advance topics in topology - point-setvml2113/teachers college, columbia... · 2012-04-24 ·...

ADVANCE TOPICS IN TOPOLOGY - POINT-SET

NOTES COMPILED BY KATO LA

19 January 2012

Background

Intervals:pa, bq “ tx P R | a ă x ă buÓ Ó

calc. notation set theory notation

pa,8q

p´8, bq

,

/

/

/

/

/

/

/

/

/

/

.

/

/

/

/

/

/

/

/

/

/

-

“Open” intervals

ra, bs, ra,8q: Closed

pa, bs, ra, bq: Half-openzHalf-closed

Open Sets: Includes all open intervals and union of open intervals. i.e., p0, 1q Y p3, 4q.

Definition: A set A of real numbers is open if @ x P A, D an open interval contain-ing x which is a subset of A.

Question: Is Q, the set of all rational numbers, an open set of R?

- No. Consider1

2. No interval of the form

ˆ

1

2´ ε,

1

2` ε

˙

is a subset of Q. We can

ask a similar question in R2.

Is R open in R2? - No, because any disk around any point in R will have points above andbelow that point of R.

Date: Spring 2012.1

2 NOTES COMPILED BY KATO LA

Definition: A set is called closed if its complement is open.

In R, p0, 1q is open and p´8, 0s Y r1,8q is closed. R is open, thus Ø is closed. r0, 1qis not open or closed. In R, the set t0u is closed: its complement is p´8, 0q Y p0,8q. InR2, is tp0, 0qu closed? - Yes.

Chapter 2 - Topological Spaces & Continuous Functions

Definition: A topology on a set X is a collection T of subsets of X satisfying:

(1) Ø, X P T(2) The union of any number of sets in T is again, in the collection(3) The intersection of any finite number of sets in T , is again in T

Alternative Definition: ¨ ¨ ¨ is a collection T of subsets of X such that Ø, X P T andT is closed under arbitrary unions and finite intersections.

Definition: A set X for which a topology has been defined is called a topological space.

Example: X “ ta, b, cu

T1 “ tØ, X, tau , ta, buu is a topology on X.T2 “ tØ, Xu is known as the “trivial” topology.T3 “ tall subsets of Xu is known as the “discrete” topology.T4 “ tØ, X, tb, cu , tcuu has complements of sets in T1.

Notice which Tx are subsets of Ty and vice versa.

Definition: A set which belongs to T is called an open set.

Example: If we let T contain all the sets which, in a calculus sense, we call open - Wehave “R with the standard [or usual] topology.”

Example: [Example 3, Page 77 in the text]X is a set.Tf contains all sets whose complements is either X or finite OR contains Ø and all setswhose complement is finite.

Recall: pAXBqA “ AA YBA and pAYBqA “ AA XBA

ADVANCE TOPICS IN TOPOLOGY - POINT-SET 3

Checking condition (2): Suppose tUαu is a collection of sets in Tf .

Taking the complement ofď

Uα ñ´

ď

Uα

¯A

“č

`

U Aα˘

ñ which must be finite because

each of the U Ai is finite.

Checking condition (3):

˜

nč

i“1

Ui

¸A

“

nď

i“1

`

U Ai˘

each U Ai is finite. ñď

`

U Ai˘

must

be finite.

Basis for a Topology

Definition: If X is a set, a basis for a topology T on X is a collection B of subsets ofX [called “basis elements”] such that:

(1) Every x P X is in at least one set in B(2) If x P X and x P B1 X B2 [where B1, B2 are basis elements], then there is a basis

element B3 such that x P B3 Ă B1 XB2

Question: How in fact do you know that you get a topology from basis elements?

Examples: [of bases]

(i) Open intervals of the form pa, bq are a basis for the standard topology on R.(ii) Interior of circle are a basis for the standard topology in R2.(iii) All one-point subsets of X are a basis for the discrete topology.

26 January 2012

Examples:

(i) X “ ta, b, c, d, eu, T1 “ tØ, X, tau , ta, buu, T2 “ tØ, X, tau , tcu , ta, bu , ta, b, cu , ta, cuu.T2 Ą T1, so T2 is “finer” than T1. i.e., The set that is “finer” has more open sets.We say, “tau is open in T1” which is equivalent to tau P T1.

In T2, tau is open. tau is not closed since tauA “ tb, c, d, eu is not open.(ii) On the real number line, T is referred to as the “usual” or “standard” topology

on R:(a) Sets in T are sets that are “open” in the calculus sense - i.e., p1, 3q, p0,8q,

and p1, 3q Y p4, 6q.(b) Sets in T are those that are unions of all sets of the from pa, bq where a ă b.


Recall the definition of a basis: A basis for a topology pX,T q is a collection of subsets Bsuch that:

(1) @ x P X is in some B P B(2) If x P X and x P B1 XB2, then there is B3 such that x P B3 Ă B1 XB2

Note: Every basis “generates” a topology this way: A subset U is “open” if every elementin U satisfies: x P B Ă U for some B P B.

Three Classic Lemmas

Lemma 1: Every element in T is a union of basis elements.

Lemma 2: If pX,T q is a topological space and C is a collection of open sets suchthat for every open set A, if x P A, then x P C Ă A for some C P C , then C is a basis.

Lemma 3: Let B and B1 be bases for T and T 1 on X. Then T 1 is finer than T ifand only if for each x and each B P B containing x, there is B1 P B1 such that x P B1 Ă B.

Further examples:

(i) R` is the “lower limit topology” on R where a basis is the set of all intervals of theform ra, bq where a ă b.

Is (0, 1) open in R`? i.e., Can we write (0, 1) as a union of sets of the formra, bq?

- Yes. (0, 1) =

„

1

2, 1

˙

Y

„

1

3, 1

˙

Y ¨ ¨ ¨ Y

„

1

n, 1

˙

Y ¨ ¨ ¨

How do R` and R compare? i.e., Can we write [0, 1) as a union of sets of theform pa, bq?

- No. Consider using the same method as above. If we attempt to include zero, say

„

´1

100, 1

˙

Y ¨ ¨ ¨ , will include zero, but as well as erroneous values less than zero.

Every set open in R is open in R`, but not the converse. Thus, R` is finer than R.


(ii) Rk is the so-called “k-topology” where the basis elements are intervals of the form

pa, bq OR pa, bq z k where k “

"

1,1

2,1

3,1

4, ¨ ¨ ¨

*

. How do R and Rk compare?

Well, all sets open in R are open in Rk. Are sets open in Rk open in R as well?

i.e., p´1, 1q z k?“ p´1, 0q Y

ˆ

1

2, 1

˙

Y

ˆ

1

3,1

2

˙

Y

ˆ

1

4,1

3

˙

Y ¨ ¨ ¨ .

This is very close because 0 P left-hand side, but 0 R right-hand side. So Rk isfiner than R.

How do R` and Rk compare?

r0, 1q ‰

ˆ

1

2, 1

˙

Y

ˆ

1

3, 1

˙

Y ¨ ¨ ¨ .

r0, 1q is open in R`, but not in Rk and p´1, 1q is open in Rk, but not in R`.

So R` and Rk are not comparable . i.e., Just like skew lines!

A topology on Z:

Open sets in the basis have two forms:

"

tnu if n is oddtn´ 1, n, n` 1u if n is even

This is the “Digital Line Topology .”

Is p3, 6q open? - No. p3, 6q “ t3u Y t3, 4, 5u Y t5u Y t5, 6, 7u. In fact, is any set liket6u [“even,” in other words] open? - No.

But r3, 7s is open: r3, 7s “ t3, 4, 5u Y t5, 6, 7u

Note: t3u and t5u are open sets in the basis, but not necessary as they are includedin r3, 7s by t3, 4, 5u and t5, 6, 7u respectively in the digital line topology.


2 February 2012

Basis ñ Topology

If we have a basis, we get a topology this way:

Definition - Basic Property [BP]: A set A is called open if every element in A isin some basis element which is a subset of A. i.e., Definition of open when given a basis.

Note: If B is a basis element, is B open? - Yes! If x P B, then x P Bljhn

basis element

Ă B.

Do all sets U that satisfy the basic property form a topology?

(1) Does Ø satisfy BP? - Yes, vacuously. Does X satisfy BP? i.e., If x P X, is there aB so that x P B Ă X? - Yes, by definition of a basis.

(2) If tUαu is a collection of sets satisfying BP, doesď

α

Uα? Since all Uα satisfy BP,

for each Uα and x P Uα, there is a B such that x P B Ă Uα. So if x Pď

α

Uα, then

x P Uα1 for at least one set. So there is a B such that x P B Ă Uα Ăď

α

Uα.

(3) If U1, ¨ ¨ ¨ , Un satisfy BP, does U1 X U2 X ¨ ¨ ¨ X Un satisfy BP?Proof for Two Sets [Easily extended to n sets]:U1, U2 satisfy BP. ñ x P U1, then x P B1 Ă U1 and similarly, x P U2, thenx P B2 Ă U2 and Does U1 X U2 satisfy BP? i.e., Is the intersection open?If x P U1 X U2, then x P B1 Ă U1 and x P B2 Ă U2. So x P B3 Ă U1 X U2 byproperty (2) by the definition of as basis.

Proof of Lemma 1:

T generated by basis?“ Unions of basis elements

pðq A set in the right-hand side is a union of open sets. So it is open, and thus, in theleft-hand side.

pñq A set in the left-hand side os open, so it satisfies BP. Form a union of all thosebasis elements over all X in the set. Thus, it is a union of basis elements. ♣


The Order Topology

We digress to discuss order relations [Page 24 in the text]. :

Definition: A relation R on a set A is called an order relation or linear relation orsimple order if and only if

(1) If x, y P A with x ‰ y, then xRy or yRx(2) xRx is never true(3) If xRy and yRz, then xRz

Suppose we have two linear orders on ta, b, cu. Are they essentially the same relation ornot? In other words, are the following the same?

aRb, bRc, aRccRb, bRa, cRa

*

Same Order Type

One classic linear order on Rˆ R:

pa, bq ă pc, dq ðñ

"

a ă c ora “ c and b ă d

This is usually referred to as the “Dictionary Order .” Notice there is no “smallest”element.

An order relation on the unit circle:

pa, bq ă pc, dq ðñ

"

b ă d orb “ d and a ă c

Is there a “smallest” one? - Yes, p0,´1q. End of digression. :

Definition: Let X be a set with a linear order. Assume X has ě two elements. Thenthe collection B is a basis for the order topology on X if B consists of all sets of thefollowing type:

(1) All open intervals pa, bq(2) If X has a smallest element a0, then all sets ra0, bq(3) If X has a largest element b0, then all sets pa, b0s

B is a basis. Check for confirmation [Check it out!] Is the order topology on R the usualtopology? - Yes! Parts (b) and (c) do not apply on R, so there is only (a) to follow!


9 February 2012

Order Topology [Continued...]

Examples:

(1) Order Topology on R(2) The dictionary order on Rˆ R. There is no first or last element.(3) ta, bu ˆ Z` “ ta1, a2, a3, ¨ ¨ ¨ ; b1, b2, b3, ¨ ¨ ¨ u

pa1, a4q “ ta2, a3upa6, b1q “ ta7, a8, ¨ ¨ ¨ uThis topology has a first element, but no last element. We can even look at sets ofthe form:ra1, a4q “ ta1, a2, a3ura1, a2q “ ta1u

The Product Topology

Definition: If X and Y are topological spaces, the product topology, X ˆ Y , is thetopology having as a basis all sets of the form U ˆV , where U is open in X and V is openin Y .

Example: Rˆ RThe product topology on R ˆ R has basis elements of the form U ˆ V which are openrectangles. The usual topology on R has basis elements which are open disks. It turns outthese two topologies are equivalent. By Lemma 3, to compare topologies, we can comparetheir respective basis elements. i.e., Given a point in one basis element of a topology, can Ifit a basis element from a different topology in the original topology basis? In our example,Given a basis element in RˆR [an open rectangle], given any point in our open rectangle,can I fit an open disk around that point that is completely contained in the open rectangle?- Yes! And vice versa.

Another example is to ask ourselves,“What is an example of a basis given a topology?”

Question: Is “a basis all sets of the form U ˆ V , where U is open in X and V is open inY ” really a basis? That is,

(1) Is every element x in some set

open in Xhnlj

U ˆ

open in Yhnlj

V ?x P X ˆ Y

(2) If x P pU1ˆV1qX pU2ˆV2q, then x P pU1 X U2q

some U -type set

ˆ

some V -type set

pV1 X V2q Ă pU1ˆV1qX pU2ˆV2q


Theorem: If B1 is a basis for X1,B2 a basis for Y , then a basis for X ˆ Y is all sets ofthe form

tB1 ˆB2 | B1 P B1, B2 P B2u

Proof is left to the reader.

Subspaces

Definition: Let pX,T q be a topological space. If Y Ă X, then the subspace topologyon Y consists of open sets of the form U X Y , where U is open in X.

Question: Is the above definition really a topology?

(1) Ø “ Øljhn

open in X

X Y ; Yljhn

open in X

X Y .

(2) If tUα X Y u are all in the topology, isď

α

UαXY open? ñď

pUαXY q “

˜

ď

α

Uα

¸

open in X

X Y .

(3) If U1 X Y,U2 X Y, ¨ ¨ ¨ , Un X Y are open, isnč

i“1

pUi X Y q open? Butč

pUi X Y q “

˜

nč

i“1

Ui

¸

open in X

X Y .

Examples:

(1) Consider R with the usual topology and subset Z. What topology does Z have asa subspace of this R?

t3u is open because t3u “

ˆ

21

2, 3

1

2

˙

X Z.

In fact, every point-set is open. Every set is open. This is the discrete topology.(2) Let T “ R2 with the usual topology. Let S “

px1, x2q P R2 | x21 ` x22 “ 1

(

. LetA “ tpx1, x2q P S | x1 ą 0u.Is A open in T? - No, it cannot be written as the union of interior of circles.Equivalently, a point in A does not lie inside a circle, which is completely in A.Is S open in T? - No, because x21 ` x22 “ 1, if the equality was less than or equal,then yes.Is A open in S if S has the subspace topology in relation to T? - In subspace S of

T , an open set is any normal open set inč

S. Thus, A is open in the subspace

topology.


16 February 2012

Lemma: Let Y be a subspace of X. If U is open in Y and Y is open in X, then U is openin X.

Proof : Suppose U is open Y . i.e., Y intersect with something originally open in theoriginal set. Then U “ Y X U 1, which is open in X. Remember: Any open set in asubspace is the intersection of the subspace with an open set in X. So U is open in Xbecause it is the intersection of two sets, both open in X.

Theorem: If A is a subspace of X,B is a subspace of Y , then A ˆ B has the sametopology as AˆB inherits as a subspace of X ˆ Y .Proof is left to the reader.

Examples:

(1) Z ˆ Z in the order topology on Z ô Z with the discrete topology. p2, 1qpoint “p1, 3qinterval ˆ p0, 2qinterval.But consider the set Zˆ Z in R2. What topology does it have if we think of it asa subspace of R2 with the usual topology? - Any point p can be written as theintersection of an open set in R2 with ZˆZ. So any tpu is open. This is the discretetopology.

(2) Consider R with the usual topology and Y “ r0, 1s. Y is not open, which maycause a problem.(a) What is Y in the order topology? - Open sets are of the form: pa, bq, r0, bq, and pa, 1s.

Not

ˆ

1

2,3

4

.

(b) What is Y in the subspace topology that it gets [inherits] from R, the ordertopology? - Here, the open sets are normal open sets intersected with Y :pa, bq, r0, bq, and pa, 1s. We get precisely the same sets! Note this is notalways the case.

(3) Y “ r0, 1s Y t2u. What is Y in the order topology? - Basic open sets are of theform: pa, bq, r0, bq, and pa, 2s. t2u is not open in the order topology. What is Y inthe topology it inherits as a subspace of R with the usual topology? - t2u is open:

t2u “

ˆ

11

2, 2

1

2

˙

X Y . t2u in both topologies stated above are not the same.

(4) Do t1, 2u ˆ Z` and Z` ˆ t1, 2u have the same order type in the dictionary ordertopology? - No. It is helpful to simply state where the topologies differ. Forexample, the first topology has order type ω2, intervals where there is an infiniteamount of points between the endpoints, and two points have no immediate prede-cessors. While the second topology has order type ω, intervals where there is a finitenumber of elements between the end points, and every element has an immediatepredecessor except (1, 1).


(5) Consider I “ r0, 1s and I ˆ I. Is I ˆ I as a subspace of Rˆ R with the dictionaryorder topology the same as I ˆ I with its own dictionary order topology? - Theshort answer is no. Consider the first topology as T1 and the second topology asT2. Everything open in T2 is open in T1. Though not proven, it appears convincingthat T2 Ă T1.

Closed Sets

Definition: A set in a topological space is called closed if and only if its complement isopen.

Examples:

(1) In R with the usual topology, all singleton sets are closed: t0u is closed, and itscomplement is p´8, 0q Y p0,8q, which is open.

(2)

px, yq | x2 ` y2 “ 1(

in R2 with the usual topology. - Take any point not in thecircle, we can make an open set around said point that does not touch the circle.

(3)

"

1

2,1

3,1

4, ¨ ¨ ¨

*

in R with the usual topology. - The complement is not open. So the

set, call it K, is not closed. Is K open? - No.(4) r0, 1q is not open or closed.(5) Y “ r0, 1s Y p2, 3q in the subspace topology of R with the usual topology. What

is p2, 3q, open or closed? - Open. Y X p2, 3q. r0, 1s is open as well: r0, 1s “

Y X

ˆ

´1, 11

2

˙

. In Y , the complement of r0, 1s is p2, 3q. So, r0, 1s being open, p2, 3q

must be closed. In fact, both r0, 1s and p2, 3q are both open and closed.

Question: Give an example of a topology with a closed set A which becomes open whenone point is removed.

Using R with the usual topology: A “ txu Ñ Ø or A1 “ r1,8q Ñ p1,8q or A2 “RÑ p´8, xq Y px,8q.


23 February 2012

Facts about Closed Sets

Theorem: In a topological space X,

(1) Ø, X are closed(2) An arbitrary intersection of closed sets, is closed.(3) The union of a finite number of closed sets, is closed.

Theorem: If Y is a subspace of X, A is closed in Y , Y is closed in X, then A is closed in X.

Theorem: If Y is a subspace of X and A is closed in Y , then A “ B X Y where Bis closed in X [and conversely].

Note: It is noteworthy to compare the above theorems with what we have already learnedabout open sets and see how they are alike and differ.

Closure and Interior

Definition: The closure of a set A, denoted A, is the intersection of all the closed setsthat contain A.Note: The above definition is “nice” because we see at once that A is closed. It is a “poor”definition because envisioning all the closed sets containing A can be difficult.

Definition: The interior of A is the union of all open subsets of A.Note: The interior of A is open. A open ô A = interior of A, A closed ô A “ A.

Examples:

(1) A “

"

1

2,1

3,1

4, ¨ ¨ ¨

*

; Closure: AY t0u; Interior: Ø.

(2) A “ Q; Closure: R; Interior: Ø.(3) Weird topological space - X “ ta, b, cu; Open sets: Ø, X, tau, ta, bu; If A “ tau;

Closed sets: X,Ø, tb, cu, tcu; If A “ tau, B “ tbu; A “ X,B “ X X tb, cu “ tb, cu.

Theorem: If Y is a subspace of X, A a subset of Y , A is the closure of A in X, then theclosure of A in Y is AX Y .

Question: If A Ă B, is A Ă B? - A “

ˆ

1

2, 1

˙

, B “ p0, 1q. A “

„

1

2, 1

Ć B. Now, if

A Ă B, is A Ă B?


“Big” Theorem: Let A be a subset of X, a topological space:

(1) x P Aô Every open set containing x intersects A.(2) x P Aô Every basis element containing x intersects A.

Note: The implication of this writing style is to ensure the intersections are nonempty.

Alternative Wording: x P A ô Every neighborhood of x intersects A. Why isthis the case?

A Word on Logic: P ñ Q

conditional statement

ô

contrapositive

Qñ P ; P ô Q is P ñ Q and Qñ P

The contrapositive of the alternative wording is as follows: x R A ô D a neighborhoodof x that does not intersect A.

Proof : Suppose x R A is the intersection of all closed sets containing A. Then x isnot in at least on closed set B that contains A : x R B. So x P BA. Since B is closed,it contains A. Then BA is open not containing A. ñ A neighborhood of x that does notintersect A. ♣.

Limit Points

Definition: x is a limit point of A if and only if every neighborhood of x intersects A ina point other than x itself.

Examples:

A A Set of Limit Points pA1q

r0, 1s r0, 1q r0, 1s

t0u t0u Ø

R Q R"

0,1

2,1

3, ¨ ¨ ¨

* "

1

2,1

3, ¨ ¨ ¨

*

t0u


Main Result: A “ AYA1

Proof :

pñq x P Añ Every neighborhood of x intersects A.If x P A, then x P AYA1.If x R A, then every neighborhood of x intersects A in a point other than x itself -So x P A1.

pðq x P AYA1.If x P A, then x P A because A Ă A.If x P A1, then every neighborhood of x intersects A - So x P A. ♣

If R has the discrete topology, then whatever A is, Aljhn

Will equal A in this topology

“č

Bα All closed sets con-

taining A.

The minimal separation axiom is T0.T0: Given two distinct points x and y, there is either an open set containing x and not yor an open set containing y and not x.T1: The same condition as T0, but the “or” condition becomes and .T2 [Hausdorff Space]: Given two distinct points x and y, there are disjoint open sets A andB such that x P A and y P B.

What do we mean when we say that a sequence tx1, x2, x3, ¨ ¨ ¨ u in a topological spaceconverges to a point x? - It means that every neighborhood of x contains all points inthe sequence from some xn on.

Odd behavior in a non-T2 space:X “ ta, b, cu, Open sets: Ø, X, ta, bu, tb, cu, tbu. What does tb, b, b, ¨ ¨ ¨ u converge to?Claim: The sequence converges to b.Neighborhoods of b: X, ta, bu, tb, cu, tbu. i.e., Every term is in this neighborhood of b.Some for all neighborhoods. So the sequence converges to b.In fact, tb, b, b, ¨ ¨ ¨ u converges to a, to b, and ti c!But this is not a Hausdorff space because we can find an open set around b, but any openset around a will contain b.

Munkres explains in a Hausdorff space, sequences have unique limits and singleton setsare always closed. Why is, say txu, always closed in a Hausdorff space?


22 March 2012

Functions

f : Aljhn

domain

ÞÝÑ B. We can speak of fpT q where T is a subset of the domain.

Example: f : R ÞÝÑ R by fpxq “ x2. If A “ p1, 2q, then fpAqljhn

image set of f on everything in A

“ p

“12

1 ,

“22

4 q.

We also write f´1pAq to represent all elements that f maps to A. Above, for exam-ple, f´1pp1, 4qq “ p´2,´1q Y p1, 2q.

Continuous Functions

Calculus: F is continuous at “a” if: @ ε ą 0, D δ ą 0 such that |x´a| ă δ ñ |fpxq´fpaq| ăε.

“f is continuous.” ðñ “f is continuous for every ‘a’ in the domain.”

Topologically:

f is continuous ðñ

$

’

’

’

’

&

’

’

’

’

%

f´1pAq is open whenever A is open

OR

the inverse image of every open set is open

If f : pX,T1q ÞÝÑ pY,T2q, then f is continuous if and only if the inverse image of ev-ery open set in Y is an open set in X.

Question: What does it mean to claim that a function f : X ÞÝÑ Y is not continu-ous? - For at least one open set in Y , f´1pY q is not open.


Suppose fpxq “

$

’

’

&

’

’

%

x´1

2if x ă 2

2

3x´

2

3if x ě 2

in R2

f´1pp0, 3qq “

ˆ

1

2, 3

1

2

˙

. Now, let A “

ˆ

11

2, 3

˙

open

. Then f´1pAq “

„

2, 31

2

˙

not open

.

What about fpxq “

"

0 if x ă 21 if x ě 2

in R2

What set in A provides a counter-example of showing f is not continuous?

- Let A “

ˆ

1

2, 1

1

2

˙

open

, then f´1pAq “ r2,8s

not open

Next, consider R in the usual topology and X “ ta, b, cu with open sets: tX,Ø, tau, ta, buu.f : R ÞÝÑ X this way: fp1q “ b, fpxq “ a, if x ‰ 1. Is f´1pAq open when A is open?

f´1pØq “ Ø

f´1pXq “ R

f´1ptauq “ R z t1u “ p´8, 1q Y p1,8q

f´1pta, buq “ R

,

/

/

/

/

/

/

/

/

.

/

/

/

/

/

/

/

/

-

all open!

Note that a maps to all numbers except 1.


Suppose we have the same whole space X on the usual topology on R and

fpxq “

$

&

%

a ÞÑ 1b ÞÑ 2c ÞÑ 3

Is f continuous?

f´1ˆˆ

1

2, 1

1

2

˙˙

“ tau which is open, so f´1ˆˆ

1

2, 1

1

2

˙˙

is open.

f´1ˆˆ

21

2, 3

1

2

˙˙

open

“ tculjhn

not open

ñ So f is not continuous.

Identity Maps

Definition: The so-called identity map is defined as ipxq “ x.

Suppose i : X ÞÝÑ X. Will it always be continuous? Note: If i : R ÞÝÑ R, then i iscontinuous. In fact, it is our usual y “ x.

Example: Let X be the domain R in the usual topology. Let Y be the range Rl, thelower limit topology ra, bq.

r0, 1q

“f´1pAq

ÞÝÑ r0, 1q

A

Notice f´1pAq is not open in R under the usual topology. A is open under the lower limittopology. Thus, i is not continuous.

Now, suppose i : Rl ÞÝÑ R. Then i is an identity function as it takes pa, bq

open

ÞÝÑ pa, bq

open

.

If we have one set, two topologies, and i : pX,T1q ÞÝÑ pX,T2q and i is continuous [asin our preceding example], what can we deduce about T1 and T2? - T1 Ą T2. i.e., T1

must be finer than T2.

Theorem 18.1: Several statements equivalent to “f is continuous.”

We will prove the following: f : X ÞÝÑ Y is continuous ðñ @ x P X and each neigh-borhood V of fpxq, there is a neighborhood U of fpxq such that fpUq Ă V .


Proof :

(ñ) Suppose f is continuous. Let x P X and let V be a neighborhood of fpxq. f iscontinuous, so f´1pV q is open in X. Let U “ f´1pV q.Now fpUq “ fpf´1pV qq

f´1pfpV qqdoes not always equal V

“ V . So, fpUq Ă V . [In fact, equality holds here.]

(ð) Suppose V is open in Y . Why is f´1pV q open in X? For each x P f´1pV q D an openneighborhood that maps into V . Take the union of all these open neighborhoods.The union will be open and equal to f´1pV q. So f´1pV q is open.

Definition: A homeomorphism is one-to-one and onto. Two “homeomorphic” spacesare ones that are topologically equivalent. Given f : X ÞÝÑ Y , f continuous, and f´1

continuous. f is a homeomorphism and X,Y are homeomorphic.

Example: p0, 1q and p10, 20q as interval subspaces of R under the usual topology are home-omorphic. What is a one-to-one and onto map: p0, 1q ÞÝÑ p10, 20q which is continuous andfor which the inverse function is continuous?

29 March 2012

Metric Spaces

Definition: A metric on a set X is a function d : X ˆX ÞÝÑ R such that

(1) dpx, yq ě 0 and dpx, yq “ 0 ô x “ y(2) dpx, yq “ dpy, xq(3) dpx, zq ď dpx, yq ` dpy, zq

Every metric gives rise to a topology on X through the use of “epsilon-balls,” [ε-balls]denoted Bpx, εq “ ty P X | dpx, yq ă εu which is the set of all points whose distance fromx is less than epsilon.

Once we know that ε-balls form a basis, we can get all the open sets by taking unionof the ε-balls.


Examples:

(i)

If dpx, yq “

"

0 if x “ y1 otherwise

B

ˆ

x,1

2

˙

“ txu This is open. Thus, B

ˆ

x,1

2

˙

is open. ñ txu is open.

This metric generates the discrete topology.

(ii) Let P “ px1, y1q, Q “ px2, y2q, and dpP,Qq “a

px1 ´ x2q2 ` py1 ´ y2q2 . Theseε-balls generate the usual topology on R2.

Do these ε-balls indeed for a basis?

(1) If x P X, then x P Bpx, 1q since dpx, xq ă 1.

Before proving the next [and last] basis property, we would like to justify thefollowing statement:

If y P Bpx, εq, then there is a Bpy, δq such that Bpy, δq Ă Bpx, εq

For δ, choose something that is less than ε´ δpx, yq. Then, Bpy, δq Ă Bpx, εq.Why? If z P Bpy, δq, then is z P Bpx, εq?

dpx, zq ď dpx, yq ` dpy, zq

ă dpx, yq ` δ

ă dpx, yq ` pε´ dpx, yqq

“ ε

(2) If x P Bypy, ε1q XBzpz, ε2q, then x P Bxpx, ?q Ă Bypy, ε1q XBzpz, ε2q. What is “?”.Look at ε1 ´ dpx, yq and ε2 ´ dpx, zq. Use the smaller of the two for “?”.

Definition: A metric space is a topological space that has a metric whose ε-balls gener-ate the space in question. A space for which no metric can generate the open sets is callednon-metrizable .


Examples:

(i) R with dpx, yq “ |x´ y|. Open balls are open intervals. e.g., B

ˆ

1,1

2

˙

“`

12 , 1

12

˘

.

The metric topology is the usual topology on R.

(ii) dpP,Qq “a

px1 ´ x2q2 ` py1 ´ y2q2 on R2 gives the usual topology on R2. Thismetric is sometimes called the 2-metric and can be extended to Rn.

(iii) On R2: dpP,Qq “ |x1 ´ x2| ` |y1 ´ y2|. This is known as the 1-metric or the“taxicab metric.”

(iv) On R2: dpP,Qq “ max t|x2 ´ x1|, |y2 ´ y1|u. This is the “infinity metric,” de-noted d8.

One way to begin understanding a strange metric is to look at a few specificε-balls. For (ii), (iii), and (iv), what are the points which are, say one unitaway from p0, 0q?

It turns out (ii) yields the usual open unit circle, (iii) yields an open rhombus,and (iv) yields an open square.

If Cr0, 1s is the set of all continuous functions: r0, 1s ÞÝÑ R, then

(v) d1pf, gq “

ż 1

0|fpxq ´ gpxq| dx

(vi) d2pf, gq “

„ż 1

0pfpxq ´ gpxqq2 dx

1

2

(vii) d8pf, gq “ max0ďxď1

|fpxq ´ gpxq|


Theorem: Every metric space is Hausdorff.

Proof : Let p, q P X where p ‰ q and X is a metric space. Let ε “1

2dpp, qq and consider

Bpp, εq and Bpq, εq. Is Bpp, εq XBpq, εq “ Ø?

Suppose r P Bpp, εq XBpq, εq, then

dpp, qq ď dpp, rq ` dpr, qq

ă ε` ε

“ 2ε

“ dpp, qq

ñ dpp, qq ă dpp, qq ñð

But this is a contradiction! Thus, there must have been nothing in the intersection fromthe beginning. ♣

It is common [or as Smith stated: “It is not uncommon”] to define a new metric in termsof a familiar metric, particularly if the familiar metric has some undesirable property.

Suppose we have an unbounded metric. We can turn it into a bounded metric thatgenerates the same topology.

Note: A bounded metric is one which always gives values between M and ´M for someM P Z.

Example: If d is unbounded, the define d˚px, yq “ mintdpx, yq, 1u. The above is a metric[Check it out!], and its values are less than or equal to one. But d˚ will generate the sametopology as d.


12 April 2012

Chapter 3 - Connectedness & Compactness

Connected Topological Spaces

Consider R under the usual topology. r0, 1s Y r2, 3s is not connected; R is connected;p´8, 0q Y p0,8q is not connected; and p´8, 0s Y r0,8q “ R.

In R2: (The y-axis) Y The graph of y “

ˇ

ˇ

ˇ

ˇ

1

x

ˇ

ˇ

ˇ

ˇ

is not connected. This is the usual as-

ymptotic graph of1

xreflected along the y-axis.

Consider (The y-axis) Y The graph of y “ sin

ˆ

1

x

˙

where x ą 0. Is this connected?

Definition: A topological space is separated [or not connected or disconnected] if itcan be written as the union to two disjoint, nonempty, open sets:

X “ U Y Vopen and nonempty

A topological space is connected if it cannot be written as a separation.

Facts:

(1) X is connected ô the only sets that are both open and closed are X and Ø.

(1.5) X is not connected ô D proper, nonempty subsets that are both open and closed.

(2) X is not connected ô There are sets U and V which are disjoint, nonempty, andneither set contains a limit point of the other.

Reiteration: X “ U Y V . V is closed, so it contains all its limit points. Sonone of these limit points an be in U since U X V “ Ø.

Note: Addressing (The y-axis) Y The graph of y “ sin

ˆ

1

x

˙

, this actually is connected

by fact (2) [or the failure of fact (2)].


19 April 2012

Lemma: If U, V are a separation of X and if Y is a connected subspace of X, then eitherY Ă U or Y Ă V .

Proof : Suppose X “ U YV where U, V are open, non-empty, disjoint sets of X. ConsiderY X U, Y X V . These sets are open in Y and are disjoint. Both these sets cannotboth benon-empty or Y would not be connected. Thus, Y X U, Y X V is empty - Say Y X U “ Ø.This means that Y X V “ Y which tells use that Y Ă V . ♣

Theorem: The image of a connected space under a continuous map is connected.

Proof : Let f : X ÞÝÑ Y be continuous. Suppose fpXq “ Z. Then f : X ÞÝÑ Z isonto and continuous. [Use the face that inverse image of open sets is open].

Compact Spaces

Definition: An open covering of a subspace or space is a collection of open sets whoseunion is the subspace or space. e.g., Let X “ R, Y “ p0, 1q. If we consider all intervals of

the form

ˆ

a´1

3, a`

1

3

˙

where 0 ă a ă 1, we have an open covering of Y .

Some spaces have the following property: For every open covering, there is a finite subcol-lection which is still a covering. Such spaces are called compact .

Example: Y “ p0, 1q is not compact subset of R under the usual topology - Althoughthe covering just given does have a finite subcovering, not every covering of p0, 1q has a

finite subcover. e.g.,

ˆ

1

2, 1

˙

,

ˆ

1

3, 1

˙

,

ˆ

1

4, 1

˙

, ¨ ¨ ¨ “8ď

n“1

ˆ

1

n, 1

˙

Ă p0, 1q is an open cover-

ing, but there does not exist a finite number of elements that will still cover p0, 1q.

However, r0, 1s is compact.

Proof : Suppose r0, 1s has an open covering. D an a P r0, 1s such that r0, as is contained ina finite number of these open sets. What is the largest a such that r0, as is covered by afinite number of these open sets? Could a ă 1? - No, because a would be “inside” someopen interval that covers a and that means we could extend r0, as even further to theright . Contradiction. 6 a “ 1. ♣


Theorem: The image of a compact space under a continuous map is compact.

Proof : Proof is on page 116 in the text.

In R with the usual topology:

(a) A subspace Y is connected ô Y is in an interval(b) A subspace Y is compact ô Y is closed and bounded

Why is (a) true?

Proof :

pñq Suppose Y Ă R is connected. Is it an interval?

connected ñ interval ô interval ñ connected

A is not an interval ñ There is a, b, t P A and a ă t ă b such that t R A.e.g., U “ p´8, tq, V “ pt,8q and Y X U, Y X V is a separation of Y .

pðq Suppose Y is an interval. Is it connected?Suppose Y “ p´8, aq. If Y is not connected, then Y “ U Y V where U, V areopen, non-empty, and disjoint sets. Let t “ supV . Can t “ a? Suppose t “ a. Ist P U or t P V ? If t P U , then D pa´ ε, as Ă U , etcetera.

Think about this: One can argue the entire interval p´8, as must be in U . So V ‰ Ø.Then, what if t P V ? What about other interval types?

advance topics in topology - point-setvml2113/teachers college, columbia... · 2012-04-24 ·...

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