip, puncture, shred, or pulverize M in the process. The basic questions to ask about metric spaces are:

(a) Given M, N, are they homeomorphic? (b) What are the continuous functions from M to N? A major goal of this chapter is to show you how to answer these questions

in many cases. For example, is the circle homeomorphic to the interval? To the sphere? etc. Figure 25 indicates that the circle and the (perimeter of the) triangle are homeomorphic, while Figure 14 shows that (a, b), the semicircle, and lR are homeomorphic.

Figure 2S The circle and triangle are homeomorphic.

t This is a rare case in mathematics in which spelling is important. Homeomorphism f. homomorphism.


A natural question that should occur to you is whether continuity of f- I is actually implied by continuity of a bijection f. It is not. Here is an instructive example.

Consider the interval [0,2n) = {x E lR : 0 .:s x < 2n} and define f : [0,2n) ---+ SI to be the mapping f(x) = (cosx, sinx) where SI is the unit circle in the plane. The mapping f is a continuous bijection, but the inverse bijection is not continuous. For there is a sequence of points (Zn) on SI in the fourth quadrant that converges to p = (1, 0), and f- I (Zn) does not converge to f-I(p) = O. Rather it converges to 2n. Thus, f is a continuous bijection whose inverse bijection fails to be continuous. See Figure 26.

[ ) o Z1t

Figure 26 f wraps [0, 2n) bijectively onto the circle.

To build your intuition about continuous mappings and homeomorphisms, consider the following examples shown in Figure 27: the unit circle in the plane, a trefoil knot in lR3, the perimeter of a square, the surface of a donut (the 2-torus), the surface of a ceramic coffee cup, the unit interval [0, 1], the unit disc including its boundary. Equip all with the inherited metric. Which should be homeomorphic to which?

Closed Sets and Open Sets

Now we come to two basic concepts in a metric space theory - closedness and openness. Let M be a metric space and S be a subset of M. A point p E M is a limit of S if there exists a sequence (Pn) in S that converges to it. t

t A limit of S is also sometimes called a limit point of S. Take care though: some mathematicians require that a limit point of S be the limit of a sequence of distinct points of S. They would say that a


ow D

o

Figure 27 Seven metric spaces.

Definition S is a closed set if it contains all its limits. t

Definition S is an open set if for each PES there exists an r > 0 such that

d(p, q) < r =} q E S.

A little thought will convince you that in the metric space JR., the closed interval [a, b) is a closed set and the open interval (a, b) is an open set. See also Theorem 7 below.

4 Theorem Openness is dual to closedness: the complement of an open set is closed and the complement of a closed set is open.

Proof Suppose that SCM is an open set. We claim that SC is closed. If Pn --+ P and Pn E SC we must show that p E SC. Well, if p ~ SC then pES and, since S is open, there is an r > 0 such that

d(p, q) < r =} q E S.

finite set has no limit points. We will not adopt their point of view. Another word used in this context. especially by the French, is "adherence." A point p adheres to the set S if and only if p is a limit of S. In more general circumstances, limits are defined using "nets" instead of sequences. You can read more about nets in graduate level topology books such as Topology by James Munkres.

t Note how similarly algebraists used the word "closed." A field (or group or ring, etc.) is closed under its arithmetic operations: sums, differences, products, and quotients of elements in the field still lie in the field. In our case it is limits. Limits of sequences in S must lie in S.


Since Pn ---+ p, d(p, Pn) < r for all large n, which implies that Pn E S, contrary to the sequence being in se. Since the supposition that P lies in S leads to a contradiction, p actually does lie in se, proving that se is closed.

Suppose that S is closed. We claim that se is open. Take any p E se. If there fails to exist an r > 0 such that

d(p, q) < r ::::} q Ese,

then for each r = lin, n = 1,2, ... , there exists a point Pn E S such that d(p, Pn) < lin. This sequence in S converges to P Ese, contrary to closededness of S. Therefore there actually does exist an r > 0 such that

d (p, q) < r ::::} q E Se,

which proves that se is open. D

Most sets, like doors, are neither open nor closed, but ajar. Keep this in mind. For example, neither (a, b] nor its complement is closed in lR; (a, b] is neither closed nor open. Unlike doors, however, sets can be both open and closed at the same time. For example, the empty set 0 is a subset of any metric space and it is always closed. There are no sequences and no limits to even worry about. Similarly the full metric space M is a closed subset of itself: it certainly contains the limit of any sequence that converges in M. Thus, 0 and M are closed subsets of M. Their complements, M and 0, are therefore open: 0 and M are both closed and open.

Subsets of M that are both closed and open are elopen. See also Exercise 92. It turns out that in lR the only clopen sets are 0 and R In Q!, however, things are quite different, sets such as {r E Q! : -y'2 < r < y'2} being clopen in Q!. To summarize,

A subset of a metric space can be closed, open, both, or neither.

You should expect the "typical" subset of a metric space to be neither closed nor open.

The topology of M is the collection T of all open subsets of M.

5 Theorem T has three properties:t as a system it is closed under union, finite intersection, and it contains 0, M. That is,

t Any collection T of subsets of X that satisfies these three properties is called a topology on X, and X is called a topological space. Topological spaces are more general than metric spaces: there exist topologies that do not arise from a metric. Think of them as pathological. The question of which topologies can be generated by a metric and which cannot is discussed in Topology by Munkres.

Section 1 Metric Space Concepts

(a) Every union of open sets is an open set. (b) The intersection offinitely many open sets is an open set. (c) 0 and M are open sets.

61

Proof (a) If {U,,} is any collection of open subsets of M and V = UU" then V is open. For if p E V then p belongs to at least one U" and there is an r > 0 such that

d(p, q) < r :::} q E U".

Since U" c V, this implies that all such q lie in V, proving that V is open. (b) If UI, ... , Un are open sets and W = nUk then W is open. For if

pEW then for each k, 1 ::: k ::: n, there is an rk > 0 such that

Take r = min{r1, ... , rn}. Then

d(p, q) < r :::} q E Ub

for each k; i.e., q E W = nUb proving that W is open. (c) It is clear that 0 and M are open sets. D

6 Corollary The intersection of any number of closed sets is a closed set; the finite union of closed sets is a closed set; 0 and M are closed sets.

Proof Take complements and use DeMorgan's Laws. If {K,,} is a collection of closed sets then U" = (K"Y is open and

K = nK" = (Uu,,)C Since UU" is open, its complement K is closed. Similarly, a finite union of closed sets is the complement of the finite intersection of their complements, and is a closed set. D

What about an infinite union of closed sets? Generally, it is not closed. For example, the interval [1/ n, 1] is closed in JR, but the union of these intervals as n ranges over N is the interval (0, 1] which is not closed. Neither is the infinite intersection of open sets open in general.

Two sets whose closedness/openness properties are basic are:

lim S = {p EM: p is a limit of S}

M r p = {q EM: d (p, q) < r}.

The former is the limit set of S; the latter is the r-neighhorhood of p.

62 A Taste of Topology

7 Theorem lim S is a closed set and Mr P is an open set.

8 Lemma The following are equivalent. (a) p is a limit of S. (b) Vr > 0, Mr(p) n S i- 0.

Chapter 2

Proof Assume (a): there is a sequence (Pn) in S with Pn -+ p. For each r > 0 there is an N such that for all n ~ N, Pn E Mr(P), which verifies (b).

Assume (b): each Mr(P) contains a point of S. Take r = 1, r = 1/2, ... , r = lin, .... There exists Pn E M1/n(p) n S. The sequence (Pn) converges to p, which verifies (a). D

Proof of Theorem 7 Simple but not immediate! Suppose that Pn -+ P and each Pn lies in lim S. Since Pn is a limit of S there is a point qn E S such that

Then

implies that qn -+ p, so P E lim S and lim S is a closed set. To check that Mr P is an open set, take any q E Mr P and observe that

s = r - d(p, q) > O.

By the triangle inequality, if d (q, x) < s then

d(p, x) :::: d(p, q) + d(q, x) < r,

and Msq C Mr p. See Figure 28. Since each q E Mr P has some Msq that is contained in Mr p, Mr P is an open set. D

A neighborhood of a point p in M is any open set V that contains p. Theorem 7 implies that V = Mr P is a neighborhood of p. Eventually, you will run across the phrase "closed neighborhood" of p, which refers to a closed set that contains an open set that contains p. Until further notice, all neighborhoods are open.

Usually, sets defined by strict inequalities are open while those defined by equalities or non-strict inequalities are closed. Examples of closed sets in lR are: finite sets, [a, b], N, and the set {OJ U {lin: n EN}. Each contains all its limits. Examples of open sets in lR are open intervals.


.. p

Figure 28 Why the r-neighborhood of p is an open set.

Open subsets of IR

Besides bounded open intervals (a, b) with a, b E JR, we consider unbounded open intervals

(-00, b), (a, (0), and JR = (-00, (0).

9 Theorem Every open set U C ~ can be uniquely expressed as a countable union of disjoint open intervals. The endpoints of the intervals do not belong to U.

Proof For each x E U define

ax = inf{a: (a, x) C U}

bx = sup{b: (x, b) C U}.

Then Ix = (ax, bx) is an open interval, possibly unbounded, and

x E Ix C U.

It is the maximal open interval in U that contains x. If the endpoint b lies in U then, by openness of U, there is some open interval J such that

bE J c U.

But Ix U J is then a larger open interval in U that contains x, contrary to maximality. Thus the endpoints of Ix do not lie in U.


If y E U is a second point and if Ix n Iy =1= 0 then Ix U Iy is an open interval in U that contains both x and y. By maximality,

We conclude that for any x, y E U, the intervals Ix, Iy are either equal or disjoint. This shows that U is the disjoint union of such maximal open intervals.

There can be only countably many disjoint open intervals in R For, given any collection of disjoint intervals in JR, choose a rational number in each. Disjointness implies that the chosen rational numbers are distinct, so there are only countably many of them, and only countably many open intervals. Thus U is the countable disjoint union of the maximal open intervals Ix.

Uniqueness of this division of U into disjoint open intervals is left for the reader to check. D

Topological description of continuity

A property of a metric space or of a mapping between metric spaces that can be described solely in terms of open sets (or equivalently, in terms of closed sets) is called a topological property. The next result describes continuity topologically.

Let f : M -+ N be given. The pre-image t of a set V c N is

fpre(V) = {p EM: f(p) E V}.

For example, if f : JR2 -+ JR is the function defined by the formula

then the pre-image of the interval [3,6] in JR is the annulus in the plane with inner radius 1 and outer radius 2. Figure 29 shows the domain of f as JR2 and the target as R The range is the set of real numbers ::: 2. The graph of f is a paraboloid with lowest point (0,0,2). The second part of the figure shows the portion of the graph lying above the annulus. You will find it useful to keep in mind the distinctions among the concepts: function, range, and graph.

t The pre-image of V is also called the inverse image of V and is denoted by f-I(V). Unless f is a bijection, this notation leads to confusion. There is no map f- I and yet expressions like V ::l f(f-I (V)) are written that mix maps and non-maps. By the way, if f sends no point of Minto V then fpre(V) is the empty set.


6

o

Figure 29 The function f : (x, y) f-+ x 2 + y2 + 2 and its graph over the pre-image of [3,6].

10 Theorem The following are equivalent for continuity of f : M ---+ N. (i) The E, 8 condition.

(ii) The closed set condition: the pre-image of each closed set in N is closed in M.

(iii) The open set condition: the pre-image of each open set in N is open inM.

Proof Totally natural! Assume (i). Let K be closed in N and let (Pn) be a sequence in fpre(K) that converges to p in M. We must show that p E fpre(K). According to Theorem 2, f preserves limits so

lim f(Pn) = fp· n--+oo

By assumption f(Pn) E K. Since K is closed, fp E K. Hence p E

fpre(K), which gives (ii). Since (fpre(u))c = fPre(u c), (ii) =} (iii). Assume (iii). Let p E M and E > 0 be given. According to Theorem 7,

NE(fp) is open in N. By (iii), fpre(NE(fp)) is open in M, and p belongs to it, so there is some M8(P) C fpre(NE(fp)). Thus,

d(x, p) < 8 =} d(fx, fp) < E,

and f is continuous. See Figure 30. D

I hope that you find the closed and open set characterizations of continuity elegant. Note that no explicit mention is made of the metric. The open set condition is purely topological. It would be perfectly valid to take as a definition of continuity that the pre-image of each open set is open. In fact this is exactly what's done in general topology.


Figure 30 The E, 8 - condition for a continuous function f : M ---+ N.

11 Corollary A homeomorphism f : M ---+ N bijects the collection of open sets in M to the collection of open sets in N. It bijects the topologies.

Proof Let V be an open set in N. By Theorem 10, since f is continuous, the pre-image of V is open in M. Since f is a bijection, this pre-image V = {p EM: f(p) E V} is exactly the image of V by the inverse bijection, V = f-I(V). The same thing can be said about f- I since f- I

is also a homeomorphism. That is, V = f(V) . Thus, sending V to f(V) bijects the topology of M to the topology of N. 0

Because of this corollary, a homeomorphism is also called a topological equivalence.

In general, continuous maps do not need to send open sets to open sets. For example, the squaring map x f--* x 2 from ~ to ~ sends the open interval (-1 , 1) to the non-open interval [0, 1). See also Exercise 34.

Closure, Interior, and Boundary

Let S be a subset of the metric space M. Its closure, interior, and boundary are defined as follows:

closure S = nK where K ranges through the collection of all closed sets that contain S. Equivalently,

S = {x EM: if K is closed and S C K then x E K}.

interior int(S) = UV where V ranges through the collection of all open sets contained in S. Equivalently,

int S = {x EM: for some open V C S, x E V} .

boundary as = S n sc.


Since the intersection of closeds is closed, S and a S are closed. By construction, S is the smallest closed set that contains S because it is contained in every closed set that contains S. Similarly the interior of S is the largest open set contained in S, and the boundary is the closure minus the interior.

Alternate notations for closure, interior, and boundary are

S

int(S)

as

cl(S) o

S . S = bd(S) = fr(S).

The last stands for "frontier." Several useful facts about closure, interior, and boundary are presented in the exercises. See Exercise 92 in particular.

12 Proposition S = lim S.

Proof By Theorem 7, lim S is a closed set. It contains S, because for each s E S, the constant sequence (s, s, ... ) converges to s. By minimality of the closure,

S C limS.

On the other hand, lim S consists of limits of S, and because S is closed it must contain these limits, so

limS C S,

and the two sets are equal. D

Inheritance

Suppose that S is a subset of the metric space M and also of the metric space N. If S is open or closed in M, is the same true when S is judged in N? Not necessarily. The interval (a, b) is closed in itself, as is any metric space, but it is not closed in R It fails to contain its limits a, b in R

Even more striking is the situation concerning the set

S = {x E Q : -J2 < x < J2}.

As a subset of Q, S is clopen (both closed and open) but as a subset of JR it is neither closed nor open. It contains all its limits in Q but not all its limits in JR, and the same is true of its complement Q \ S.

The notation used above for closure, S, does not take into account the ambient metric space M. To clarify the metric space / metric subspace


situation we can write clM(S) to indicate the closure of S relative to the metric space M. It is the set of all limits in M of S. Similarly we can write aMeS) and intM(S) to indicate that we are working in M.

Let N be a metric subspace of M. The r-neighborhood of pEN, NrP, is the set of points in N whose distance to p is less than r. It is exactly the same as Mr p n N. Closures behave the same way. If SeN then clN(S) = clM(S) n N. See Figure 31.

Figure 31 The closure of S in N is its closure in M intersected with N.

13 Inheritance Principle If KeN c M where M is a metric space and N is a metric subspace then K is closed in N if and only if there is some subset L of M such that L is closed in M and K = L n N. That is, N inherits its closeds from M.

Proof Assume that K is closed in N , and consider the set L = clM(K). It is a closed subset of M and consists of K together with its limits in M. See Figure 32.

M

L M \ L D Figure 32 The relations among K, L, M, N.


None of these limits in M lies in N \ K since K is closed in N. Hence K = L n N. On the other hand if L is any closed set in M then L n N contains all its limits in Nand L n N is closed in N. 0

14 Corollary Dually, a metric subspace inherits its opens from the big metric space.

Proof Take complements. o 15 Corollary Assume that N is a metric subspace of M and also is a closed subset of M. A set KeN is closed in N if and only if it is closed in M.

Proof Assume that K is closed in N. If p is a limit of K then p is also a limit of N and lies in N since N is closed in M. Since K is closed in N, p also lies in K. Thus K is closed in M. On the other hand if KeN is closed in M then it contains all its limits in M and thus contains all its limits in N.

o 16 Corollary Assume that N is a metric subspace of M and also is an open subset of M. A set U C N is open in N if and only if it is open in M.

Proof The proof is left to the reader as Exercise 25. o

Clustering and Condensing

Two concepts similar to limits are often used in metric spaces - clustering and condensing. The set S clusters at p (and p is a cluster pointt of S) if each Mr P contains infinitely many points of S. The set S condenses at p (and p is a condensation point of S) if each Mr P contains uncountably many points of S. Thus, S limits at p, clusters at p, or condenses at p according to whether each Mr P contains some, infinitely many, or uncountably many points of S. See Figure 33.

17 Theorem The following are equivalent conditions to S clustering at p. (i) There is a sequence of distinct points in S that converges to p.

(ii) Each neighborhood of p contains infinitely many points of S. (iii) Each neighborhood of p contains at least two points of S. (iv) Each neighborhood of p contains at least one point of S other

than p.

Proof Clearly (i) ::::} (ii) ::::} (iii) ::::} (iv), and (ii) is the definition of clustering. It remains to check (iv) ::::} (i).

t Cluster points are also called accumulation points. As mentioned above, they are also sometimes called limit points, a usage that conflicts with the limit idea. A finite set S has no cluster points, but of course, each of its points p is a limit of S since the constant sequence (p, p, p, ... ) converges to p.


Figure 33 Limiting, clustering, and condensing behavior.

Assume (iv) is true: each neighborhood of P contains a point of S other thanp.lnM1Pchooseapointpl E (S\{pn.Setr2 = min(l/2,d(Pl,p», and in the smaller neighborhood Mr2 P, choose P2 E (S \ {pn. Proceed inductively: set rn = min(l/n, d(Pn-l, p» and in MrnP, choose Pn+l E

(S \ {p n. The points Pn are distinct since they have different distances to P,

d(Pl, p) ~ r2 > d(P2, p) ~ r3 > d(P3, p) ~ ....

Thus (iv) :::::} (i) and the four conditions are equivalent. See Figure 34. D

u • P2

Figure 34 The sequence of distinct points that converge to p.

Condition (iv) is the form of the definition of clustering most frequently used, although it is the hardest to grasp. It is customary to denote by S' the set of cluster points of S.

18 Proposition SUS' = s. Proof A cluster point is a type of limit of S, so S' C lim Sand

SUS' c limS.


On the other hand, if P E lim S then either pES or else each neighborhood of P contains points of S other than p. This implies that PES U S',

limS C SUS',

and the two sets are equal. o

19 Corollary S is closed if and only if S' c S.

Proof S is closed if and only if S = S. Since S = SUS', equivalent to S' C Sis S = S. 0

20 Corollary The least upper bound and greatest lower bound of a nonempty bounded subset S C lR belong to the closure of S. Thus, if S is closed, they belong to S.

Proof If b = l.u.b. S then each interval (b - r, b] contains points of S. The same is true for intervals [a, a + r) where a = g.l.b. S. 0

Product Metrics

We next define a metric on the Cartesian product M = MI X M2 of two metric spaces. There are three natural ways to do so:

dEep, q) = Jdl(PI, ql)2 + d2(P2, q2)2

dmax(p, q) = max{d l (PI, ql), d2(P2, q2)}

dsum(p, q) = dl (PI, ql) + d2(P2, q2)

where P = (PI, P2) and q = (ql, q2) belong to M. (dE is the Euclidean product metric.) The proof that these expressions actually define metrics on M is left as Exercise 83.

21 Theorem (Convergence in a product space) Thefollowing are equiv

alent for a sequence Pn = (Pin, P2n) in M = MI x M 2: (a) (Pn) converges with respect to the metric dmax.

(b) (Pn) converges with respect to the metric dE. (c) (Pn) converges with respect to the metric dsum.

(d) (Pin) and (P2n) converge in MI and M2 respectively.

Proof The issue is comparability of the metrics. We claim that


Dropping the smaller term inside the square root shows that dmax ::::: dE; comparing the square of dE and the square of dsum shows that the latter has the terms of the former and the cross term besides, so dE ::::: dsum ; and clearly dsum is no larger than twice its greater term, so dsum ::::: 2dmax . From comparability, equivalence of (a) - (d) is immediate. 0

22 Corollary A sequence in a Cartesian product of m metric spaces converges with respect to the sum metric if and only if it converges with respect to the max metric if and only if each component sequence converges.

Proof dmax ::::: dE::::: dsum ::: m dmax . o

23 Corollary (Convergence in ]Rm) A sequence of vectors (vn ) in]Rm converges in]Rm if and only if each of its component sequences (Vin) converges, I ::: i ::: m. The limit of the vector sequence is the vector

v lim Vn n-+oo ( lim Vln , lim V2n, ..• , lim Vmn ).

n~oo n--+-oo n---+oo

Proof This a special case of the previous corollary. o

Use of dmax and dsum often simplifies proofs by avoiding square root manipulations. The sum metric is also called the Manhattan metric or the taxicab metric. Figure 35 shows the "unit discs" with respect to these metrics in ]R2.

(0, 1) (1, 1)

v·

:y •• • -1" •• :

max (p;I, Iyl) < 1

(1,0) (1,0)

Ixl + Iyl < 1

Figure 35 The unit disc in the max metric is a square, and in the sum metric it is a rhombus.


Continuity of Arithmetic in IR

Addition is a mapping lEt x lEt --+ lEt that assigns to (x, y) the real number x + y. Subtraction and multiplication are also such mappings. Division is a mapping lEt x (lEt \ {On --+ lEt that assigns to (x, y) the number x/Yo

24 Theorem The arithmetic operations oflEt are continuous.

Proof Let (xo, Yo) E lEt x lEt be given and let E > 0 be given. Set So =

IXol + IYol· (+) Take 8 = E. If Ix - xol + Iy - Yol < 8 then

I(x + y) - (xo + Yo)1 :s Ix - xol + Iy - Yol < 8 = E.

( -) Take 8 = E. If Ix - Xo I + I Y - Yo I < 8 then

I(x - y) - (xo - Yo)1 :s Ix - xol + Iy - yol < 8 = E.

(x) Take 8 = min(l, E/(l + so)). If Ix - xol + Iy - Yol < 8 then

I(x . y) - (xo . Yo)1 :s IxllY - Yol + Ix - xollYol < E.

(-:-) Take 8 = min(IYol /2, 1, EY5/(2so + 2)). If Ix - xol + Iy - Yol < 8 then

Ixyo - xYI + Ixy - xoyl I(x -:- y) - (xo -:- Yo)1 :s < E.

Iyyol

25 Corollary The sums, differences, products, and quotients of real valued continuous functions are continuous. (The denominator functions should not equal zero.)

Proof Take, for example, the sum f + g where f, g : M --+ lEt are continuous. It is the composite of continuous functions

M ~ lEt x lEt +) lEt

x ~ (f(x),g(x)) ~ Sum(f(x), g(x)),

and is therefore continuous. The same applies to the other operations. D

Cauchy Sequences

In Chapter 1 we discussed the Cauchy criterion for convergence of a sequence of real numbers. There is a natural way to carry these ideas over to a metric space M. The sequence (Pn) in M satisfies a Cauchy condition provided that for each E > 0 there is an integer N such that for all k, n 2: N, d(Pb Pn) < E, and (Pn) is said to be a Cauchy sequence. In symbols,

"'IE > 0 3N such that k, n 2: N ::::} d(Pb Pn) < E.


The terms of a Cauchy sequence "bunch together" as n ---+ 00. Each convergent sequence (Pn) is Cauchy. For if (Pn) converges to pas n ---+ 00

then, given E > 0, there is an N such that for all n 2: N,

E d(Pn, p) < 2·

By the triangle inequality, if k, n 2: N then

so convergence =} Cauchy. Theorem 1.5 states that the converse is true in the metric space REvery

Cauchy sequence in R converges to a limit in R In the general metric space, however, this need not be true. For example, consider the metric space Ql of rational numbers, equipped with the standard distance d(x, y) = Ix - yl, and consider the sequence

(rn) = (l.4, 1.41, 1.414,1.4142, ... ).

It is Cauchy. Given E > 0, choose N > -loglO E. If k, n 2: N then irk - rnl ::::: lO-N < E. Nevertheless, (rn) refuses to converge in tQl. After all, as a sequence in R it converges to ,J2, and if it also converges to some r E tQl, then by uniqueness of limits in R, r = ,J2, something we know is false. In brief, convergence =} Cauchy but not conversely.

A metric space M is complete if each Cauchy sequence in M converges to a limit in M. Theorem 1.5 implies that R is complete.

26 Theorem Rm is complete.

Proof Let (Pn) be a Cauchy sequence in Rm. Express Pn in components as

Pn = (PIn, ... , Pmn)·

Because (Pn) is Cauchy, each component sequence (Pjn)neN is Cauchy. Completeness of R implies that the component sequences converge, and therefore the vector sequence converges. 0

27 Theorem Every closed subset of a complete metric space is a complete metric space.

Proof Let A be a closed subset of the complete metric space M and let (Pn) be a Cauchy sequence in A. It is of course also a Cauchy sequence in M and therefore it converges to a limit P in M. Since A is closed, pEA.

o


28 Corollary Every closed subset of Euclidean space is a complete metric space.

Proof Obvious from the previous theorem and completeness of ffi.m . 0

Boundedness

A subset S of a metric space M is bounded if for some p E M and some r > 0,

S C Mrp.

A set which is not bounded is unbounded. For example, the elliptical region 4x2 + y2 < 4 is a bounded subset of ffi.2, while the hyperbola xy = 1 is unbounded.

Distinguish the word "bounded" from the word "finite." The first refers to physical size, the second to the number of elements. The concepts are totally different. Also, boundedness has little connection to the existence of a boundary - a clopen subset of a metric space has empty boundary, but some clopen sets are bounded.

It is easy to check that the terms of a Cauchy sequence (Pn) in M form a bounded subset of M. Simply take E = 1, and apply the Cauchy definition: there is an N such that for all n, m ~ N, d(PI, Pm) < 1. Choose

By the triangle inequality, the Cauchy sequence lies entirely in Mr(pd. Since a convergent sequence is Cauchy, its terms also form a bounded set.

Boundedness is not a topological property. For example, consider ( - 1, 1) and R They are homeomorphic, although (-1, 1) is bounded and ffi. is unbounded. The same example shows that completeness is not a topological property.

A function from M to another metric space N is a bounded function if its range is bounded. That is, there exist q E Nand r > ° such that

fM C Nrq.

Note that a function can be bounded even though its graph is not. For example, x f-+ sin x is a bounded function ffi. ~ ffi. although its graph, {(x, y) E ffi.2 : y = sin x}, is an unbounded subset of ffi.2.


2 Compactness

Compactness is the single most important concept in real analysis. It is what reduces the infinite to the finite.

A subset A of a metric space M is (sequentially) compact if every sequence (an) in A has a subsequence (ank ) that converges to a limit in A.

The empty set and finite sets are trivial examples of compact sets. For a sequence (an) contained in a finite set repeats a term infinitely often, and the corresponding constant subsequence converges.

Compactness is a good feature of a set. We will develop criteria to decide whether a set is compact. The first is the most often used, but beware! -its converse is generally false.

29 Theorem Every compact set is closed and bounded.

Proof Suppose that A is a compact subset of the metric space M and that p is a limit of A. There is a sequence (an) in A converging to p. By compactness, some subsequence (ank ) converges to some q E A, but every subsequence of a convergent sequence converges to the same limit as does the mother sequence, so q = p and pEA. Thus, A is closed.

To see that A is bounded, choose and fix any point p EM. Either A is bounded or else for each n E N there is a point an E A such that d(p, an) :::: n. Compactness implies that some subsequence (ank ) converges. Convergent sequences are bounded, which contradicts the fact that d(p, ank ) ---+ 00 as k ---+ 00. Therefore (an) can not exist and A is bounded.

30 Theorem The closed interval [a, b] c lR is compact.

Proof Let (xn) be a sequence in [a, b]. Let

C = {x E [a, b] : Xn < x only finitely often}.

o

Since a E C, C i= 0. Clearly b is an upper bound for C. By the least upper bound property of lR, c = l.u.b. C exists, c E [a, b]. We claim that a subsequence of (xn) converges to c. Suppose not, i.e., no subsequence of (xn) converges to c. Then for some E > 0, Xn lies in (c - E, C + E) only finitely often, which implies that c + E E C, contrary to c being an upper bound for C. Hence some subsequence of (xn) does converge to c, and [a, b] is compact. 0

To pass from lR to lRm , think about compactness in terms of Cartesian products.

31 Theorem The Cartesian product of two compact sets is compact.

Section 2 Compactness 77

Proof Let (an, bn) E A x B be given where A C M and BeN are compact. There exists a subsequence (ank ) that converges to some point a E A as k ~ 00. The subsequence (bnk ) has a sub-subsequence (bnk(e)) that converges to some b E B as £ ~ 00. The sub-subsequence (ank(e)) continues to converge to the point a. Thus

as £ ~ 00. This implies that A x B is compact. o 32 Corollary The Cartesian product of m compact sets is compact.

Proof Write Al x A2 X ... x Am = Al X (A2 X ... X Am) and perform induction on m. (Theorem 31 handles the bottom case m = 2.) 0

33 Corollary A box [aI, btl x ... x [am, bml is compact.

Proof Obvious from Theorem 30 and the previous corollary. o

An equivalent formulation of these results is the

34 Bolzano-Weierstrass Theorem Any bounded sequence in ]Rm has a convergent subsequence.

Proof A bounded sequence is contained in a box, which is compact, and therefore the sequence has a subsequence that converges to a limit in the ~x. 0

Here is a simple fact about compacts.

35 Theorem A closed subset of a compact set is compact.

Proof If A is a closed subset of the compact set C and if (an) is a sequence of points in A, then clearly (an) is also a sequence of points in C, so by compactness of C, there is a subsequence (ank ) converging to a limit p E C. Since A is closed, p lies in A, which proves that A is compact. 0

Now we come to the first partial converse to Theorem 29.

36 Heine-Borel Theorem Every closed and bounded subset of]Rm is compact.

Proof Let A C ]Rm be closed and bounded. Boundedness implies that A is contained in some box, which is compact. Since A is closed, Theorem 35 implies that A is compact. 0


The Heine-Borel Theorem states that closed and bounded subsets of Euclidean space are compact, but it is vital to remember that a closed and bounded subset of a general metric space may fail to be compact. For example, the set S of rational numbers in [0, 1] is a closed, bounded, noncompact subset of Q. The set S is noncompact because it contains sequences which have no subsequences that converge in S.

Ten Examples of Compact Sets

1. Any finite subset of a metric space; for instance, the empty set. 2. Any closed subset of a compact set. 3. The union of finitely many compact sets. 4. The Cartesian product of finitely many compact sets. 5. The intersection of arbitrarily many compact sets. 6. The unit ball in ffi.3. 7. The boundary of a compact set, for instance the unit 2-sphere in ffi.3. 8. The set {x E ffi.: 3n EN and x = lin} U {O}. 9. The Hawaiian earring. See page 53.

10. The Cantor set. See Section 5.

Nests of Compacts

If Al :) A2 :) ... :) An :) An+! :) ... then (An) is a nested sequence of sets. Its intersection is

nAn = {p: foreachn, p E An}.

See Figure 36.

Figure 36 A nested sequence of sets.


For example, we could take An to be the disc {z E ]R2 : Izl ::: lin}. The intersection of all the sets An is then the singleton {OJ. On the other hand, if An is the ball {z E ]R3 : Izl ::: 1 + lin} then nAn is the closed unit ball B3.

37 Theorem The intersection of a nested sequence of compact non-empty sets is compact and non-empty.

Proof Let (An) be such a sequence. By Theorem 29, An is closed. The intersection of closed sets is always closed. Thus, nAn is a closed subset of the compact set AI, and is therefore compact. It remains to show that the intersection is non-empty.

An is non-empty, so for each n E N we can choose an E An. The sequence (an) lies in Al since the sets are nested. Compactness of Al implies that (an) has a subsequence (ank ) converging to some point pEA 1. The limit p also lies in the set A2 since except possibly for the first term, the subsequence (ank ) lies in A2 and A2 is a closed set. The same is true for A3 and for all the sets in the nested sequence. Thus, p E nAn and nAn is shown to be non-empty. 0

The diameter of a non-empty set ScM is the supremum of the distances d (x, y) between points of S.

38 Corollary If in addition to being nested, non-empty, and compact, the sets An have diameter that tends to 0 as n -+ 00, then A = nAn is a single point.

Proof For each kEN, A is a subset of Ab which implies that A has diameter zero. Since any distinct points lie at positive distance from each other, A consists of at most one point, while by Theorem 37 it consists of at least one point. See also Exercise 27. 0

A nested sequence of non-empty noncompact sets can have empty intersection. For example, the open discs with center (lIn, 0) on the x-axis and radius lin are nested, n EN, but contain no common point. (Their closures do intersect at a common point, the origin.) See Figure 37.

Continuity and Compactness

Next we discuss how compact sets behave under continuous transformations.

39 Theorem If f : M -+ N is continuous and A is a compact subset of M then fA is a compact subset of N. That is, the continuous image of a compact is compact.


Figure 37 This nested sequence has empty intersection.

Proof Suppose that (bn ) is a sequence in fA. For each n E N choose a point an E A such that f(an) = bn. By compactness of A there exists a subsequence (ank ) that converges to some point pEA. By continuity of f it follows that

bnk = f(ank ) -+ fp E fA.

Thus, any sequence (bn ) in f A has a subsequence converging to a limit in f A, and fA is compact. 0

From Theorem 39 follows the natural generalization of the minimax theorem in Chapter 1 which concerned continuous real-valued functions defined on an interval [a, b]. See Theorem 23 in Chapter 1.

40 Corollary A continuous real-valued function defined on a compact set is bounded; it assumes maximum and minimum values.

Proof Let f : M -+ lR. be continuous and let A be a compact subset of M. Theorem 39 implies that f A is a compact subset of R so by Theorem 29 it is closed and bounded. Thus, the greatest lower bound, v, and least upper bound, V, of f A exist and belong to fA; there exist points p, PEA such that for all a E A, v = fp:::; fa:::; fP = v. 0

Homeomorphisms and Compactness

A homeomorphism is a bi-continuous bijection. Originally, compactness was called bi-compactness. This is reflected in the next theorem.


41 Theorem If M is compact and M is homeomorphic to N then N is compact. Compactness is a topological property.

Proof If f : M ---+ N is a homeomorphism then Theorem 39 implies that f M is compact. 0

It follows that [0, 1] and]R are not homeomorphic. One is compact and the other is not.

42 Theorem If M is compact then a continuous bijection f : M ---+ N is a homeomorphism - its inverse bijection f- l

: N ---+ M is continuous.

Proof Suppose that qn ---+ q in N. Since f is a bijection Pn = f- l (qn) and p = f-l(q) are well defined points in M. To check continuity of f- l

we must show that Pn ---+ p. If (Pn) refuses to converge to p then there is a subsequence (Pnk) and

a 8 > ° such that for all k, d(Pnk' p) ::: 8. Compactness of M gives a sub-subsequence (Pnk(e») that converges to a point p* E Mas £. ---+ 00.

Necessarily, d(p, p*) ::: 8, which implies that p i- p*. Since f is continuous

f(Pnk(e») ---+ f(p*)

as £. ---+ 00. The limit of a convergent sequence is unchanged by passing to a subsequence, and so f(Pnk(e) = qnk(e) ---+ q as £. ---+ 00. Thus, f(p*) = q = f (p ), contrary to f being a bijection. It follows that Pn ---+ P and therefore that f- I is continuous. 0

If M is not compact then Theorem 42 can fail. For example, the bijection f : [0,2rr) ---+ ]R2 defined by f(x) = (cosx, sinx) is a continuous bijection onto the unit circle in the plane, but it is not a homeomorphism. This useful example was discussed on page 58. Not only does this f fail to be a homeomorphism, but there is no homeomorphism at all from [0, 2rr) to SI. The circle is compact while [0, 2rr) is not. Therefore they are not homeomorphic.

Absolute Closedness

Not only is a compact space M closed in itself, as is every metric space, but it is also a closed subset of each metric space in which it is embedded. More precisely, we say that h : M ---+ N embeds Minto N if h is a homeomorphism from M onto hM. (The metric on hM is the one it inherits from N.) Topologically M and hM are equivalent. A property of M that holds


also for every embedded copy of M is an absolute or intrinsic property ofM.

43 Theorem A compact is absolutely closed and absolutely bounded.

Proof Obvious from Theorems 29 and 39. D

For example, no matter how the cirele is embedded in ~3, it is a elosed and bounded set. See also Exercises 28 and 87.

Uniform Continuity and Compactness

In Chapter 1 we defined the concept of uniform continuity for functions of a real variable. The definition in metric spaces is analogous. A function f : M -+ N is uniformly continuous if for each E > 0 there exists a 8 > 0 such that

p, q EM and dM(p, q) < 8 :::} dN(fp, fq) < E.

44 Theorem Every continuous function defined on a compact is uniformly continuous.

Proof Suppose not, and f : M -+ ~ is continuous, M is compact, but f is not uniformly continuous. Then there is some E > 0 such that no matter how small 8 is, there exist points p, q E M with d (p, q) < 8 but Ifp - fql ::: E. Take 8 = lin and let Pn, qn be sequences of points in M such thatd(Pn, qn) < lin while If(Pn) - f(qn)1 ::: E. Compactness of M implies that there is a subsequence Pnk which converges to some P E M as k -+ 00. Since d(Pn, qn) < lin -+ 0 as n -+ 00, (qnk) converges to the same limit as does (Pnk) as k -+ 00; namely, qnk -+ p. Continuity implies that f(Pnk) -+ fp and f(qnk) -+ fp. If k is large then

If(Pnk) - f(qnk) I ::: If(Pnk) - fpl + Ifp - f(qnk) I < E,

contrary to the supposition that If(Pn) - f(qn)1 ::: E for all n. D

Theorem 44 gives a second proof that continuity implies uniform continuity on an interval [a, b]. For [a, b] is compact.

3 Connectedness

As another application of these ideas, we consider the general notion of connectedness. Let A be a subset of M, a metric space. If A is neither the empty set nor is it M, then A is a proper subset of M. Recall that if A is both elosed and open in M it is said to be elopen. The complement of a elopen set is elopen. The complement of a proper subset is proper.

Section 3 Connectedness 83

If M has a proper elopen subset A, M is disconnected. For there is a separation of M into proper, disjoint elopen subsets,

M = A u A C•

(The notation U indicates disjoint union.) M is connected if it is not disconnected - it contains no proper elopen subset. Connectedness of M does not mean that M is connected to something, but rather that M is one unbroken thing. See Figure 38.

Figure 38 M and N illustrate the difference between being connected and being disconnected.

45 Theorem If M is connected, f : M -+ N is continuous, and f is onto then N is connected. The continuous image of a connected is connected.

Proof Simple! If A is a elopen proper subset of N then, according to the open and elosed set conditions for continuity, fPre(A) is a elopen subset of M. Since f is onto and A i= 0, fpre(A) i= 0. Similarly, fpre(A c ) i= 0. Therefore fpre(A) is a proper elopen subset of M, contrary to M being connected. It follows that A can not exist and N is connected. 0

46 Corollary If M is connected and M is homeomorphic to N then N is connected. Connectedness is a topological property.

Proof N is the continuous image of M. o 47 Corollary (Generalized Intermediate Value Theorem) Everycontinuous real-valued function defined on a connected domain has the intermediate value property.

Proof Assume that f : M -+ IR is continuous and M is connected. If f assumes values a < fJ in IR and if it fails to assume some value Y with a < Y < fJ, then

M = {x EM: f(x) < Y} u {x EM: f(x) > Y}

is a separation of M, contrary to connectedness. o 48 Theorem IR is connected.


Proof If U C JR. is open, closed, and non-empty, we claim that U = R By Theorem 9, U is the countable disjoint union of open intervals whose endpoints do not belong to U. If (a, b) is one of these intervals and b < 00,

then closedness of U implies that b E U, a contradiction, so b = 00.

Similarly, a = -00, and U = R Since JR. contains no proper clopen set, it is connected. D

49 Corollary (Intermediate Value Theorem for JR.) Every continuous function f : JR. -+ JR. has the intermediate value property.

Proof Immediate from the Generalized Intermediate Value Theorem and connectedness of R D

Thus we have a second proof of the Intermediate Value Theorem 1.24.

50 Corollary The following metric spaces are connected: the intervals (a, b), [a, b], the circle, and all capital letters of the Roman alphabet.

Proof The interval (a, b) is homeomorphic to JR., while [a, b] is the continuous image of JR. under the map whose graph is shown in Figure 39. The circle is the continuous image of JR. under the map t f-+ (cos t, sin t). Connectedness of the letters A, ... , Z is equally clear. D

!(x)=b

!(x)=a

• • a b

Figure 39 The function f surjects JR. continuously to [a, b].

Connectedness properties give a good way to distinguish non-homeomorphic sets.

Example The union of two disjoint closed intervals is not homeomorphic to a single interval. One set is disconnected and the other is connected.


Example The closed interval [a, b] is not homeomorphic to the circle Sl. For removal of a point x E (a, b) disconnects [a, b] while the circle remains connected upon removal of any point. More precisely, suppose that h : [a, b] -+ Sl is a homeomorphism. Choose a point x E (a, b), and consider X = [a, b] \ {x}. The restriction of h to X is a homeomorphism from X onto Y, where Y is the circle with one point, hx, removed. But X is disconnected, while Y is connected. Hence h can not exist and the segment is not homeomorphic to the circle.

Example The circle is not homeomorphic to the figure eight. Removing any two points of the circle disconnects it, but this is not true of the figure eight. Or, removing the crossing point disconnects the figure eight, but removing any point of the circle leaves it connected.

Example The circle is not homeomorphic to the disc. For removing two points disconnects the circle but does not disconnect the disc.

As you can see, it is useful to be able to recognize disconnected subsets S of a metric space M. By definition, S is a disconnected subset of M if it is disconnected when considered in its own right as a metric space with the metric it inherits from M; it has a separation S = A u B such that A and B are proper closed subsets of S. According to the inheritance principle, closedness in S can be analyzed in M. The facts that no point of A is a limit of B and no point of B is a limit of A imply that the closure of B in M misses A and the closure of A in M misses B. (Note: "in M.") In symbols,

AnB = 0 = AnB.

The separated sets A, B must be disjoint but they need not be closed in M nor need their closures in M be disjoint.

Example The punctured interval X = [a, b] \ {c} is disconnected if a < c < b. For X = [a, c) U (c, b] is a separation of X. Observe that the closures in lR of the separated sets have non-empty intersection.

Example Any subset Y of the punctured interval is disconnected if it meets both [a, c) and (c, b). For Y = ([a, c) n Y) u «c, b) n Y) is a separation of Y.

51 Theorem The closure of a connected set is connected. More generally, if SeT c Sand S is connected, then so is T.

Proof It is equivalent to show that if T is disconnected then S is disconnected. Disconnectedness of T implies that


T=AuB

where A, B are clopen and proper. The set K = A n S is clopen in S according to the Inheritance Theorem 13, but can K be improper? If K = 0 then A eSc. Since A is proper there exists pEA. Since A is open in T, there exists a neighborhood Mr P such that

TnMrP cAe SC.

The neighborhood Mr P contains no points of S, which is contrary to p belonging to S. Thus, K =I 0. Similarly, L = B n S =I 0, so S = K u L is a separation of S, proving that S is disconnected. 0

Example The outward spiral expressed in polar coordinates as

S = {(r, €i) : (1 - r)B = 1 and €i :::: nj2}

has closure S = S U Sl, Sl being the unit circle. Since S is connected, so is the closure. See Figure 22.

52 Theorem The union of connected sets sharing a common point is connected.

Proof Let S = USa where each Sa is connected and p E nSa. If Sis disconnected then it has a separation S = A u A C where A, A C are proper and clopen. One of them contains p, say it is A. Then A n Sa is a non-empty clopen subset of Sa. Since Sa is connected, A n Sa = Sa for each ex, and A = S. This implies that A C = 0, a contradiction. 0

Example The 2-sphere S2 is connected. For S2 is the union of great circles, each passing through the poles.

Example Every convex set C in ]Rm (or in any metric space with a compatible linear structure) is connected. If we choose a point pEe then each q E C lies on a line segment [p, q] c C. Thus, C is the union of connected sets sharing the common point p. It is connected.

A path joining p to q in a metric space M is a continuous function f: [a, b] -+ M such that fa = p and fb = q. If each pair of points in M can be joined by a path in M then M is path-connected. See Figure 40.

53 Theorem Path-connected implies connected.

Proof Assume that M is path-connected but not connected. Then M A u A C for some proper clopen A eM. Choose pEA and q E A c. There is a path f : [a, b] -+ M from p to q. The sets fPre(A) and fpre(A c )

contradict connectedness of [a, b]. 0


a b

Figure 40 A path f in M that joins p to q.

Example All connected subsets of]R are path-connected. See Exercise 65.

Example Every open connected subset of]Rm is path-connected. See Exercises 58, 64.

Example The topologist's sine curve is a compact connected set that is not path-connected. It is M = GUY where

G = {(x, Y) E]R2 : y = sin l/x and 0 < x ::s l/n}

Y = {CO, y) E]R2 : -1 ::s y ::s 1}.

See Figure 41. The metric on M is just Euclidean distance. Is M connected? Yes! The graph G is connected and M is its closure. By Theorem 51, Mis connected.

0.5

o

--{).5

0.05 0.1 0.15 0.2 0.25 0.3 0.35

Figure 41 The topologist's sine curve M is a closed set. It includes the vertical segment Y at x = O.


4 Coverings

For the sake of simplicity we have postponed discussing compactness in terms of open coverings until this point. Typically, students find coverings a challenging concept. It is central, however, to much of analysis, for example measure theory. A collection U of subsets of M covers A c M if A is contained in the union of the sets belonging to U. The collection U is a covering of A.

If U and V both cover A and if V c U in the sense that each set V E

V belongs also to U then we say that U reduces to V, and that V is a subcovering. If all the sets in U are open, U is an open covering of A. If every open covering of A reduces to a finite subcovering then we say that A is covering compact t .

The idea is that if A is covering compact and U is an open covering of A then just a finite number of the open sets are actually doing the work of covering A. The rest are redundant.

A covering U of A is also called a cover of A. The members ofU are not called covers. Instead, you could call them scraps. Imagine the covering as a patchwork quilt that covers a bed, the quilt being sewn together from overlapping scraps of cloth. See Figure 42.

Figure 42 A covering of A by nine scraps.

The mere existence of a finite open covering of A is trivial and utterly worthless. Every set A has such a covering, namely the single open set M. Rather, for A to be covering compact, each and every open covering of A

t You will frequently find it said that an open covering of A has a finite subcovering. "Has" means "reduces to."

Section 4 Coverings 89

must reduce to a finite subcovering of A. Deciding directly whether this is so is daunting. How could you hope to verify the finite reducibility of all open coverings of A? There are so many of them. For this reason we concentrated on sequential compactness; it is relatively easy to check by inspection whether every sequence in a set has a convergent subsequence.

To check that a set is not covering compact it suffices to find an open covering which fails to reduce to a finite sUbcovering. Occasionally this is simple. For example, the set (0, 1] is not covering compact in lR because its covering

u = {( I / n, I + 1/ n) : n E N}

fails to reduce to a finite subcovering.

54 Theorem For a subset A of a metric space M the following are equivalent:

(a) A is covering compact. (b) A is sequentially compact.

Proof that (a) implies (b) We assume that A is covering compact and prove that it is sequentially compact. Suppose not. There is a sequence (Pn) in A, no subsequence of which converges in A. Each point a E A therefore has some neighborhood Mra such that Pn E Mra only finitely often. (The radius r may depend on the point a.) The collection {Mra : a E A} is an open covering of A and by covering compactness it reduces to a finite subcovering

{Mn (a1), Mr2 (a2), ... , Mrk(ak)}

of A. Since Pn appears in each of these finitely many neighborhoods Mri (ai)

only finitely often, it follows from the pigeon-hole principle that (Pn) has only finitely many terms, a contradiction. Thus, (Pn) can not exist and A is sequentially compact. D

The following presentation of the proof that (b) implies (a) appears in Royden's book, Real Analysis. A Lebesgue number for a covering U of A is a positive real number A such that for each a E A there is some U E U with MAa C U. Of course, the choice ofthis U depends on a. It is crucial, however, that the Lebesgue number A is independent of a E A.

The idea of a Lebesgue number is that we know each point a E A is contained in some U E U, and if A is extremely small then MAa is just a slightly swollen point - so the same should be true for it too. No matter where in A the neighborhood MAa is placed, it lies wholly in some member of the covering. See Figure 43.

If A is noncompact then it may have open coverings with no positive Lebesgue number. For example, let A be the open interval (0, 1). It is


Figure 43 Small neighborhoods are like swollen points; U has a positive Lebesgue number.

covered by itself, (0, 1) C (0, 1) = U. Then for each r > 0, the rneighborhood of a, (a - r, a + r), fails to lie in U when 0 < a < r. See Exercise 45.

55 Lebesgue Number Lemma Every open covering of a sequentially compact set has a Lebesgue number A > O.

Proof Suppose not: U is an open covering of a sequentially compact set A, and yet for each A > 0 there exists an a E A such that no U E U contains M;.,.a. Take A = lin and let an E A be a point such that no U E U contains M 1/ n (an). By sequential compactness, there is a subsequence (ank ) converging to some point pEA. Since U is an open covering of A, there exist r > 0 and U E U with MrP C U. If k is large then d(ank , p) < rl2 and 1 Ink < r 12, which implies by the triangle inequality that

MI/nk(ank) C MrP C U,

contrary to the supposition that no U E U contains M I/n (an). We conclude that, after all, U does have a Lebesgue number A > O. See Figure 44. D

Proofthat (b) implies (a) in Theorem 54 Let U be an open covering of the sequentially compact set A. We want to reduceU to a finite sUbcovering. By the Lebesgue number Lemma, U has a Lebesgue number A > O. Choose anyal E A and some UI E U such that

M;.,.(al) CUI.

Section 4 Coverings

Q ........ . u

Figure 44 The neighborhood Mr P engulfs the smaller neighborhood M I / nk (ank )·

91

If UI :J A then U reduces to the finite subcovering {Ud consisting of a single set, and the implication (b) =} (a) is proved. On the other hand, as is more likely, if UI does not contain A then we choose an uncovered point a2 E A and a set U2 E U such that

MA(a2) C U2.

Either U reduces to the finite subcovering {UI, U2} (and the proof is finished) or else we can continue, eventually producing a sequence (an) in A and a sequence (Un) in U such that

MA(an) C Un and an+1 E (A \ (UI U ... U Un)).

We will show that such sequences (an), (Un) lead to a contradiction. By sequential compactness, there is a subsequence (ank ) that converges to some pEA. For a large k, d(ank , p) < A and

P E MA(ank ) C Unk ·

See Figure 45. All ani with e > k lie outside Unk , which contradicts their convergence

to p. Thus, at some finite stage the process of choosing points an and sets Un terminates, and U reduces to a finite subcovering {UI, ... , Un} of A, which implies that A is covering compact. 0

Upshot In light of Theorem 54, the term "compact" may now be applied equally to any set obeying (a) or (b).


Figure 45 The point ank is so near p that the neighborhood M)..(ank )

engulfs p.

Total Boundedness

The Heine-Borel Theorem states that a subset of ~m is compact if and only if it is closed and bounded. In more general metric spaces, such as Q, the assertion is false. But what if the metric space is complete? It is still false. For example the discrete metric on N makes N closed and bounded in N but it is noncompact: after all, what subsequence of (l, 2, 3, ... ) converges?

But mathematicians do not quit easily. The Heine-Borel Theorem ought to generalize beyond ~m somehow. Here is the concept we need: a set A C M is totally bounded if for each E > 0 there exists a finite covering of A by E-neighborhoods. No mention is made of a covering reducing to a subcovering. How close total boundedness is to the worthless fact that every metric space has a finite open covering!

56 Theorem (Generalized Heine-Borel Theorem) AsubsetoJacomplete metric space is compact if and only if it is closed and totally bounded.

Proof Let A be a compact subset of M. Therefore it is closed. To see that it is totally bounded, let E > 0 be given and consider the E-neighborhood covering of A,

Compactness of A implies that this covering reduces to a finite subcovering and therefore A is totally bounded.

Conversely, assume that A is a closed and totally bounded subset of the complete metric space M. We claim that A is sequentially compact: any

Section 4 Coverings 93

sequence (an) in A has a subsequence that converges in A. Set Ek = 1/ k, k = 1, 2, .... Since A is totally bounded we can cover it by finitely many E I-neighborhoods

MEl (qI), ... , MEl (qm).

By the pigeon-hole principle, terms ofthe sequence an lie in at least one of these neighborhoods infinitely often, say it is MEl (PI). Choose

ani E Al = A n MEl (PI).

Any subset of a totally bounded set is totally bounded, so we can cover Al by finitely many E2-neighborhoods. For one of them, say ME2 (P2), an lies in A2 = Al n ME2 (P2) infinitely often. Choose an2 E A2 with n2 > nl.

Proceeding inductively, cover Ak-I by finitely many Ek -neighborhoods, one of which, say MEk (N), contains terms of the sequence (an) infinitely often. Then choose ank E Ak = Ak- 1 n MEk(Pk) with nk > nk-l. Then (ank ) is a subsequence of (an). It is Cauchy because for k, £ ::: K,

. 2 dlamA K < 2EK = -. - K

Completeness of M implies that (ank ) converges to some P E M and since A is closed, pEA. Hence A is compact. D

57 Corollary A metric space is compact if and only if it is complete and totally bounded.

Proof Every compact metric space M is complete. This is because, given a Cauchy sequence (Pn) in M, compactness implies that some subsequence converges in M, and if a Cauchy sequence has a convergent subsequence then the mother sequence converges too. As observed above, compactness immediately gives total boundedness.

Conversely, assume that M is complete and totally bounded. Any metric space is closed in itself. By Theorem 56, M is compact. D

Perfect Metric Spaces

A metric space M is perfect if M' = M: each P E M is a cluster point of M. Recall that M clusters at P if each Mr P is an infinite set. For example [a, b] is perfect and tQl is perfect. N is not perfect; none of its points are cluster points.

58 Theorem Every non-empty, peifect, complete metric space is uncountable.


Proof Suppose not: assume M is non-empty, perfect, complete, and countable. Since M consists of cluster points it must be denumerable and not finite. Say

M = {XI,X2, ... }

is a list of all the elements of M. Define

MrP = {q EM: d(p, q) :s r}. It is the "closed neighborhood" of radius r at p. If x ~ Y we say that Y excludes x. Proceeding as in the proof of Theorem 56 we inductively choose a nested sequence of these closed neighborhoods to exclude more and more points in the sequence (xn). They nest down to a point in M that is different from all the Xn , a contradiction.

Specifically take YI = X2, rl = min (1 , d (Xl, X2) /2), and set ~

YI = MrJ (YI).

Then YI excludes Xl. Since M clusters at YI, infinitely many Xn lie in MTJ (YI), so we can choose Y2 E MrJ (YI) such that Y2 =1= X2· The choice of

r2 = min{1/2, d(Y2, x2)/2, (rl - d(YI, Y2»}

causes Y2 = Mr2 (Y2) to be a subset of YI and to exclude Xl, X2. See Figure 46.

•

Figure 46 The exclusion of successively more points of the sequence (Xn).

Nothing stops us from continuing indefinitely, and we get a nested sequence of closed neighborhoods

Section 5 Cantor Sets 95

such that Yn excludes Xl, ... , Xn , and has radius r n ::: 1/ n. Thus the center points Yn form a Cauchy sequence. Completeness of M implies that

lim Yn = Y EM n-+oo

exists. Since the sets Yn are closed and nested, Y E Yn for each n. Does Y equal Xl? No, for Y I excludes Xl. Does it equal X2? No, for Y2 excludes X2. In fact, for each n, Y i= Xn . The point Y is nowhere in the supposedly complete list of elements of M, a contradiction. 0

59 Corollary lR and [a, b] are uncountable.

Proof lR is complete and perfect, while [a, b] is compact, therefore complete, and perfect. Neither is empty. 0

60 Corollary A non-empty peifect complete metric space is everywhere uncountable in the sense that each r-neighborhood is uncountable.

Proof The r/2-neighborhood M r / 2(p) is perfect: it clusters at each of its points. The closure of a perfect set is perfect. Thus, Mr /2 (p) is perfect. Being a closed subset of a complete metric space, it is complete. According to Theorem 58, Mr/2(p) is uncountable. Since Mr/2(p) C MrP, MrP is uncountable. 0

5 Cantor Sets

Cantor sets are fascinating examples of compact sets that are maximally disconnected. (To emphasize the disconnectedness, one sometimes refers to a Cantor set as "Cantor dust.") Here is how to construct the standard Cantor set. Start with the unit interval [0, 1] and remove its open middle third, (1/3,2/3). Then remove the open middle third from the remaining two intervals, and so on. This gives a nested sequence eO => el => e2 => ... where eO = [0, 1], el is the union of the two intervals [0, 1/3] and [2/3,1], e2 is the union of four intervals [0, 1/9], [2/9, 1/3], [2/3,7/9], and [8/9, 1], e3 is the union of eight intervals, and so on. See Figure 47.

In general en is the union of2n intervals, each oflength 1/3n . Each en is compact. The standard middle thirds Cantor set is the nested intersection

We refer to e as "the" Cantor set. Clearly it contains the endpoints of each of the intervals comprising en. Actually, it contains uncountably many more points than these endpoints! There are other Cantor sets defined by


CO -------------------------------------------

c 1 --------------

c 2 --

c 3 - - An endpoint

c4 --

c' ----

c

Figure 47 The construction of the standard middle thirds Cantor set e.

removing, say, middle fourths, pairs of middle tenths, etc. All Cantor sets turn out to be homeomorphic to the standard Cantor set. See Section 6.

A metric space M is totally disconnected if each point p E M has arbitrarily small clopen neighborhoods. That is, given E > 0 and p E M, there exists a clopen set U such that

p E U C MEP.

For example, any discrete space is totally disconnected. So is <Q.

61 Theorem The Cantor set is a compact, non-empty, peifect, and totally disconnected metric space.

Proof The metric on e is the one it inherits from JR, the usual distance Ix - YI. Because e ::J E, the set of endpoints of the en-intervals, e is nonempty and infinite. It is compact because it is the intersection of compacts.

To show e is perfect and totally disconnected, take any x E e and any E > O. Fix n so large that 1/3n < E. The point x lies in one of the 2n intervals I of length 1/3n comprising en. The set of e -endpoints in I, En I, is infinite and contained in the interval (x - E, X + E). Thus e clusters at x and e is perfect. See Figure 48.

I

• X-E x X+E

Figure 48 The endpoints of e cluster at x.

The interval I is closed in JR and therefore in en. The complement J = en \ I consists of finitely many closed intervals and is therefore closed too. Thus

Section 5 Cantor Sets 97

C (C n I) u (C n J)

exhibits C as the disjoint union of two elopen sets, C n I being a subset of (x - E, X + E) which contains x. Hence C is totally disconnected. 0

62 Corollary The Cantor set is uncountable.

Proof Being compact, C is complete, and by Theorem 58, every complete, perfect, non-empty metric space is uncountable. 0

A more direct way to see that the Cantor set is uncountable involves a geometric coding scheme. Take the code: 0 = left and 2 = right. Then

Co = left interval = [0, 1/3] C2 = right interval = [2/3, 1],

and C I = Co U C2. Similarly, the left and right subintervals of Co are coded Coo and CO2 , while the left and right subintervals of C2 are C20 and C22 .

This gives C2 Coo U CO2 U C20 U C22 .

The intervals that comprise C3 are specified by strings of length 3. For instance, C220 is the left subinterval of C22 . In general an interval of Cn is coded by an address string of n symbols, each a 0 or a 2. Read it like a zip code. The first symbol gives the interval's gross location (left or right), the second symbol refines the location, the third refines it more, and so on.

Imagine now an infinite address string w = WI W2W3 .•. of O's and 2's. Corresponding to w, we form a nested sequence of intervals

the intersection of which is a point p = p (w) E C. Specifically,

where win = WI ... Wn truncates W to an address of length n. See Theorem 37.

As we have observed, any infinite address string defines a point in the Cantor set. Conversely, any point p E C has an address W = w(p): its first n symbols a = win are specified by the interval Ccx of Cn in which plies. A second point q has a different address, since there is some n for which p and q lie in distinct intervals Ca and C f3 of C n .

In sum, the Cantor set is in one-to-one correspondence with the collection of addresses. Each address w defines a point p(w) E C and each point p E C has a unique address w (p).


If each 2 is replaced by 1, an address w becomes a base-two expansion of somex E [0, 1]. There areuncountablymanyx E [0, 1],uncountablymany base-two expansions, and uncountably many addresses w, all of which reestablishes the uncountability of C. In the same vein, W can be interpreted directly as the base-three expansion of p. See Exercise 16 in Chapter 1. In Section 6 we will make more use of this geometric coding.

If SCM and S = M then S is dense in M. For example, Q is dense in R The set S is somewhere dense if there exists an open non-empty set U C M such that S n U :::J U. If S is not somewhere dense then it is nowhere dense.

63 Theorem The Cantor set contains no interval.

Proof Suppose not: C contains some interval (a, b). Choose n such that 1 j3n < b - a. The Cantor set is contained in a set Cn consisting of finitely many closed intervals, all shorter than b - a. Hence Cn can not contain (a, b), and neither can C. 0

64 Corollary The Cantor set is nowhere dense in ~.

Proof If C is dense in an open set U then U C C. Since C is closed, U C C. Any non-empty open set U C ~ contains an interval (a, b), but C never contains an interval. Hence U = 0 and C is nowhere dense. 0

The existence of an uncountable nowhere dense set is astonishing. Even more is true: the Cantor set is a zero set: it has "outer measure zero." By this we mean that, given any E > 0, there is a countable covering of C by open intervals (ab bk), and the total length of the covering is

(Outer measure is one of the central concepts of Lebesgue Theory. See Chapter 6.) After all, C is a subset of C n , which consists of 2n closed intervals, each oflength I j3n . If n is large enough then 2n j3n < E. Enlarging each of these closed intervals to an open interval keeps the sum of the lengths < E, and it follows that C is a zero set.

If we discard subintervals of [0, 1] in a different way, we can make a fat Cantor set - one that has positive outer measure. Instead of discarding the middle thirds of intervals at the nth stage in the construction, we discard only the middle ljn! portion. The discards are grossly smaller than the remaining intervals. See Figure 49. The total amount discarded from [0, 1]

Section 6* Cantor Set Lore 99

is < 1, and the total amount remaining, the outer measure of the fat Cantor set, is positive. See Exercise 3.32.

Figure 49 In forming a fat Cantor set, the gap intervals occupy a progressively smaller proportion of the Cantor set intervals.

6* Cantor Set Lore

In this section, we explore some arcane features of Cantor sets. Although the continuous image of a connected set is connected, the con

tinuous image of a disconnected set may well be connected. Just crush the disconnected set to a single point. Nevertheless, I hope you find the following result striking, for it means that the Cantor set C is the universal compact metric space, of which all others are merely shadows.

65 Cantor Surjection Theorem Given a compact non-empty metric space M, there is a continuous surjection of C onto M.

See Figure 50. Exercise 107 suggests a direct construction of a continuous surjection C -+ [0, 1], which is already an interesting fact. The proof of Theorem 65 involves a careful use of the address notation from Section 5 and the following simple lemma about dividing a metric space M into small pieces. A piece of M is any compact, non-empty subset of M.

c ----

Figure 50 (J' surjects C onto M.


66 Piece Lemma A compact metric space M is the union of dyadically many small pieces. Specifically, given E > 0, there exist 2k pieces of M, each with diameter::::: E, whose union is M.

Proof (Recall that "dyadic" refers to powers of 2.) Cover M with neighborhoods of radius E /2. By compactness, finitely many of them suffice, say UI , ... , Urn cover M. Then M is the union of the m pieces

each of diameter ::::: E. Choose n with m ::::: 2n and set Ui = Urn for m ::::: i ::::: 2n. (That is, repeat the last piece in the list 2n - m times.) Then M is the union of the pieces UI , .•• , U2n, and each has diameter::::: E. D

Let us denote the length of an address string a = al ... an by

lal =n.

There are exactly 2n such a. We refer also to a as a word using the letters 0,2. A dyadic filtration of M is a collection ?It = {Ma} of pieces of M such that

(a) a varies freely in the collection of all finite words formed with the letters 0, 2.

(b) For each n EN, M = U1al=nMa. (c) If a is expressed as a compound word a = f3 8 then Ma C MfJ. (d) max{diamMa: lal =n} ~ Oasn ~ 00.

We call ?It a filtration because its nth level?ltn = {Ma : lal = n} consists of many small sets. Think of the sets Ma as the holes in a sieve that filters a liquid. (The analogy is imperfect since the Ma may overlap.) As n ~ 00,

we filter M more and more finely.

67 Dyadic Filtration Lemma Every non-empty compact metric space has a dyadic filtration.

Proof By Lemma 66 there is an integer nl and 2n1 pieces, each with diameter::::: I, whose union is M. Since there are 2n

1 dyadic words of length n I,

we use them to label these pieces as Ma. Then

If Y is a word of length k < n I, define My to be the union of all the pieces Ma such that Y = alk. (Recall that alk is the truncation of a to its first


k letters.) This defines the filtration ?ltn for 1 ::: n ::: n] and makes the nesting condition (c) automatic.

In the same way, each Ma can be expressed as a union of 2n2 sub-pieces

Ma = U (Ma)f3, 1f3I=n2

each of diameter::: 1/2. Define Maf3 = (Ma)f3 when I.BI = n2. If Y is a word of length k < n2 define MaY to be the union of all the Ma fJ with Y = .Blk. See Figure 51.

M

Figure 51 The piece Ma divided into sub-pieces Ma fJ .

This defines pieces corresponding to all words 8 of length::: n1 + n2,

and

M = U (U (Ma)fJ) = U Mo. lal=1I1 IfJl=n2 lol=nl + n2

For each word 8 of length n 1 + n2 is expressed uniquely as a compound word

8=a.B where lal = nl and I.BI = n2. This extends the previous filtration to ?ltn ,

I ::: n ::: nl + n2, and again (c) is automatic. Nothing stops us from passing to pieces of ever smaller diameter, which produces the desired dyadic filtration ?It = U?ltn • 0

Consider the intervals Ca that define the standard Cantor set C. Their intersections with C, Ca n C, give a dyadic filtration of C . An infinite address W = WI W2 ... specifies the precise location of a point p E C as

pew) = n Cvln.

nEN


As we saw in Section 5, points of C are in one-to-one correspondence with these infinite words w. For each p E C there is an w with p = pew), and for each w, the point pew) belongs to C.

Proof of Theorem 65 We are given a compact, non-empty metric space M, and we must find a continuous smjection a : C -+ M, C being the standard Cantor set. Let {Ma} be a dyadic filtration of M and define

q(w) = nMw,n nEN

where w is an infinite dyadic address. Then define a : C -+ M by

a(p) = q(w)

where p = pew). Every point of M has at least one address. (It has several addresses if

the pieces Ma overlap.) Hence a is a surjection, and it remains to check continuity.

Let E > ° be given and choose n so that

max{diam Ma : lal = n} < E.

Then choose 8 so small that the intervals Ca in Cn lie farther apart than 8; i.e. 8 < 1j3n . If p, p' E C and Ip - p'l < 8 then p, p' belong to a common interval Ca with lal = n, so their infinite addresses w(p), w(p') both begin with the same string of length n, namely a. Therefore a(p), a(p') both belong to the same piece Ma , and

d(a(p), a(p'» < E

since Ma has diameter < E. This implies that a is continuous. 0

Peano Curves

68 Theorem There exists a Peano curve, a continuous path in the plane which is space filling in the sense that its image has non-empty interior. In fact, there is a Peano curve whose image is the closed unit disc B2.

Proof Let a : C -+ B2 be a continuous surjection supplied by Theorem 65. Extend a to a map r : [0, 1] -+ B2 by setting

{

a (x)

rex) = (1 - t)a(a) + ta(b)

if x E C

if x = (l - t)a + tb E (a, b)

and (a, b) is a gap interval.


A gap interval is an interval (a, b) c CC such that a, b E C. Because a

is continuous, la(a) - a(b)1 -+ ° as la - bl -+ 0. Hence r is continuous. Its image includes the disc B2 and thus has non-empty interior. In fact the image of r is exactly B2, since the disc is convex and r just extends a via linear interpolation. See Figure 52. D

\ cr

gap interval

-- - .... 0 b .. - --

~~~: ~:) .......•. .. ' .. ,

c

cr

Figure 52 Filling in the Cantor surjection a to make a Peano space filling curve r.

This Peano curve can not be one-to-one since C is not homeomorphic to B2. (C is disconnected while B2 is connected.) In fact no Peano curve r can be one-to-one since the removal of a point from [0, 1] usually disconnects it, while this not true for open subsets of JR2 .

Cantor Spaces

We say that M is a Cantor space if, like the standard Cantor set C, it is compact, non-empty, perfect, and totally disconnected.

69 Moore-Kline Theorem Every Cantor space is homeomorphic to the standard middle-thirds Cantor set C.

A Cantor piece is a non-empty clopen subset S of a Cantor space M. It is easy to see that S is also a Cantor space. Since a Cantor space is totally


disconnected, each point has a small elopen neighborhood N. Thus, a Cantor space can always be divided into two disjoint Cantor pieces, M = NuNc.

70 Dyadic Partition Lemma A Cantor space is the disjoint dyadic union of small Cantor pieces.

Proof A partition of a set is a division of it into disjoint subsets. In this case the small Cantor pieces form a partition of the Cantor space M.

Let E > 0 be given. Since M is totally disconnected and compact, we can cover it with finitely many elopen neighborhoods VI, ... , Vm having diameter:::: E. To make the sets Vi disjoint, define

VI = VI

V2 = VI \ V2

Vm = Vm \ (VI U ... U Vm-I).

If any of these Vi is empty, discard it. The Vi are elopen and therefore are Cantor pieces.

Choose n with m < 2n. Divide Vm into two disjoint Cantor pieces, WI, XI, divide XI into two disjoint Cantor pieces W2, X2, etc. Eventually this gives 2n disjoint Cantor pieces

whose union is M and whose diameters are :::: E. o 71 Dyadic Cantor Filtration Lemma A Cantor space M has a dyadic filtration {Ma} by Cantor pieces. For each fixed n E N, the pieces Ma with 10: I = n are disjoint.

Proof Using Lemma 70 in place of Lemma 66, the proof of Lemma 67 gives a partition

M= U M a ,

lal=n

where the symbol U indicates disjoint union. o

Proof of the Moore-Kline Theorem 69 If we use the filtration supplied by Lemma 71, the continuous surjection (j : C --+ M defined in the proof of Theorem 65 is one-to-one. For disjointness of the Cantor pieces Ma with lal = n implies that distinct points of M have distinct addresses. A continuous one-to-one surjection from one compact metric space to another is a homeomorphism. 0


72 Corollary The fat Cantor set is homeomorphic to the standard Cantor set.

Proof Immediate from the Moore-Kline Theorem. D

73 Corollary A Cantor set is homeomorphic to its own Cartesian square, C:::::::C x C.

Proof It is enough to check that C x C is a Cantor space. It is. See Exercise 109. D

The fact that a non-trivial space is homeomorphic to its own Cartesian square is disturbing, is it not?

Ambient Topological Equivalence

Although all Cantor spaces are homeomorphic to each other when considered as abstract metric spaces, they can present themselves in very different ways as subsets of Euclidean space. Two sets A, B in lRm are ambiently homeomorphic if there is a homeomorphism of lRm to itself that sends A onto B. For example, the sets

A={O}U[1,2]U{3} and B={O}U{1}U[2,3]

are homeomorphic when considered as metric spaces, but there is no ambient homeomorphism of lR that carries A to B. Similarly, the trefoil knot is homeomorphic but not ambiently homeomorphic to a planar circle. See also Exercises 125, 126.

74 Theorem Any two Cantor sets in lR are ambiently homeomorphic.

Let M be a Cantor space contained in R According to Theorem 69, M is homeomorphic to the standard Cantor set C. We want to find a homeomorphism of lR to itself that carries C to M.

The convex hull of S c lRm is the smallest convex set H that contains S. When m = 1, H is the smallest interval that contains S.

75 Lemma A Cantor space M c lR can be divided into two Cantor pieces whose convex hulls are disjoint.

Proof Obvious from one-dimensionality of lR: choose a point x E lR \ M such that some points of M lie to the left of x and others lie to its right. Then

M Mn(-oo,x) U (x,oo)nM


divides M into disjoint Cantor pieces whose convex hulls are disjoint closed intervals. 0

Proof of Theorem 74 Let M C JR be a Cantor space. We will find a homeomorphism r : JR -+ JR sending C to M. Lemma 75 leads to a dyadic filtration {Ma} by Cantor pieces Ma whose convex hulls are disjoint when lal = n. With respect to the left/right order of JR, label the sets Ma in the same way that the Cantor middle third intervals are labeled: Mo and M2 are the left and right pieces of M, Moo, M02 are the left and right pieces of Mo, and so on. Then a : C -+ M automatically is monotone. Extend a across the gap intervals affinely as was done in the proof of Theorem 68, and extend it to JR \ [0, 1] in any affine increasing fashion such that r(O) = a(O) and r(l) = a(1). Then r : JR -+ JR extends a to R Monotonicity of a implies that r is one-to-one, while continuity of a implies that r is continuous. r : JR --+ JR is a homeomorphism that carries C onto M. 0

As an example, one may construct a Cantor set in JR by removing from [0, 1] its middle third, then removing from each of the remaining intervals two symmetrically placed subintervals; then removing from each of the remaining six intervals, four symmetrically placed subintervals; and so forth. In the limit we get a nonstandard Cantor set M. According to Theorem 74, there is a homeomorphism of JR to itself sending the standard Cantor set C toM.

Another example is the fat Cantor set mentioned on page 98.

Theorem Every two Cantor spaces in JR2 are ambiently homeomorphic.

The key step is to show that M has a dyadic disc partition. That is, M can be divided into a dyadic number of Cantor pieces, each piece contained in the interior of a small topological disc Di, the Di being mutually disjoint. (A topological disc is any homeomorph of the closed unit disc B2. Smallness refers to diam D i .) The proofs I know of the existence of such dyadic partitions are tricky cut-and-paste arguments and are beyond the scope of this book. See Moise's book, Geometric Topology in Dimensions 2 and 3. See also Exercise 111.

Antoine's Necklace

A Cantor space M C JRm is tame if there is an ambient homeomorphism h : JRm -+ JRm that carries the standard Cantor set C (imagined to lie on the xl-axis in JRm) onto M. If M is not tame it is wild. Cantor spaces contained in the line or plane are tame. In 3-space, however, there are wild ones,


Cantor sets A so badly embedded in ]R3 that they act like curves. It is the lack of a "ball dyadic partition lemma" that causes the problem.

The first wild Cantor set was discovered by L. Antoine, and is known as Antoine's Necklace. The construction involves the solid torus or anchor ring, which is homeomorphic to the Cartesian product B2 x 51. It is easy to imagine a necklace of solid tori: take an ordinary steel chain and modify it so its first and last links are also linked. See Figure 53.

Figure 53 A necklace of sixteen solid tori.

Antoine's construction then goes like this. Draw a solid torus A o. Interior to A 0 , draw a necklace A 1 of several small solid tori, and make the necklace encircle the hole of A o. Repeat the construction on each solid torus T comprising A I . That is, interior to each T, draw a necklace of very small solid tori so that it encircles the hole of T. The result is a set A 2 C A I which is a necklace of necklaces. In Figure 53, A2 would consist of 256 solid tori. Continue indefinitely, producing a nested decreasing sequence A 0 => A I => A 2 => .... The set An is compact and consists of a large number (l6n ) of extremely small solid tori arranged in a hierarchy of necklaces. It is an nth order necklace. The intersection A = nAn is a Cantor space, since it is compact, perfect, non-empty, and totally disconnected. See Exercise 110.

Certainly A is bizarre, but is it wild? Is there no ambient homeomorphism h of lR 3 that sends the standard Cantor set C onto A? The reason that h can not exist is explained below.


Referring to Figure 54, the loop K passing through the hole of A 0 can not be continuously shrunk to a point in ]R3 without hitting A. For if such a motion of K avoids A then, by compactness, it also avoids one of the high order necklaces An. In]R3 it is impossible to continuously de-link two linked loops, and it is also impossible to continuously de-link a loop from a necklace of loops. (These facts are intuitively believable but hard to prove. See Dale Rolfsen's book, Knots and Links.)

Figure 54 K loops through A 0, which contains the necklace of solid tori.

On the other hand, each loop)." in ]R3 \ C can be continuously shrunk to a point without hitting C. For there is no obstruction to pushing)." through the gap intervals of C.

Now suppose that there is an ambient homeomorphism h of]R3 that sends C to A. Then)." = h-1 (K) is a loop in]R3 \ C, and it can be shrunk to a point in ]R 3 \ C, avoiding C. Applying h to this motion of)." continuously shrinks K to a point, avoiding A, which we have indicated is impossible. Hence h can not exist, and A is wild.

7* Completion

Many metric spaces are complete (for example, all closed subsets of Euclidean space are complete), and completeness is a reasonable property to require of a metric space, especially in light of the following theorem.

76 Completion Theorem Every metric space can be completed.

Section 7* Completion 109

This means that just as ffi. completes Q, we can take any metric space M and find a complete metric space if containing M whose metric extends the metric of M. To put it another way, M is always a metric subspace of a complete metric space. In a natural sense the completion is uniquely determined by M.

77 Lemma Given four points p, q, x, y EM, we have

Id(p, q) - d(x, y)1 :s d(p, x) + d(q, y).

Proof The triangle inequality implies that

and hence

d(x, y) :s d(x, p) + d(p, q) + d(q, y)

d(p, q) :s d(p, x) + d(x, y) + dey, q),

-(d(p, x) + d(q, y)) :s d(p, q) - d(x, y) :s (d(p, x) + d(q, y)).

A number sandwiched between -k and k has magnitude :s k, which completes the proof. D

Proof of the Completion Theorem 76 We consider the collection e of all Cauchy sequences in M, convergent or not, and convert it into the completion of M. (This is a bold idea, is it not?) Cauchy sequences (Pn) and (qn), are co-Cauchy if d(Pn, qn) -+ 0 as n -+ 00. Co-Cauchyness is an equivalenc~ relation on e. (This is easy to check.)

Define,!:! to be e modulo the equivalence relation of being co-Cauchy. Points of M are equivalence cla~es P = [(Pn)] such that (Pn) is a Cauchy sequence in M. The metric on M is

where P = [(Pn)], Q = [(qn)]. It on!}' remains to verify three things: (a) D is a ~ell defined metric on M.

(b) Af.. c M. (c) M is complete.

None of these assertions is really hard to prove, although the details are somewhat messy because of possible equivalence class/representative ambiguity.

(a) By Lemma 77


Thus (d(Pn, qn» is a Cauchy sequence in~, and because ~ is complete,

exists. Let (p~) and (q~) be sequences that are co-Cauchy with (Pn) and (qn), and let

L' = lim d(p~, q~). n-+oo

Then

As n -+ 00, the first and third terms tend to O. By Lemma 77, the middle term is

Id(Pn, qn) - d(p~, q~)1 :s d(Pn, p~) + d(qn, q~),

which also tends to 0 as n -+ 00. Hence L = L' and D is well defined ~

on M. The d distance on M is symmetric and satisfies the triangle inequal-ity. Taking limits, these properties carry over to D on M, while positive definiteness follows directly from the co-Cauchy definition.

(b) Think of each P E M as a constant sequence, p = (p, P, p, p, ... ). Clearly it is Cauchy and clearly the D-distance between two constant sequences p and q is the same as the d-distance between the points P and q. In this way M is naturally a metric subspace of M.

(c) Let (PkhEN be a Cauchy sequence in M. We must find Q E M to which Pk converges as k -+ 00. (Note that (Pk ) is a sequence of equivalence classes, not a sequence of points in M, and convergence refers to D not d.) Because D is well defined we can use a trick to shorten the proof. We claim that there exists an element (Pk,n) E Pk such that for all m, n EN,

(1) 1

d (Pk,m, Pk,n) < k'

All the terms of this representative are closely bunched, not merely those in the tail. Choose and fix some sequence (P:'n)nEN in Pk. Since it is Cauchy, there is some N = N(k) such that for all m, n :::: N,

* * 1 d(Pk m' Pk n) < -. , , k

The sequence Pk,n = P:'n+N is co-Cauchy with (P:'n) and it satisfies (1). For each k, choose a representative (Pk,n) E Pk that satisfies (1), and

define qn = Pn n' We claim that (qn) is a Cauchy sequence in M and that , ~

Pk converges to its equivalence class Q as k -+ 00. That is, M is complete.


Let E > 0 be given. There exists N ::: 3/ E such that if k,.e ::: N

If k,.e ::: N then then by (I),

d(qk, qe) = d(Pk,k, Pe,e)

~ d(pk,k, Pk,n) + d(Pk,n, Pe,n) + d(Pe,n, Pe,e)

I I ~ k + d(Pk,n, Pe,n) + "C

2E ~ 3 + d(Pk,n, Pe,n)'

The inequality is valid for all n and the left-hand side, d(qk, qe), does not depend on n. The limit of d(Pk,n, Pe,n) as n ---+ 00 is D(Pk, Pe), which we know to be < E/3. Thus, if k,.e ::: N then d(qk, qe) < E and (qn) is Cauchy. Similarly we see that Pk ---+ Q as k ---+ 00. For, given E > 0, we choose N::: 2/E such that ifk, n ::: Nthend(qk,qn) < E/2,fromwhichitfollows that

d(Pk,n, qn) ~ d(Pk,n, Pk,k) + d(Pk,k, qn)

= d(Pk,n, Pk,k) + d(qk, qn)

I E < - + - < E. - k 2

The limit of the left-hand side of this inequality, as n ---+ 00, is D(Pk, Q). Thus

lim Pk = Q k-HXJ

~

and M is complete. D

Uniqueness of the completion is not surprising, and is left as Exercise 101. A different proof of the Completion Theorem is sketched in Exercise 4.37.

A Second Construction of lR from Q

In the particular case that the metric space M is Q, the Completion Theorem leads to a construction of lR from Q via Cauchy sequences. Note, however, that applying the theorem as it stands involves circular reasoning, for its proof uses completeness of lR to define the metric D. Instead, we use only the Cauchy sequence strategy.


Convergence and Cauchyness for sequences of rational numbers are concepts that make perfect sense without a priori knowledge ofR Just take all E'S and 8's in the definitions to be rational. The Cauchy completion ij of Q is the collection e of Cauchy sequences in Q modulo the equivalence relation of being co-Cauchy.

We claim that ij is a complete ordered field. The arithmetic on ij is defined by

P + Q = [(Pn +qn)]

PQ = [(Pnqn)]

P - Q = [(Pn - qn)]

P/Q = [(Pn/qn)]

where P = [(Pn)] and Q = [(qn)]. Of course Q i= 0 = [(0,0, ... )] in the fraction P / Q. Exercise 102 asks you to check that these natural definitions make ij a field. Although there are many things to check -well definedness, commutativity, and so forth - all are effortless. There are no sixteen case pro~s as with cuts. Also, just as with metric spaces, Q is naturally a subfield of Q when we think of r E Q as the constant sequence r=[(r,r, ... )].

That's the easy part; now the res.!: To define the order relation on Q we re-work some of the cut ideas. If

P E ij has a representative [(Pn)], such that for some E > 0, we have Pn ::: E for all n then P is positive. If - P is positive, P is negative.

Then we define P -< Q if Q - f is positive. Exercise 103 asks you to check that this defines an order on Q, consistent with the standard order < on Q in the sense that for all P, q E Q, P < q {::::::} P -<]". In particular, you are asked to prove the trichotomy property: each P E Q is either positive, negative, or zero, and these possibilities are mutually exclusive.

Combining Cauchyness with the definition of -< gives

P = [(Pn)] -< Q = [(qn)] {::::::} there exist E > ° and N E N

(2) such that for all m, n ::: N,

Pm + E < qn·

It remains to check the least upper bound property. Let P be a non-empty subset of ij that is bounded above. We must find a least upper bound for P. We assert that the least upper bound for P is the equivalence class Q of the following Cauchy sequence (qQ, ql, q2, ... ).

(a) qQ is the smallest integer such that qQ is an upper bound for P. (b) ql is the smallest fraction with denominator 2 such that ql is an upper

bound forP.


(c) q2 is the smallest fraction with denominator 4 such that q2 is an upper bound forP.

(d) ... (e) qn is the smallest fraction with denominator 2n such that qn is an

upper bound for P. Since P i= 0 we can choose p* = [(P:)] E P. For some N* and all m,n::: N*,

Ip! - p:1 < 1.

Hence PN* - 1 ~ P*. Integers smaller than PN* - 1 do not give upper bounds for P, so qo is well defined. The other qn are also well defined and form a monotone decreasing sequence

in Q. By construction Iqn - qn-ll ::: 1/2n. Thus, if m ::: n then

0::: qm - qn = qm - qm+l + qm+l - qm+2 + ... + qn-l - qn 1 1 1

< --+ ... +- <-. - 2m+1 2n 2m

~

It follows that (qn) is Cauchy and Q = [(qn)] E Ql. Suppose that Q is not an upper bound for P. Then there is some P =

[(Pn)] E P with Q -< P. By (2), there is an E > 0 and an N such that for alln:::N,

qN + E < Pn·

It follows that q N -< P, a contradiction to q N being an upper bound for P. On the other hand suppose that there is a smaller upper bound for p,

R = (rn) -< Q. By (2) there are E > 0 and N such that for all m, n ::: N,

Fix a k ::: N with 1/2k < E. Then for all m ::: N,

1 r m < qk - E < qk - 2k ·

By (2), R -< qk - 1 12k. Since R is an upper bound for p, so is qk - 1 12k, a contradiction to qk being the smallest fraction with denominator 2k such that qk is an upper bound for P.

This completes the verification that the Cauchy completion of Q is a complete ordered field. Uniqueness implies that it is isomorphic to the


complete ordered field lR constructed by means of Dedekind cuts in Section 2 of Chapter 1. Decide for yourself which of the two constructions of the real number system you like better - cuts or Cauchy sequences. Cuts make least upper bounds straightforward and algebra awkward, while with Cauchy sequences, it is the reverse.

Exercises 115

Exercises

1. (0, 1) is an open subset of]R but not of]R2, when we think of]R as the x-axis in ]R2. Prove this.

2. For which intervals [a, b] in ]R is the intersection [a, b] n Q a clopen subset of the metric space Q?

3. Prove directly from the definition of closed set that each single point is a closed subset of a metric space. Why does this imply that a finite set of points is also a closed set?

4. Prove that S clusters at p if and only if for each r > ° there is a point q E Mr(P) n S, such that q =1= p.

5. Prove that a set U C M is open if and only if none of its points are limits of its complement.

6. If S, T eM, a metric space, and SeT, prove that (a) SeT. (b) int(S) c int(T).

7. Construct a set with exactly three cluster points. 8. If A c B C C, A is dense in B, and B is dense in C prove that A is

dense in C. 9. Is the set of dyadic rationals (the denominators are powers of2) dense

in Q? In ]R? Does one answer imply the other? (Recall that A is dense in B if A C Band A ~ B.)

10. (a) Find a metric space in which the boundary of Mr P is not equal to the sphere ofradius rat p, {x EM: d(x, p) = r}.

(b) Need the boundary be contained in the sphere? 11. Let T be the collection of open subsets of a metric space M, and /C

the collection of closed subsets. Show that there is a bijection from T onto /C.

12. Let M be a metric space with the discrete metric, or more generally a homeomorph of M.

(a) Prove that every subset of Mis clopen. (b) Prove that every function defined on M is continuous. (c) Which sequences converge in M?

13. Show that every subset of N is clopen. What does this tell you about every function f : N -+ M, where M is a metric space?

14. The distance from a point p in a metric space M to a non-empty subset SCM is defined to be dist(p, S) = inf{d(p, s) : s E S}.

(a) Show that p is a limit of S if and only if dist(p, S) = 0. (b) Show that p t---+ dist(p, S) is a uniformly continuous function

of p E M.


15. What is the set of points in JR.3 at distance exactly 1/2 from the unit circle C in the plane,

T = {p E JR.3 : 3q E C and d(p, q) = 1/2

and for all q' E C, d(p, q) ::s d(p, q')}?

16. ShowthatS C MissomewheredenseinMifandonlyifint(S) i= 0. That is, S is nowhere dense in M if and only if its closure has empty interior.

17. Assume that f : M -+ N is a function from one metric space to another which satisfies the following condition: if a sequence (Pn) in M converges then the sequence (f (Pn)) in N converges. Prove that f is continuous.

18. The simplest type of mapping from one metric space to another is an isometry. It is a bijection f : M -+ N that preserves distance in the sense that for all p, q EM,

If there exists an isometry from M to N then M and N are said to be isometric, M == N. You might have two copies of a unit equilateral triangle, one centered at the origin and one centered elsewhere. They are isometric. Isometric metric spaces are indistinguishable as metric spaces.

(a) Prove that every isometry is continuous. (b) Prove that every isometry is a homeomorphism. (c) Prove that [0, 1] is not isometric to [0,2].

19. Prove that isometry is an equivalence relation: if M is isometric to N, show that N is isometric to M; show that any M is isometric to itself (what mapping of M to M is an isometry?); if M is isometric to N and N is isometric to P, show that M is isometric to P.

20. Is the perimeter of a square isometric to the circle? Homeomorphic? Explain.

21. Which capital letters of the roman alphabet are homeomorphic? Are any isometric? Explain.

22. Is lR homeomorphic to Q? Explain. 23. Is Q homeomorphic to N? Explain. 24. An ant walks on the floor, ceiling, and walls of a cubical room. What

metric is natural for the ant's view of its world? What metric would a spider consider natural? If the ant wants to walk from a point p to a point q, how could it determine the shortest path?

Exercises 117

25. Assume that N is an open metric subspace of M and that U eN. ( a) Prove that U is open in N if and only if it is open in M. (b) Conversely, prove that if openness of SeN is equivalent to

openness in M then N is open in M. (c) Do the same for closedness. (d) Deduce that a clopen metric subspace N is the only example in

which the concepts of openness and closedness in the subspace agree exactly with the concepts in the big space.

26. Consider a sequence (xn) in the metric space R (a) If (xn) converges in JR prove that the sequence of absolute values

(Ixn I) converges in R (b) Prove or disprove the converse.

27. Let (An) be a nested descreasing sequence of non-empty closed sets in the metric space M.

(a) If M is complete and diam An ---+ 0 as n ---+ 00, show that nAn is exactly one point.

(b) To what assertions do the sets [n, (0) provide counterexamples? 28. Prove that there is an embedding of the line as a closed subset of

the plane, and there is an embedding of the line as a bounded subset of the plane, but there is no embedding of the line as a closed and bounded subset of the plane.

29. (a) Prove that every convergent sequence is bounded. That is, if (Pn) converges in the metric space M, prove that there is some neighborhood Mrq containing the set {Pn : n EN}.

(b) Is the same true for a Cauchy sequence in an incomplete metric space?

30. A sequence (xn) in JR increases if n < m implies Xn ~ X m . It strictly increases if n < m implies Xn < X m . It decreases or strictly decreases if n < m always implies Xn :::: Xm or always implies Xn > Xm .

A sequence is monotone if it increases or it decreases. (a) Prove that every sequence in JR which is monotone and bounded

converges in JR. (b) Prove that this monotone sequence condition is equivalent to

the least upper bound property. 31. Let (xn) be a sequence in R

*(a) Prove that (xn) has a monotone subsequence. (b) How can you deduce that any bounded sequence in JR has a

convergent subsequence?


32. Let (Pn) be a sequence and f : N -+ N a bijection. The sequence (qkhEN with qk = P f(k) is a rearrangement of (Pn).

(a) Are limits of a sequence unaffected by rearrangement? (b) What if f is an injection? (c) A surjection?

33. If f : A -+ Band g : C -+ B such that ACe and for each a E A, f(a) = g(a) then f extends to g. Assume that f : S -+ IR is a uniformly continuous function defined on a subset S of a metric space M.

(a) Prove that f extends to a uniformly continuous function f : S-+R

(b) Prove that f is the unique continuous function defined on S such that f(x) = f(x) for all XES.

(c) Prove the same things when N is a complete metric space and f: S -+ N.

34. A map f : M -+ N is open if for each open set U eM, the image set feU) is open in N.

(a) If f is open, is it continuous? (b) If f is a homeomorphism, is it open? (c) If f is an open, continuous bijection, is it a homeomorphism? (d) If f : IR -+ IR is a continuous surjection, must it be open? (e) If f : IR -+ IR is a continuous, open surjection, must it be a

homeomorphism? (f) What happens in (e) if IR is replaced by the unit circle S?

35. Fold a piece of paper in half. (a) Is this a continuous transformation of one rectangle into an

other? (b) Is it injective? (c) Draw an open set in the target rectangle, and find its pre-image

in the original rectangle. Is it open? (d) What if the open set meets the crease?

The baker's transformation is a similar mapping. A rectangle of dough is stretched to twice its length and then folded back on itself. Is the transformation continuous? A formula for the baker's transformation in one variable is f(x) = 1 - 11 - 2xl. The nth iterate of f is rn = f 0 f 0 . ··0 f, n times. The orbit of a point x is

{x, f(x), f 0 f(x), ... , rn(x), ... }.

( e) If x is rational prove that the orbit of x is a finite set. (f) If x is irrational what is the orbit?

Exercises 119

36. Rotate the unit circle C by a fixed angle a, say R : C --+ C. (In polar coordinates, the transformation R sends (1, e) to (1, e + a).)

(a) If afT[ is rational, show that each orbit of R is a finite set. *(b) If afT[ is irrational, show that each orbit is infinite and has

closure equal to C. 37. Consider the identity map id : Cmax --+ ent where Cmax is the met

ric space C ([a, b], JR) of continuous real valued functions defined on [a, b], equipped with the max metric dmax(f, g) = max If(x) - g(x)l, and Cint is C([a, b], JR) equipped with the integral metric,

dint(f, g) = lb If(x) - g(x)1 dx.

Show that id is a continuous linear bijection (an isomorphism) but its inverse is not continuous.

38. Let II II be any norm on JRm and let B = {x E JRm : IIxll :::: I}. Prove that B is compact. [Hint: It suffices to show that B is closed and bounded with respect to the Euclidean metric.]

39. Assume that the Cartesian product of two non-empty sets A C M

and BeN is compact in M x N. Prove that A and B are compact. 40. Consider a function f : M --+ JR. Its graph is the set

{(p, Y) E M x JR : Y = f p}.

(a) Prove that if f is continuous then its graph is closed (as a subset of M x JR).

(b) Prove that if f is continuous and M is compact then its graph is compact.

(c) Prove that if the graph of f is compact then f is continuous. (d) What if the graph is merely closed? Give an example of a dis

continuous function f : JR --+ JR whose graph is closed. 41. Prove that the 2-sphere is not homeomorphic to the plane. 42. Draw a Cantor set C on the circle and consider the set A of all chords

between points of C. (a) Prove that A is compact.

***(b) Is A convex? 43. Suppose that (Kn) is a nested sequence of compact non-empty sets,

K1 ~ K2 ~ ... ,andK = nKn.lfforsomeJL > O,eachdiamKn :::

JL, is it true that diam K ::: JL? 44. Suppose that M is compact and that U is an open covering of M

which is "redundant" in the sense that each p E M is contained in at


least two members of U. Show that U reduces to a finite subcovering with the same property.

45. Suppose that every open covering of M has a positive Lebesgue number. Give an example of such an M that is not compact. Exercises 46- 53 treat the basic theorems in the chapter, avoiding the use of sequences. The proofs will remain valid in general topological spaces.

46. Give a direct proof that [a, b] is covering compact. [Hint: Let U be an open covering of [a, b] and consider the set

C = {x E [a, b]: finitely many members ofU cover [a, xl}.

Use the least upper bound principle to show that b E C.] 47. Give a direct proof that a closed subset A of a covering compact set

K is covering compact. [Hint: If U is an open covering of A, adjoin the set W = M \ A to U. Is W = U U {W} an open covering of K? If so, so what?]

48. Give a proof of Theorem 39 using open coverings. That is, assume that A is a covering compact subset of M and f : M --+ N is continuous. Prove directly that f A is covering compact. [Hint: What is the criterion for continuity in terms of pre-images?]

49. Suppose that f : M --+ N is a continuous bijection and M is covering compact. Prove directly that f is a homeomorphism.

50. Suppose that M is covering compact and that f : M --+ N is continuous. Use the Lebesgue number lemma to prove that f is uniformly continuous. [Hint: Consider the covering of N by E j2-neighborhoods {N€/2(q) : q EN} anditspre-imageinM, {fpre(N€/2(q» : q EN}.]

51. Give a direct proof that the nested decreasing intersection of nonempty covering compact sets is non-empty. [Hint: If Al ::J A2 ::J ...

are covering compact, consider the open sets Un = A~. If nAn = 0, what does {Un} cover?]

52. Generalize Exercise 51 as follows. Suppose that M is covering compact and e is a collection of closed subsets of M such that every intersection of finitely many members of e is non-empty. (Such a collection e is said to have the finite intersection property.) Prove that the grand intersection n AC is non-empty. [Hint: Consider

CE(;-,

the collection of open sets U = fCC : C E e}.] 53. If every collection of closed subsets of M which has the finite inter

section property also has a non-empty grand intersection, prove that

Exercises 121

M is covering compact. [Hint: Given an open covering U = {Ua}, consider the collection of closed sets e = {U a C}.]

54. If S is connected, is the interior of S connected? Prove this or give a counter-example.

55. Theorem 51 states that the closure of a connected set is connected. (a) Is the closure of a disconnected set disconnected? (b) What about the interior of a disconnected set?

*56. Prove that every countable metric space (not empty and not a singleton) is disconnected. [Astonishingly, there exists a countable topological space which is connected. Its topology does not arise from a metric.]

57. (a) Prove that a continuous function f : M --* lR, all of whose values are integers, is constant provided that M is connected.

(b) What if all the values are irrational? 58. Prove that the (double) cone {(x, y, z) E ffi.3 : x 2 + y2 = Z2} is path

connected. 59. Prove that the annulus A = {z E ffi.2 : r :s Izl :s R} is connected. 60. What function (given by a formula) is a homeomorphism from ( -1, 1)

to ffi.? Is every open interval homeomorphic to (0, I)? Why or why not?

61. A subset E of ffi.m is starlike if it contains a point Po (called a center for E) such that for each q E E, the segment between Po and q lies in E.

(a) If E is convex it is starlike. (b) Why is the converse false? (c) Is every starlike set connected? (d) Is every connected set starlike? Why or why not?

*62. Suppose that E C ffi.m is open, bounded, and starlike, and Po is a center for E.

(a) Is it true or false that all points PI in a small enough neighbor-hood of Po are also centers for E?

(b) Is the set of centers convex? (c) Is it closed? (d) Can it consist of a single point?

63. Suppose that A, B C ffi.2 are convex, closed, and have non-empty interiors.

(a) Prove that A, B are the closure of their interiors. (b) If A, B are compact, prove that they are homeomorphic.

[Hint: Draw a picture.]


64. (a) Prove that every connected open subset of]Rm is path connected. (b) Is the same true for open connected subsets of the circle? (c) What about connected non-open subsets of the circle?

65. List the convex subsets of]R up to homeomorphism. How many are there and how many are compact?

66. List the closed convex sets in ]R2 up to homeomorphism. There are nine. How many are compact?

*67. Generalize Exercises 63 and 66 to ]R3; to ]Rm. 68. Prove that (a, b) and [a, b) are not homeomorphic metric spaces. 69. Let M and N be non-empty metric spaces.

(a) If M and N are connected prove that M x N is connected. (b) What about the converse? (c) Answer the question again for path-connectedness.

70. Let H be the hyperbola {(x, y) E ]R2 : xy = 1 and x, y > O}, and let X be the x-axis.

(a) Is the set S = XU H connected? (b) What if we replace H with the graph G of any continuous

positive function f : ]R --+ (0,00); is X U G connected? (c) Give a counter-example if f is everywhere positive but discon

tinuous at just one point. 71. Assume that A, B are compact, disjoint, non-empty subsets of M.

Prove that there are ao E A, bo E B such that for all a E A, b E B,

d(ao, bo) ::: dCa, b).

*72. An arc is a path with no self-intersection. Define the concept of arc connectedness and prove that a metric space is path-connected if and only if it is arc-connected.

73. The Hawaiian earring is the union of circles of radius 1 j n and center x = ±ljn on the x-axis, for n EN. See Figure 22 on page 53.

(a) Is it connected? (b) Path-connected? (c) Is it homeomorphic to the one-sided Hawaiian earring?

74. (a) The intersection of connected sets need not be connected. Give an example.

(b) Suppose that Sl, S2, S3, ... is a sequence of connected, closed subsets of the plane and Sl ~ S2 ~ .... Is S = nSn connected? Give a proof or counter-example.

*(c) Does the answer change if the sets are compact? (d) What is the situation for a nested decreasing sequence of com

pact path-connected sets?

Exercises 123

75. Let S = JR2 \ Q2. (Points (x, y) E S have at least one irrational coordinate.) Is S connected? Prove or disprove.

*76. The topologist's sine curve is the set

{ (x, y) : x = 0 and I y I :::; 1, or 0 < x :::; 1 and y = sin 1/ x}.

See Figure 41. The topologist's sine circle is shown in Figure 55. (It is the union of a circular arc and the topologist's sine curve.) Prove that it is path-connected, but not locally path-connected.

Figure 55 The topologist's sine circle.

77. If a metric space M is the union of path-connected sets SOl' all of which have the path-connected set K in common, is M path-connected?

78. (PI, ... , Pn) is an f-chain in a metric space M if for each i, Pi E M

and d(Pi, PHI) < E. The metric space is chain-connected if for each E > 0 and each pair of points p, q E M there is an E -chain from P to q.

(a) Show that every connected metric space is chain-connected. (b) Show that if M is compact and chain-connected then it is con

nected. (c) Is JR \ Z chain-connected? (d) If M is complete and chain-connected, is it connected?


79. Prove that if M is nonempty compact, locally path-connected and connected then it is path-connected. (See Exercise 115, below.)

80. The graph of f : M -+ ]R is the set {(x, y) E M x ]R : y = f(x)}. Since M x ]R is a Cartesian product of two metric spaces it has a natural metric.

(a) If M is connected and f is continuous, prove that the graph of f is connected.

(b) Give an example to show that the converse is false. (c) If M is path-connected and f is continuous, show that the graph

is path-connected. (d) What about the converse?

81. Consider the distance function d : M x M -+ R (a) Prove that d is continuous with respect to the natural sum metric

defined on M x M, namely dsum«p, q), (pi, q')) = d(p, pi) + d(q, q').

(b) Is the same true for the other natural metrics dE and dmax? [You will find continuity of d useful in other exercises.]

82. Why is the sum metric on ]R2 called the Manhattan metric and the taxicab metric?

83. Let M I , M2 be metric spaces with metrics d I , d2, and let M = MI X

M 2 be their Cartesian product. Either (a) prove directly that the three natural metrics dE, dmax , and dsum

on M are metrics, or (b) prove in general that if II II is any norm on ]R2 and we define

a distance vector in]R2 as dpq = (dl (PI, ql), d2(P2, q2)) then d(p, q) = Ildpq II is a metric on M. (The points p = (PI, P2),

q = (qI, q2) belong to M.) (c) Does (b) imply (a)?

[Hint: For (a) use Cauchy-Schwarz. For (b) use Euclidean geometry in ]R2 and the triangle inequality for the norm II II to show that

where p, q, rEM.] 84. A metric space M with metric d can always be re-metrized so the

metric becomes bounded. Simply define the bounded metric

d(p,q) pep, q) = 1 + d(p, q)

(a) Prove that p is a metric. Why is it obviously bounded?

Exercises 125

(b) Prove that the identity map M ~ M is a homeomorphism from M with the d-metric to M with the p-metric.

(c) Infer that boundedness of M is not a topological property. (d) Find homeomorphic metric spaces, one bounded and the other

not. * 85. The implications of compactness are frequently equivalent to it. Prove

(a) If every continuous function I : M ~ 1Ft is bounded then M is compact.

(b) If every continuous bounded function I : M ~ 1Ft achieves a maximum or minimum, then M is compact.

(c) If every continuous function I : M ~ 1Ft has compact range 1M, then M is compact.

(d) If every nested decreasing sequence of non-empty closed sub-sets of M has non-empty intersection, then M is compact.

Together with Theorems 54 and 56, (a) - (d) give seven equivalent definitions of compactness. [Hint: Reason contrapositively. If M is not compact it contains a sequence (Pn) that has no convergent subsequence. It is fair to assume that the points Pn are distinct. Find radii rn > 0 such that the neighborhoods Mrn (Pn) are disjoint and no sequence qn E Mrn (Pn) has a convergent subsequence. Using the metric define a function In : Mrn (Pn) ~ 1Ft with a spike at Pn, such as

In(x) = rn - d(x, Pn) an + d(x, Pn)

where an > O. Set I(x) = In(x) if x E Mrn (Pn), and I(x) = 0 if x belongs to no Mrn (Pn). Show that I is continuous. With the right choice of an show that I is unbounded. With a different choice of an, it is bounded but achieves no maximum, and so on.]

86. Let M be a metric space on which the metric d is bounded by 1, d(p, q) < 1 for all P, q EM. The cone over M is the set

C C(M) {Po} U (0, 1] x M

with the polar metric

peer, p), (s, q)) = Ir - sl + min{r, s}d(p, q)

peer, p), Po) = r.

The point Po is the vertex of the cone. Prove that p is a metric on C. 87. Recall that if for each embedding of M, h : M ~ N, hM is closed

in N then M is said to be absolutely closed. If each h M is bounded


then M is absolutely bounded. Theorem 43 implies that compact sets are absolutely closed and absolutely bounded. Prove:

(a) If M is absolutely bounded, then M is compact. *(b) If M is absolutely closed, then M is compact. Thus, these are two more conditions equivalent to compactness. [Hint: From Exercise 85(a), if M is noncompact there is a continuous function f : M -+ JR. that is unbounded. For Exercise 87(a), show that F (x) = (x, f (x)) embeds M onto a nonbounded subset of M x R For 87(b), justify the additional assumption that the metric on M is bounded by 1. Then use Exercise 85(b) to show that if M is noncompact there is a continuous function g : M -+ (0, 1] such that for some nonclustering sequence (Pn), we have g(Pn) -+ ° as n -+ 00. Finally, show that G(x) = (g(x), x) embeds M onto a nonclosed subset of the cone C(M) discussed in Exercise 86.]

88. (a) Prove that every function defined on a discrete metric space is uniformly continuous.

(b) Infer that it is false to assert that if every continuous function f : M -+ JR. is uniformly continuous, then M is compact.

(c) Prove, however, that if M is a subset of a compact metric space K and every continuous function f : M -+ lR. is uniformly continuous, then M is compact.

89. Recall that P is a cluster point of S if each Mr P contains infinitely many points of S. The set of cluster points of S is denoted as S'. Prove:

(a) IfS C T then S' C T'. (b) (S U T)' = S' U T'. (c) s' = (S),. (d) S' is closed in M; that is, S" C S' where S" = (S')'. (e) Calculate N', Ql', JR.', (JR. \ Ql)', Q". (f) Let T be the set of points {I / n : n E N}. Calculate T' and T". (g) Give an example showing that S" can be a proper subset of S'.

90. Recall that P is a condensation point of S if each Mr P contains uncountably many points of S. The set of condensation points of S is denoted as S*. Prove:

(a) If SeT then S* C T*. (b) (SUT)*=S*UT*. (c) S* C S* where S* = (S)* (d) S* is closed in M; that is, S*' C S* where S*, = (S*)'. (e) S** C S* where S** = (S*)*.

Exercises 127

(t) Calculate N*, Q*, lR*, (lR \ Q)*, Q**. (g) Give an example showing that S* can be a proper subset of

(s) *. Thus, (c) is not in general an equality. **(h) Give an example that S** can be a proper subset of S*. Thus,

(e) is not in general an equality. [Hint: Consider the set M of all functions f : [a, b] -+ [0, l], continuous or not, and let the metric on M be the sup metric, d(j, g) = sup{lf(x) - g(x)1 : x E [a, b]}. Consider the set S of all "8-functions with rational values."]

**(i) Give examples that show in general that S* neither contains nor is contained in S'* where S'* = (S')*. [Hint: 8-functions with values lin, n EN.]

91. Recall that p is an interior point of SCM if some Mr p is contained in S. The set of interior points of S is the interior of S and is denoted int S. For all subsets S, T of the metric space M prove:

(a) int S = S \ as. (b) int S = (scy. (c) int(int S) = int S. (d) int(S n T) = int S n int T. (e) What are the dual equations for the closure? (t) Prove that int(S U T) ~ (int SUint T). Show by example that

the inclusion can be strict, i.e., not an equality. 92. A point p is a boundary point of a set SCM if every neighborhood

Mr p contains points of both S and Sc. The boundary of S is denoted as. For all subsets S, T of a metric space M prove:

(a) Sis clopen if and only if as = 0. (b) as = asc .

(c) aas Cas. (d) aaas = aas. (e) a(S U T) c as U aT.

(t) Give an example in which (c) is a strict inclusion, aas =1= as. (g) What about (e)?

*93. Suppose that E is an uncountable subset ofR Prove that there exists a point p E lR at which E condenses. [Hint: Use decimal expansions. Why must there be an interval [n, n + 1) containing uncountably many points of E? Why must it contain a decimal subinterval with the same property? (A decimal subinterval [a, b) has endpoints a = n +kllO, b = n + (k + 1)/10 for some digit k, 0 ::: k ::: 9.) Do you see lurking


the decimal expansion of a condensation point?] Generalize to ]R2 and to ]Rm.

94. The metric space M is separable if it contains a countable dense subset. [Note the confusion of language: "Separable" has nothing to do with "separation."]

(a) Prove that]Rm is separable. (b) Prove that every compact metric space is separable.

95. *(a) Prove that any metric subspace of a separable metric space is separable, and deduce that any metric subspace of]Rm or of a compact metric space is separable.

(b) Is the property of being separable topological? (c) Is the continuous image of a separable metric space separable?

96. Think up a non-separable metric space. 97. Let B denote the collection of all E-neighborhoods in ]Rm whose

radius E is rational and whose center has all coordinates rational. (a) Prove that B is countable. (b) Prove that every open subset of ]Rm can be expressed as the

countable union of members of B. (The union need not be disjoint, but it is at most a countable union because there are only countably many members of B. A collection such as B is called a countable base for the topology of lRm.)

98. (a) Prove that any separable metric space has a countable base for its topology, and conversely that any metric space with a countable base for its topology is separable.

(b) Infer that every compact metric space has a countable base for its topology.

*99. Referring to Exercise 90, assume now that M is separable, SCM, and, as before S' is the set of cluster points of S while S* is the set of condensation points of S. Prove:

(a) S* C (S')* = (5)*. (b) S** = S*, = S*. (c) Why is (a) not in general an equality?

[Hints: For (a) write S C (S \ S') U S' and 5 = (S \ S') US', show that (S \ S')* = 0, and use Exercise 90(a). For (b), Exercise 90(d) implies that S** C S*, c S*. To prove that S* C S**, write S C (S \ S*) U S* and show that (S \ S*)* = 0.]

*100. Prove that (a) An uncountable subset of]R clusters at some point of R (b) An uncountable subset of lR clusters at some point of itself.

Exercises 129

(c) An uncountable subset of lR condenses at uncountably many points of itself.

(d) What about lRm ? (e) What about any compact metric space? (f) What about any separable metric space?

101. Prove that the completion of a metric space is unique in the following natural sense: A completion of a metric space M is a complete metric X space containing M as a metric subspace such that M is dense in X. That is, every point of X is a limit of M. ~

(a) Prove that M is dense in the completion M constructed in the proof of Theorem 76.

(b) If X and X' are two completions of M prove that there is an isometry i : X -+ X' such that i (p) = p for all p EM.

(c) Prove that .£ is the unique such isometry. (d) Infer that M is unique.

*102. Prove that ij, the Cauchy sequences in Q modulo the equivalence relation of being co-Cauchy, is a field with respect to the natural arithmetic <perations defined on page 112, and that Q is naturally a sub field of Q.

103. Prove that the order on ij defined on page 112 is a bona fide order which agrees with the standard order on Q.

* 104. Let I: be the set of all infinite sequences of O's and 1 'so For example, (l 00 1110000 11111 ... ) E I:. Define the metric

where a = (an) and b = (bn) are points in I:. (a) Prove that I: is compact. (b) Prove that I: is homeomorphic to the Cantor set.

* 105. A metric on M is an ultrametric if for all x, y, Z EM,

d(x, Z) ::s max{d(x, y), dey, z)}.

(Intuitively this means that the trip from x to z can not be broken into shorter legs by making a stopover at some y.)

(a) Show that the ultrametic property implies the triangle inequality.

(b) In an ultrametric space show that "all triangles are isosceles." (c) Show that a metric space with an ultrametric is totally discon

nected.


(d) Define a metric on the set ~ of strings of O's and 1 's in Exercise 104 as

if n is the smallest index for which an i= bn

if a = b.

Show that d* is an ultrametric. (e) Prove that the identity map is a homeomorphism (~, d) ~

(~, d*). * 1 06. Q inherits the Euclidean metric from JR, but it also carries a very

different metric, the p-adic metric. Given a prime number p and an integer n, the p-adic norm of n is

1 Inlp= k

p

where pk is the largest power of p that divides n. (The norm of 0 is by definition 0.) The more factors of p, the smaller the p-norm. Similarly, if x = alb is a fraction, we factor x as

k r x = p .-s

where p divides neither r nor s, and we set

The p-adic metric on Q is

(a) Prove that dp is a metric with respect to which Ql is perfectevery point is a cluster point.

(b) Prove that d p is an ultrametric. (c) Let Qp be the metric space completion of Q with respect to

the metric dp, and observe that the extension of dp to Qp remains an ultrametric. Infer from Exercise 105 that Qp is totally disconnected.

(d) Prove that Qp is locally compact, in the sense that every point has small compact neighborhoods.

Exercises 131

(e) Infer that Qp is covered by neighborhoods homeomorphic to the Cantor set. See Gouvea's book, p-adic Numbers.

107. Prove directly that there is a continuous surjection of the middlethirds Cantor set C onto the closed interval [0, 1]. [Hint: If x E C has base 3 expansion (xn) then all the entries are O's and 2's. Write Y = (Yn) by replacing the 2's in (xn) by 1 's and interpreting the answer base 2. Show that the map x t-+ Y works.]

*108. Let P be a closed perfect subset of a separable complete metric space M. Prove that each point of P is a condensation point of P. In symbols, P = pi :::} P = P*.

109. Let M, N be non-empty metric spaces and P = M x N. (a) If M, N are perfect, prove that P is perfect. (b) If M, N are totally disconnected, prove that P is totally discon

nected. (c) What about the converses? (d) Infer that the Cartesian product of Cantor spaces is a Cantor

space. (We already know that the Cartesian product of compacts is compact.)

(e) Why does this imply that C x C = {(x, y) E ]R2 : x E

C and Y E C} is homeomorphic to C, C being the standard Cantor set?

*110. To prove that Antoine's Necklace A is a Cantor set, you need to show that A is compact, perfect, non-empty, and totally disconnected.

(a) Do so. [Hint: What is the diameter of any connected component of An, and what does that imply about A?]

(b) If, in the Antoine construction only two linked solid tori are placed inside each larger solid torus, show that the intersection A = nAn is not a Cantor set because it is not totally disconnected.

**111. Given a Cantor space M C ]R2, given a line segment [p, q] C ]R2

with p, q ~ M, and given an E > 0, prove that there exists a path A in the E -neighborhood of [p, q] that joins p to q and is disjoint from M. [Hint: Think of A as a bisector of M. From this bisection fact a dyadic disc partition of M can be constructed, which leads to the proof that M is tame.]

* 112. Consider the Hilbert cube

H = {(X\,X2, ... ) E [0,1]00: for each n EN, IXnl ~ Ij2n}.

Prove that H is compact with respect to the metric


d(x, y)) = sup IXn - Ynl n

where x = (xn), Y = (Yn). [Hint: sequences of sequences.] Remark Although compact, H is infinite-dimensional and is homeomorphic to no subset of ]Rm .

113. Prove that the Hilbert cube is perfect and homeomorphic to its Cartesian square, H ~ H x H.

***114. Assume that M is compact, non-empty, perfect, and homeomorphic to its Cartesian square, M ~ M x M. Must M be homeomorphic to the Cantor set, the Hilbert cube, or some combination of them?

115. A Peano space is a metric space M that is the continuous image of the unit interval: there is a continuous surjection r : [0, 1] ~ M. Theorem 68 states the amazing fact that the 2-disc is a Peano space. Prove that every Peano space is

(a) compact, (b) non-empty, (c) path-connected,

*(d) and locally path-connected, in the sense that for each p E M and each neighborhood U of p there is a smaller neighborhood V of p such that any two points of V can be joined by a path in U.

*116. The converse to Exercise 115 is the Hahn-Mazurkiewicz Theorem. Assume that a metric space M is a compact, non-empty, pathconnected, and locally path-connected. Use the Cantor Surjection Theorem 65 to show that M is a Peano space. [The key is to make uniformly short paths to fill in the gaps of [0, 1] \ C.]

117. One of the famous theorems in plane topology is the Jordan Curve Theorem. It states that if I : [a, b] ~ ]R2 is continuous, I(a) = I(b), and for no other pair of distinct s, t E [a, b] does I(s) equal l(t) then the complement of the path I in]R2 consists of two disjoint, connected open sets, its inside and its outside. Prove the Jordan Curve Theorem for the circle, the square, the triangle, and, if you have courage - any simple closed polygon.

118. Prove that there is a continuous sUIjection ]R ~ ]R2. What about ]Rn?

119. The utility problem gives three houses 1,2,3 in the plane and the three utilities, Gas, Water, and Electricity. You are supposed to connect each house to the three utilities without crossing utility lines. (The houses and utilities are disjoint.)

Exercises 133

(a) Use the Jordan curve theorem to show that there is no solution to the utility problem in the plane.

*(b) Show also that the utility problem cannot be solved on the 2-sphere S2.

*(c) Show that the utility problem can be solved on the surface of the torus.

*(d) What about the surface of the Klein bottle? ***(e) Given utilities UI , ... , Urn and houses HI, ... , Hn located on a

surface with g handles, find necessary and sufficient conditions on m, n, g so that the utility problem can be solved.

120. The open cylinder is (0, 1) X Sl. The punctured plane is]R2 \ {OJ. (a) Prove that the open cylinder is homeomorphic to the punctured

plane. (b) Prove that the open cylinder, the double cone, and the plane are

not homeomorphic. 121. Is the closed strip {(x, y) E ]R2 : 0 :::: x :::: I} homeomorphic to the

closed halfplane {(x, y) E ]R2 : x ?: O}? Prove or disprove. 122. Is the plane minus four points on the x-axis homeomorphic to the

plane minus four points in an arbitrary configuration? 123. Suppose that A, B C ]R2.

(a) If A and B are homeomorphic, are their complements homeomorphic?

*(b) What if A and B are compact? ***(c) What if A and B are compact and connected?

**124. Let M be a metric space and let /C denote the class of non-empty compact subsets of M. The r-neighborhood of A E /C is

MrA = {x EM: 3a E A andd(x,a) < r} = UMra. aEA

For A, B E /C define

D(A, B) = inf{r > 0: A C MrB and B C MrA}.

(a) Show that D is a metric on /C. (It is called the Hausdorff metric.)

(b) Denote by :F the collection of finite non-empty subsets of M and prove that:F is dense in /C. That is, given A E /C and given E > 0 show that there exists F E :F such that D(A, F) < E.

(c) If M is compact, prove that /C is compact. (d) If M is connected, prove that /C is connected.


***(e) If M is path-connected is K path-connected? (t) If M and M' are homeomorphic does it follow that K and K'

are homeomorphic? ***(g) What about the converse?

125. As on page 105, consider the subsets of JR,

A = {OJ U [1,2] U {3} and B = {OJ U {I} U [2, 3].

(a) Why is there no ambient homeomorphism of JR to itself that carries A onto B?

(b) Thinking of JR as the x-axis, is there an ambient homeomorphism of JR2 to itself that carries A onto B?

** 126. Consider an overhand (trefoil) knot K in JR3. It can be shown that there is no homeomorphism of JR 3 to itself that sends K to the standard unit circle Sl C JR2. (See Rolfsen's book, Knots and Links.) Thinking of JR3 as the plane X4 = 0 in (XI, X2, X3, x4)-space JR4, show that there is a homeomorphism of JR4 to itself that carries K onto Sl.

**127. Start with a set S c JR and successively take its closure, the complement ofits closure, the closure of that, and so on. S, cl (S), (cl (S) y, .... Do the same to Sc. In total, how many distinct subsets ofJR can be produced this way? In particular decide whether each chain S, cl (S), ...

consists of only finitely many sets. For example, if S = Q then we get Q, JR, 0, JR, 0, JR, 0, ... , and QC, JR, 0, JR, ... for a total of four sets.

**128. Consider the letter T. (a) Prove that there is no way to place uncountably many copies of

the letter T disjointly in the plane. [Hint: First prove this when the unit square replaces the plane.]

(b) Prove that there is no way to place uncountably many homeomorphic copies of the letter T disjointly in the plane.

(c) For which other letters ofthe alphabet is this true? (d) Let U be a set in JR3 formed like an umbrella: it is a disc

with a perpendicular segment attached to its center. Prove that uncountably many copies of U can not be placed disjointly in JR3.

(e) What if the perpendicular segment is attached to the boundary of the disc?

** 129. Let M be a complete, separable metric space such as JRm. Prove the Cupcake Theorem: each closed set K C M can be expressed uniquely as the disjoint union of a countable set and a perfect closed set, CuP = K.

Exercises

*130. Write jingles at least as good as the following.

When a set in the plane is closed and bounded, you can always draw a curve around it.

Peter Pfibik

'Tis a most indisputable fact If you want to make something compact Make it bounded and closed For you're totally hosed If either condition you lack.

Lest the reader infer an untruth (Which I think would be highly uncouth) I must hasten to add There are sets to be had Where the converse is false, fo'sooth.

Karla Westfahl

A coffee cup feeling quite dazed, said to a donut, amazed, an open surjective continuous injection, You'd be plastic and I'd be glazed.

Norah Esty

If a clopen set can be detected, Your metric space is disconnected.

David Owens

Pre-lim problems t

1. Suppose that f : JRm --+ JR satisfies two conditions: (i) For each compact set K, f(K) is compact.

(ii) For any nested decreasing sequence of compacts (Kn),

Prove that f is continuous.

135

t These are questions taken from the exam given to first year math graduate students at D.C. Berkeley.


2. Let X C jRm be compact and f : X -+ jR be continuous. Given E > 0, show that there is a constant M such that for all x, y EX,

If(x) - f(y)1 ~ Mix - yl + E.

3. Consider f : jR2 -+ R Assume that for each fixed xo, y 1--+ f(xo, y) is continuous and for each fixed Yo, x 1--+ f (x, Yo) is continuous. Find such an f that is not continuous.

4. Let f : jR2 -+ jR satisfy the following properties. For each fixed Xo E jR the function y 1--+ f(xo, y) is continuous and for each fixed Yo E jR the function x 1--+ f (x, Yo) is continuous. Also assume that if K is any compact subset of jR2 then f(K) is compact. Prove that f is continuous.

5. Let f(x, y) be a continuous real valued function defined on the unit square [0, 1] x [0, 1]. Prove that

g(x) = max{f(x, y) : y E [0, I]}

is continuous. 6. Let {Ud be a cover of jRm by open sets. Prove that there is a cover

{Vk} of jRm by open sets Vk such that Vk C Uk and each compact subset of jRm is disjoint from all but finitely many of the Vk.

7. A function f : [0, 1] -+ jR is said to be upper semi-continuous if given x E [0, 1] and E > ° there exists a 8 > ° such that Iy - xl < 8 implies that f (y) < f (x) + E. Prove that an upper semi-continuous function on [0, 1] is bounded above and attains its maximum value at some point p E [0, 1].

8. Prove that a continuous function f : jR -+ jR which sends open sets to open sets must be monotonic.

9. Show that [0, 1] can not be written as a countably infinite union of disjoint closed subintervals.

10. A connected component of a metric space M is a maximal connected subset of M. Give an example of M C jR having uncountably many connected components. Can such a subset be open? Closed? Does your answer change if jR2 replaces jR?

11. Let U C jRm be an open set. Suppose that the map h : U -+ jRm is a homeomorphism from U onto jRm which is uniformly continuous. Prove that U = jRm.

12. Let X be a non-empty connected set of real numbers. If every element of X is rational prove that X has only one element.

13. Let A c jRm be compact, x E A. Let (xn ) be a sequence in A such that every convergent subsequence of (xn ) converges to x.

Exercises 137

(a) Prove that the sequence (xn ) converges. (b) Give an example to show if A is not compact, the result in (a)

is not necessarily true. 14. Assume that f : lR ---+ lR is uniformly continuous. Prove that there

are constants A, B such that I f (x) I ::: A + B Ix I for all x E R 15. Let h : [0, 1) ---+ lR be a uniformly continuous function where [0, 1)

is the half open interval. Prove that there is a unique continuous map g : [0, 1] ---+ lR such that g(x) = hex) for all x E [0, 1).

[undergraduate texts in mathematics] real mathematical analysis || a taste of topology

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