Summary in Preparation for Test in Topology

Jacob Shapiro

August 12, 2013

Abstract

A summary (largely of Munkres topology book, and hopefully moreexamples and exercises) in preparation for a test with Prof. DamienCalaque at ETH in the summer of 2013.

Contents

0.1 Open Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.2 Misc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

I Topological Spaces 4

1 Chapter 2 Topological Spaces and Continuous Functions 41.1 12 Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . . 41.2 13 Basis for a Topology . . . . . . . . . . . . . . . . . . . . . . . 51.3 14 The Order Topology . . . . . . . . . . . . . . . . . . . . . . . 71.4 17 Closed Sets and Limit Points . . . . . . . . . . . . . . . . . . 91.5 18 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . 161.6 16 The Subspace Topology . . . . . . . . . . . . . . . . . . . . . 221.7 19 The Product Topology . . . . . . . . . . . . . . . . . . . . . . 251.8 15 The Product Topology on X ! Y . . . . . . . . . . . . . . . . 271.9 22 The Quotient Topology . . . . . . . . . . . . . . . . . . . . . 281.10 20, 21 The Metric Topology . . . . . . . . . . . . . . . . . . . . 35

2 Chapter 3 Connectedness and Compactness 412.1 23 Connected Spaces . . . . . . . . . . . . . . . . . . . . . . . . 412.2 24 Connected Subspaces of the Real Line (and path connectedness) 432.3 25 Components and Local Connectedness . . . . . . . . . . . . . 452.4 26 Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 482.5 27 Compact Subspaces of the Real Line . . . . . . . . . . . . . . 512.6 28 Limit Point Compactness . . . . . . . . . . . . . . . . . . . . 532.7 29 Local Compactness (note: not in material for testutility for

measure theory) . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

1

3 Chapter 4 Countability and Separation Axioms (not part of thecurriculum for the testfor competeness reasons) 573.1 31 The Separation Axioms . . . . . . . . . . . . . . . . . . . . . 573.2 35 Imbeddings of Manifolds . . . . . . . . . . . . . . . . . . . . 57

4 Chapter 7 Complete Metric Spaces and Function Spaces 584.1 43 Complete Metric Spaces . . . . . . . . . . . . . . . . . . . . . 584.2 44 Theorem 44.1The Peano Space-Filling Curve . . . . . . . . 614.3 45 Compactness in Metric Spaces . . . . . . . . . . . . . . . . . 614.4 46 Pointwise and Compact Convergence . . . . . . . . . . . . . . 63

II Algebraic Topology 63

5 Chapter 9 The Fundamental Group 645.1 51 Homotopy of Paths . . . . . . . . . . . . . . . . . . . . . . . 645.2 52 The Fundamental Group . . . . . . . . . . . . . . . . . . . . 675.3 53 Covering Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 695.4 54 The Fundamental Group of the Circle . . . . . . . . . . . . . 715.5 55 Retractions and Fixed Points . . . . . . . . . . . . . . . . . . 735.6 58 Deformation Retracts and Homotopy Type . . . . . . . . . . 755.7 59 The Fundamental Group of Sn . . . . . . . . . . . . . . . . . 765.8 60 Fundamental Groups of Some Surfaces . . . . . . . . . . . . . 77

6 Chapter 11The Seifert-van Kampen Theorem 796.1 67 Direct Sums of Abelian Groups . . . . . . . . . . . . . . . . . 796.2 68 Free Products of Groups . . . . . . . . . . . . . . . . . . . . 796.3 69 Free Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 816.4 70 The Seifert-van Kampen Theorem . . . . . . . . . . . . . . . 816.5 71 The Fundamental Group of a Wedge of Circles . . . . . . . . 81

7 Chapter 13Classication of Surfaces 827.1 79 Equivalence of Covering Spaces . . . . . . . . . . . . . . . . . 827.2 80 The Universal Covering Space . . . . . . . . . . . . . . . . . 827.3 81 Covering Transformations (Deck Transformations) . . . . . . 83

0.1 Open Questions

1. HWS04 Open and closed maps, question 2.

2. HWS05 ultrametrics question 3.

3. Lemma 23.1 proof.

4. Example 7 in chapter 23.

5. HWS06 application to analysis (why is A " Open(!) ?)

2

6. HWS09E01, the end of the argument.

7. HWS10E01 walk through again.

8. HWS11E05 walk through again.

0.2 Misc

1. Matrix multiplication on Mat n (R) is continuous.

2. Matrix inversion on GL (n, R) is continuous.

3. A topological space which is also a group, with composition and inversefunctions being continuous, is called a topological group.

4. Torus

(a) T1 := R2/ # where!xy

"#

!x !

y!

"$% &

!pq

"" Z2 :

!x !

y!

"=

!xy

"+

2!!pq

".

(b) Let r, R " R : R > 2r . Define p "#R3

$[0, 2! ) " [0, 2! ) by p%!

"#

"&:=

([R + r cos (#)] cos (" )[R + r cos (#)] sin (" )

r sin (#)

)

* .

Define T2 := p ([0, 2! ) ! [0, 2! )) .(c) Observe that T1 T2 S1 ! S1. This is the torus. It is connected

and compact. It is not homeomorphic to [0, 2! ) ! [0, 2! ).

(d) T1 is a topologiacl group, where m%!!

xy

""

#,!!

x !

y!

""

#

&:=

!!x !

y!

""

#.

i%!!

xy

""

#

&:=

!!( x( y

""

#, and eT1 :=

!!00

""

#.

5. If f " C#S1, R

$then &a " S1 : f (a) = f (( a).

6. Dinis theore: theorem 7.14 in Rudins PMA.

7. The least upper bound property:

Any non-empty set of that has an upper bound necessarily has a leastupper bound.

8. (HWS09E01) If X is an ordered set (equipped with the order topology) inwhich every closed interval is compact, then X has the least upper boundproperty.

3

Part I

Topological Spaces

1 Chapter 2 Topological Spaces and ContinuousFunctions

1.1 12 Topological Spaces

1.1.1 Denition of a Topology

A topology on a set X , is a set , T ) P (X ), such that:

1. ! , X " T

2. If A is some set (finite, countable, or uncountable) and if U" " T * $ " A,then

#+" $ A U"

$" T .

3. If n " N and Ui " T * i " { 1, 2, . . . , n} then (, n

i =1 Ui ) " T .

If U " T then U is called an open set. Thus T is also denoted as Open(X ).If (X \ U) " T then U is called closed set. All the closed sets are denoted byClosed(X ).

The discrete topology is defined as P (X ).

The indiscrete / trivial topology is defined as { X, ! } .

The cofinite topology is defined as Tf := { ! } + { A ) X | |X \ A | < ,} .

Apply de Morgans laws for the arbitrary union.

The cocountable topology is defined as Tc := { ! }+{ A ) X | |X \ A | = |N|} .

1.1.2 Example 1 for a Topology

X := { a, b, c} , Open(X ) := { X, ! , { a, b} , { b} , { b, c}} .

But S1 := { X, ! , { a} , { b}} (no union)

and S2 := { X, ! , { a, b} , { b, c}} (no intersection)

are not topologies on X !

1.1.3 Example for a Topology

All subsets of X which are supersets of a given set, plus the empty set.

4

1.1.4 Counter Example for a Topology

All subsets of X which intersect a given set, plus the empty set. Then de-pending on the nature of the given set, this may or may not be a topologyon X .

X = R, S " 2R and O := { ((, , b) : b " S + {,}} . Then this is atopology only if for any collection of open sets, the supremum of b of eachof those sets is in S.

1.1.5 Coarser vs. Finer

If T ! - T then T ! is finer than T , or equivalently, T is coarser than T ! . If oneset is finer or coarser than another, the two are comparable.

For any set, Tdiscrete - T cocountable - T cof inite - T trivial .

1.1.5.1 Counterexample If X := { a, b, c} , the topology { X, ! , { a, b} , { b, c} , { b}}is not comparable to the topology { X, ! , { a, b} , { a}} , because one doesnt con-tain the other and vice versa.

1.1.5.2 Counterexample The lower-limit topology is not comparable withthe K -topology.

1.2 13 Basis for a Topology

1.2.1 Denition of a Basis for a Topology

If X is a set, a basis for a topology on X is a set B . P (X ), whose elementsare called basis elements , such that:

1. * x " X &B " B : x " B .

2. IfB1, B2 " B and x " (B1 / B2) then &B3 " B such that x " B3 .(B1 / B2).

1.2.2 The Topology Generated by a Basis

If B is a basis for a topology, the topology that it generates is defined as TB ={ U " P (X ) : * x0 " U &B0 " B : x0 " B0 . U} .

Thus each basis element is itself an open set.The closed sets in a topology generated by B are Closed(X ) = { C " P (X ) : * x0 /" C &B0 " B : x0 " B0

-B0 / C = ! } .

Furthermore, it can be shown (lemma 13.1) that TB =. +

" $ A B " : B " " B * $ " A/

.But this representation is not unique (we can find many di!erent ways to writeunions of basis elements for a given open set).

1.2.2.1 Example The set of all circles in R2 is a basis for the standardtopology on R2.

The set of all rectangles in R2 is a basis for the standard topology R2.

5

1.2.2.2 Example If X is any set,+

x $ X {{ x}} is a basis for the discretetopology.

1.2.3 Example

{ (a, b) : a, b " Q} is a countable basis for the standard topology on R.

1.2.4 Example

{ [a, b) : a, b " Q} is a basis for a topology generating a topology di!erent thanthe lower limit topology on R.

1.2.5 Example

If X = R and B := { ((, , b) : b " R} + { (a, , ) : a " R} then B is not a basisfor a topology because (a, , ) / ((, , a + 2) does not contain any basis element.

1.2.6 Example

If S 0= ! is some set, define X := {{ bn } : bn " S* n " N} . If % is some nitesequence, then let B# be the set of all sequences that begin with %. ThenB := { B# : % is a finite sequence} is a basis for a topology on X .

1.2.7 Lemma 13.2 Going from Topology to Basis

Let X be a set and let T be a topology on it.Let C be a subset of T such that * U " T , * x " U &C " C : x " C . U.Then C is a basis for T .

1.2.8 Lemma 13.3Comparing Bases

Let B and B! be bases for topologies T and T ! respectively. Then T ! is nerthan T if and only if * x " X, * (B " B : x " B ) &B ! " B ! : x " B ! . B .

1.2.9 Some topologies on R

1. The Standard Topology is given by the basis { (a, b) | a, b " R 1 a < b} .

2. The Lower Limit Topology is given by the basis { [a, b) | a, b " R 1 a < b} .Rl denotes the induced topological space.

3. The K-Topology is given by the basis { (a, b) | a, b " R 1 a < b}+{ (a, b) \ K | a, b " R 1 a < b}where K :=

.1n | n " N

/. RK denotes the induced topological space.

1.2.9.1 Lemma 13.4 Open(Rl ) " Open(R) and Open(RK ) " Open(R)but Open(Rl ) \ Open(RK ) 0= ! and Open(RK ) \ Open(Rl ) 0= ! .

6

1.2.10 Denition of a Subbasis for a Topology

A subbasis for a topology is a set S . P (X ) such that#+

S$S S$

= X .The basis generated by S is given by BS = {

, ni =1 Si : Si " S , n " N} .

Note: The topology generated by BS is the coarsest topology to contain S.Note: The topology generated by some subbasis may not have as a basis thatgiven subbasis.

1.3 14 The Order Topology

Let X be a set with a total order < . That is, if a, b, c" X then:

1. Antisymmetry: ((a < b) 1 (b < a)) % a = b.

2. Transitivity: ((a < b) 1 (b < c)) % a < c .

3. Totality: ((a < b) 2 (b < a)) (this implies reflexivity).

1.3.1 Denition of the Order Topology

Let X be a set such that |X | > 1.The order topology is given by the basis

{ (a, b) | a, b " X 1 a < b}+{ [a0, b) | b " X, a 0 < x * x " X }+{ (a, b0] | a " X, x < b 0 * x " X }

Note that if X has no smallest (largest) element then there are no elements inthe second (third) set, which is still perfectly fine.

1.3.2 Example 1

The standard topology on R is the same as the order topology on R with therelation 3 on numbers.

1.3.3 Example 2The Dictionary Order on R2 (lexicographical or-der)

The dictionary order on R2, ! , is defined as:!xy

"!

!x !

y!

"$% (x < x !) 2 ((x = x !) 1 (y < y !))

Under ! , R2 has no smallest or largest element, so the order topology on R2

is given by the basis0%!

ab

",

!cd

"&|

!ab

"!

!cd

"1

!ab

",!

cd

"" R2

1.

1.3.4 Example 3The Discrete Topology on N

N has a smallest element under the order 3 , namely, the number 1. Thus theorder topology on N with 3 is the discrete topology. { 1} = [1 , 2), which is abasis element, so { 1} is open. If n " N\ { 1} then { n} = ( n ( 1, n + 1) which isalso a basis element.

7

1.3.5 Example 4The Dictionary Order on { 1, 2} ! N

The order topology in the dictionary order of this set is not discrete:0!

21

"1

is not open. Proof: Assume otherwise, thus &B2,1 " B :!21

"" B2,1 .

0!21

"1, where B is the set of basis elements in the order topology, that is, B =

0%!in

",

!im

"&| i " { 1, 2} 1 n < m

1+

0%!1n

",

!2m

"&1+

0[!11

",

!im

") | (i = 1 1 m > 1) 2 i = 2

1.

So it is clear that the only basis element that contains!21

"contains also

elements of the form!

1n

"which is a contradiction to B2,1 .

0!21

"1. Note

that any other one-point set is open, in particular,0!

11

"1is open because

0!11

"1= [

!11

",

!12

") is a basis element.

1.3.6 Rays

(HWS02E01)If X is an totally ordered set, and a is any element in X :The following sets are open in the order topology, and are called open rays:

(a, , ) , ((, , a).Proof: If X has a largest element, called b0 then (a, , ) = ( a, b0] which is a

basis element.If X has no largest element, then (a, , ) =

+x $ X, x>a (a, x), and (a, x) is a

basis element. Similarly for ((, , a).The following sets are closed in the order topology, and are called closed

rays: [a, , ), ((, , a].As it turns out, C := { (a, , ) | a " X } + { ((, , a) | a " X } is a subbasisfor

a topology on X .Claim: C generates the order topology.Proof: The basis for the topology generated by C is given by BC = {

, ni =1 Ci : Ci " C , n " N} .

Thus using Lemma 13.3 we can compare BC with the basis that generates theorder topology:

. BC generates a topology finer than the order topology.Take any x " X and any basis element B in the order topology containing

x.Case 1: B is of the form (a, b).Then (a, b) = ( a, , ) / ((, , b) is a basis element in BC.Case 2: B is of the form [a0, b), then B is itself an open ray, namely, ((, , b).Case 3: B is of the form (a, b0]. Similar to case 2.- The order topology is finer than the topology generated by BC.

8

Take any x " X and any basis element B " B C containing x. Since B iscomposed of nite intersections of open sets (we proved the Ci sthe open rays,are open in the order topology), it is itself open in the order topology. Thusby the definition of a basis for a topology &B ! in the basis generating the ordertopology such that x " B ! . B .

QED

1.4 17 Closed Sets and Limit Points

1.4.1 Denition of Closed Sets

A set A " P (X ) is said to be closed if X \ A " Open(X ).

1.4.2 ExampleClosed Intervals in R

[a, b] " Closed(R) because R\ [a, b] = ( (, , a) + (b, , ) " Open(R).

1.4.3 ExampleClosed Rays in R

[a, , ) " Closed(R) because R\ [a, , ) = ( (, , a) " Open(R).

1.4.4 ExampleNeither closed nor open intervals in R

[a, b) /" Open(R) 1 [a, b) /" Closed(R).

1.4.5 ExampleClosed in R20!

xy

"| x 4 0 1 y 4 0

1= [0 , , )! [0, , ) is closed in R2, because R2\ ([0, , ) ! [0, , )) =

(R ! ((, , 0)) + (( (, , 0) ! R) is open.

1.4.6 ExampleThe Co-Finite Topology

In the co-finite topology on a set X , Closed(X ) = { X }+{ A " P (X ) | |A | < ,} .

1.4.7 ExampleThe Discrete Topology

In the discrete topology on a set X , Open(X ) = P (X ). But * A " P (X ),(X \ A) " P (X ). Thus (X \ A) " Open(X ). Thus A " Closed(X ). SoClosed(X ) = P (X ).

1.4.8 ExampleClopen sets in the Subspace Topology

Consider Y := [0 , 1] + (2, 3) endowed with the subspace topology from R(open sets are intersections of open sets in the ambient space with the space).Thus [0, 1] " Open(Y ) because (( 0.0001, 1.0001) " Open(R) and [0, 1] =(( 0.0001, 1.0001)/ Y . (2, 3) " Open(Y ) as well because (2, 3) " Open(R) and(2, 3) = (2 , 3) / Y .

9

But [0, 1] = Y \ (2, 3), so [0, 1] " Closed(Y ).Similarly, (2, 3) = Y \ [0, 1], so (2, 3) " Closed(Y ).

1.4.9 Theorem 17.1Parallel Universe of Structures of Closed Sets

(HWS01E06)Let X be a topological space. Then the following conditions hold:

1. ! , X " Closed(X ).

2. If A is some set (finite, countable, or uncountable) and if F" " Closed(X ) * $ "A, then

#," $ A F"

$" Closed(X ).

3. If n " N and Fi " Closed(X ) * i " { 1, 2, . . . , n} then (+ n

i =1 Fi ) "Closed(X ).

Proof using de Morgans laws.Note: Kuratowskis Axioms (HWS02E02)

1.4.10 Theorem 17.2Criterion for Closed Sets in Subspace Topol-ogy

Let Y be a subspace of X . Then a set A " Closed(Y ) i! &F " Closed(X ) suchthat A = F / Y .

1.4.11 Theorem 17.3(another) Criterion for Closed Sets in Sub-space Topology

Let Y be a subspace of X . If A " Closed(Y ) and Y " Closed(X ) thenA " Closed(X ).

1.4.12 Interior and Closure

The interior of a set A . X is defined as

int (A) :=2

V %A, V $ Open (X )

V

The closure of a set A . X is defined as

closure (A) :=3

C & A, C $ C losed (X )

C

Some properties of interior and closure:

closure (A) " Closed(X )

int (A) " Open(X )

int (A) ) A ) closure (A)

A " Open(X ) % A = int (A)

A " Closed(X ) % A = closure (A)

10

1.4.13 Boundary

(HWS02E02)The boundary of A is defined as &A := closureX (A) + closureX (X \ A).Another possibility: &A = closureX (A) \ int X (A)..Note:

A " Open(X ) $% A = closureX (A) \ &A.

A " Open(X ) / Closed(X ) $% &A = ! .

X \ (int X (A)) = closureX (X \ A)

&&A0= &A. For example, with A = Q =% &A = R =% &&A= &R = ! .

closureX (int X (closureX (int X (A)))) = closureX (int X (A)) .

int X (closureX (int X (closureX (A)))) = int X (closureX (A)) .

&(closureX (A)) 0= &A. For example, A = Q, &A = R, &(closureX (A)) =&R = ! .

1.4.14 Theorem 17.4Closure in Subspace or Ambient Space

Let Y be a subspace of X . Let A " P (Y ). Then:

closureY (A) = closureX (A) / Y

1.4.15 [I] Theorem 17.5Another Characterization of Closure

Let X be a topological space and A " P (X ).

1. closure (A) = { x " X | Ux / A 0= ! * Ux " Open(X ) s.t. x " Ux } .

2. If a basis for a topology on X , B, generates Open(X ), then closure (A) ={ x " X | Bx / A 0= ! * Bx " B s.t. x " Bx } .

Proof of 1. by contradiction in both directions of containment.

1.4.16 Neighborhoods

If X is a topological space and x " X , a neighborhood of x, U, is a subset ofX such that &V " Open(X ) : x " V, V . U. This is equivalent to saying thatx " int (V ).

Some authors (Munkres for instance) define a neighborhood already to beopen, others only require it to contain an open set containing x.

From this point onwards in this text we shall follow Munkres convention forneighborhoods.

In this terminology, x " closure (A) i! every neighborhood of x intersectsA.

11

1.4.17 ExampleClosure of Subsets in R

closureR ((0, 1]) = [0 , 1] because every neighborhood of 0 intersects (0, 1],and every point outside of [0, 1] has a neighborhood that does not intersect(0, 1].

If B =.

1n | n " N

/then closureR (B ) = B + { 0} .

If C = { 0} + (1, 2) then closureR (C) = [1 , 2] + { 0} .

closureR (Q) = R

closureR (N) = N

closureR ({ x " R | x > 0} ) = { x " R | x 4 0}

If Y := (0 , 1] . R then A :=#0, 12

$" P (Y ). closureR (A) =

40, 12

5. Thus

by theorem 17.4, closureY (A) =40, 12

5/ Y = (0 , 12 ].

1.4.18 Limit Points Denition

If X is a topological space and and A " P (X ) and if x " X , we say that xis a limit point, or a cluster point, or a point of accumulation of A if everyneighborhood of x intersects A in some point other than x itself (that is, if* U " Open(X ) s.t. x " U, A / (U\ { x} ) 0= ! ). Said di!erently, x is a limitpoint of A if x " closureX (A\ { x} ).

1.4.19 ExampleLimit Points of Subsets in R

The set of limit points of (0, 1] in R is [0, 1].

B =.

1n | n " N

/has 0 as its single limit point.

C = { 0} + (1, 2) has [1, 2] as the set of its limit points.

Q has R as the set of its limit points.

N has no limit points in R.

{ x " R | x > 0} has { x " R | x 4 0} as the set of its limit points.

1.4.20 Theorem 17.6Closure and the Set of Limit Points of a Set

Let X be a topological space and let A " 2X .Define A ! := { x " X | x " closureX (A\ { x} )} (which is the set of all limit

points of A in X ).Claim: closureX (A) = A + A !Proof by inclusion of both sides.

12

1.4.21 Corollary 17.7 A " Closed(X ) 5 A ! . A

Proof: A " Closed(X ) 5 A = closureX (A), but by Theorem 17.6, closureX (A) =A + A ! .

So we get A " Closed(X ) 5 A = A + A ! 5 A ! . A.

1.4.22 Hausdor! Spaces

1.4.23 Denition of Convergence in Topological Spaces

Let X be a topological space. Let f : N 6 X be a map. We can think of f (n)as a sequence.

The sequence of points f (N) . X converges to the point a " X i! * U "Open(X ) such that a " U, &maU " N such that f ({ n " N | n 4 maU } ) . U.

If we consider N with the co-finite topology, then that translates to:What is the topology on N so that f 6 a i! f is continuous at , ?(23:53) (+zeno) You actually need the one-point compactification of N, call

it N~ which is N with the point oo adjoined, and a set U in N~ is open i! eitheroo is not in it, or else oo is in it and N\U is finite. (23:53) (+zeno) THen thesequence f(1), f(2), ... of points of X converges the point x = f(oo) i! f : N~ ->X is continuous at oo.

(HWS03E05) (b) If a sequence does not converge in a given topology, itwill also not converge in a finer topology.

1.4.23.1 Example with the Cocountable Topology (HWS02E06) (c)A sequence { xn } converges to x in (R, Tcocountable ) $% & m " N such that

* n 4 m, xn = x.

1.4.24 One-point sets are closed in R or R2

{ x0} is closed in either one of R or R2 becaues (according to 1.4.15) everypoint di!erent from x has a neighborhood that does not intersect { x0} , soclosure ({ x0} ) = { x0} , and so { x0} would be closed in either R or R2.

In R or R2, a sequence cannot convergence to more than point.

1.4.25 ExampleOne-point sets are not always closed, and conver-gence is strange

Take X := { a, b, c} with the topology Open(X ) := { X, ! , { a, b} , { b} , { b, c}} .Claim: { b} /" Closed(X ) Proof: X \ { b} = { a, c} /" Open(X ).Claim: The sequence { x i } where x i := b* i " N convergences to the point b

and to the points a and to c. Proof: The open sets that contain a are exactlyX and { a, b} . Since x i = b is in both of these * i " N, x i 6 a. Same goes for c.

13

1.4.26 Denition of Hausdor! Spaces

A topological space X is called Hausdor! i! * x1, x2 " X such that x1 0= x2,&Ui " Open(X ) : x i " Ui * i " { 1, 2} and U1 / U2 = ! .

1.4.27 Theorem 17.8 |A | < , % A " Closed(X ) for Hausdor! X

Proof:Case 1: A = { x0} . Take any y " X \ { x0} . Since X is Hausdor!, we know

that there exist two disjoint neighborhoods of x0 and y respectively. But thatmeans that every point y 0= { x0} has a neighborhood that does not intersect{ x0} , so that closureX ({ x0} ) = { x0} .

Case 2: A = { x1, x2, . . . , xn } for some n " N. Since A is the finiteunion of closed sets (by case 1), by 1.4.9 (parallel universe for closed sets)A " Closed(X ).

1.4.28 Examplenon-Hausdor! Space with point-sets being closed

Claim: R with the co-finite topology is not a Hausdor! space.Proof: Take any x, y " R such that x 0= y. Any neighborhood of x is of the

form R\ { x1, x2, x3, . . . , xn } where n " N and the same for any neighborhoodof y. The intersection of any two such neighborhoods will thus certainly not bedisjoint.

Claim: Every finite set in R with the co-finite topology is closed.Proof: The complement of every such set, by definition, is open.

1.4.29 T1 axiom

Every finite set is closed.

1.4.30 [I] Theorem 17.9Characterization of limit points in T1 spaces

Let X be a T1 space, let A " 2X , and let x " X .Claim: x " A ! 5 * U " Open(X ) s.t. x " U, |U / A | = , .Proof: $ This way works always (we dont have to assume T1).If every neighborhood of x intersects A in infinitely many points, then it

must intersect A\ { x} . Thus x " A ! .% Assume that x " A ! , and assume that the theorem does not hold, that is,

&U0 " Open(X ) such that x " U0 and |U0 / A | < , . Thus |U0 / (A\ { x} )| a [a

! , b) " Open(Rl ).

1.5.8 Denition of Continuity at a Single Point

Let X, Y be topological spaces and let f : X 6 Y be a map. Let x0 " X . f is(defined to be) continuous at x0 i! * neighborhood V of f (x0), &a neighborhoodU of x0 such that f (U) . V .

1.5.9 Theorem 18.1Characterization of Continuity

Let X, Y be topological spaces and let f : X 6 Y be a map. Then th efollowingare equivalent:

1. f is continuous (according to 1.5.1).

2. * A " 2X , f (closureX (A)) . closureY (f (A)) .

3. * B " Closed(Y ) , f ) 1 (B ) " Closed(X ).

4. f is continuous at x, * x " X (according to 1.5.8).

Proof:1. % 2.

(compare with Rudins PMA Exercise 4.2)Take some x " closureX (A).Take a neighborhood V of f (x). By 1.5.1 f ) 1 (V ) " Open(X ) and observe

that x " f ) 1 (V ). Thus f ) 1 (V ) is a neighborhood of x. But x " closureX (A),so by 1.4.16, &x ! " X such that x ! " f ) 1 (V ) / A. Thus f (x !) " V / f (A) %V / f (A) 0= ! . So we conclude that every neighborhood of f (x) intersectsf (A). So that again by 1.4.16 f (x) " closureY (f (A)) .

2. =% 3.Take some B " Closed(Y ).Claim: closureX

#f ) 1 (B )

$. f ) 1 (B )

Proof: Take some x " closureX#f ) 1 (B )

$. By the previous part, f (x) "

closureY#f

#f ) 1 (B )

$$.

17

But it is a fact of set theory that f#f ) 1 (B )

$. B (in fact if f is surjective

there will be an equality).In addition, by 1.4.12, it is clear that any closed set that contains a set is

going to contain its closure, so if f#f ) 1 (B )

$. B % closureY

#f

#f ) 1 (B )

$$.

closureY (B ) = B .So we conclude that f (x) " B % x " f ) 1 (B ).Thus f ) 1 (B ) " Closed(X ).3. % 1.

Take some V " Open(Y ). Thus Y \ V " Closed(Y ). Thus f ) 1 (Y \ V ) "Closed(X ). But f ) 1 (Y \ V ) = f ) 1 (Y ) \ f ) 1 (V ) = X \ f ) 1 (V ). Thus f ) 1 (V ) "Open(X ). Thus 1.5.1.

1. % 4.Every neighborhood of f (x), V has f ) 1 (V ) as a neighborhood of x, such

that f#f ) 1 (V )

$. V .

4. % 1.Take V " Open(Y ). If f ) 1 (V ) = ! we are done as ! " Open(X ). Other-

wise, &x " f ) 1 (V ). Then f (x) " V and so V is a neighborhood of f (x). By4., &Ux , a neighbhorhood of x such that f (Ux ) . V . Observe that this leads toUx . f ) 1 (V ). Thus f ) 1 (V ) =

+x $ f " 1 (V ) Ux " Open(X ).

1.5.10 The Initial Topology

(HWS03E03)Let Y, I be sets and let (X i , Ti ) be a topological space, and f i : Y 6 X i be

a map * i " I .Define the initial topology on Y with respect to f i as the topology generated

by the subbasis+

i $ I f) 1i (Ti ).

The initial topology is the coarsest topology such that f i : (Y, T ) 6(X i , Ti ) is continuous * i " I .

If (Z, V) is a topological space and f : Z 6 Y is a map, then f iscontinuous $% f i 7 f is continuous * i " I .

If A . X , then the initial topology on A with respect to the inclusion mapis the subspace topology on A.

If (X i , Ti ) i $ I is a family of topological spaces and f i are the projectionsf i :

7j $ I X j 6 X i , then the initial topology on

7j $ I X j with respect to

the projections f i is the product topology.

1.5.11 The Final Topology

(HWS03E04)Let Y, I be sets and let (X i , Ti ) be a topological space, and f i : X i 6 Y be

a map * i " I .Define the final topology on Y with respect to f i as T :=

.O . Y : * i " I, f ) 1i (O) " T i

/.

18

The final topology is the finest topology on Y such that f i is continuous* i " I .

If (Z, V) is a topological space and f : Y 6 Z is a map, then f iscontinuous $% f 7 f i is continuous * i " I .

If X is a set Ti are a bunch of topologies on it, and id i : (X, Ti ) 6 X , thefinal topology on X with respect to id i is

,i $ I Ti .

1.5.12 The Co-Product Topology (Disjoint Union Topology)

(HWS03E04) (c) and (d)

The disjoint union of two topological spaces X 1 and X 2 is defined as0!

x1

": x " X 1

1+

0!x2

": x " X 2

1. If f 1 : X 1 )6 X 1

8X 2 and f 2 : X 2 )6 X 1

8X 2 are the two in-

clusion maps, then the final topology with respect to f 1 and f 2 is { U18

U2 . X 18

X 2 : U1 " Open(X 1) 1 U2 " Open(X 2)} .Similarly, for any collection of topological spaces, the final topology with

respect to the inclusion maps is. 8

i $ I Ui .8

i $ I X i : Ui " Open(X i ) * i " I/

.This is dened as the coproduct topology.

Example: Take { 0, 1} with the discrete topology.8

n $ N { 0, 1} with thecoproduct topology is discrete.

1.5.13 Counter Example

(HWS03E03) (d)Take { 0, 1} with the discrete topology. Observe that { 0, 1} N with the prod-

uct topology is not discrete!

1.5.14 Homeomorphism

Let X, Y be topological spaces and let f : X 6 Y be a bijection. If both f andf ) 1 : Y 6 X are continuous, then f is called a homeomorphism.

Equivalently, a bijection f : X 6 Y is a homeomorphism if the following istrue: U " Open(X ) 5 f (U) " Open(Y ).

1.5.15 Example

(HWS02E05)(a, b) is homeomorphic to (0, 1)[a, b] is homeomorphic to [0, 1] if a, b are finite.

1.5.16 Example

(HWS03E02)The cantor set is homeomorphic to { 0, 1} N.

19

1.5.17 Example

()The map idX : (X, T ) 6 (X, T !) is continuous i! T is finer than T ! , and

idX is a homeomorphism i! T = T ! .

1.5.18 Topological Imbedding (=Embedding)

Let X, Y be topological spaces and let f : X 6 Y be an injection . If thebijection g : X 6 f (X ) defined by g (x) := f (x) * x " X happens to be ahomeomorphism, f is called a topological imbedding of X in Y .

1.5.19 CounterexampleBijective continuous non-homeomorphism

One example is g defined in 1.5.7.Another example is the following (Compare with Example 4.21 in Rudins

PMA):

Let S1 :=0!

xy

"" R2 | x2 + y2 = 1

1be a topological space endowed with

the subspace topology from R2.Let [0, 1) be a topological space endowed with the order topology.

Define f : [0, 1) 6 S1 by f (t) :=!cos (2!t )sin (2!t )

"* t " [0, 1).

Observe that f is bijective, and continuous.However, f ) 1 is not continuous:Take the set [0, 14 ) " Open([0, 1)).Claim: f

#[0, 14 )

$/" Open

#S1

$Proof: Every basis element containing the

point f (0) =!10

"is not contained inside of f

#[0, 14 )

$.

In addition, if we define a new function g : [0, 1) 6 R2 by g (t) := f (t) * t "[0, 1) then g is a counterexample for a continuous injective map that is not atopological imbedding.

1.5.20 Theorem 18.2Rules for Constructing Continuous Functions

Let X, Y, and Z be topological spaces.

ConstantFunction If y0 " Y and f : X 6 Y is defined such that f (X ) = { y0}then f is continuous.

Proof: Take any V " Open(Y ).Case 1: y0 " V , then f ) 1 (V ) = X " Open(X ).Case 2: y0 /" V , then f ) 1 (V ) = ! " Open(X ).

Inclusion If A is a subspace of X , then the inclusion function j : A 6 Xdefined by j (a) := a * a " A is continuous.

Proof: Take any V " Open(X ). Then j ) 1 (V ) = A / V . By definition of thesubspace topology, this set is " Open(A).

20

Composition If f : X 6 Y and g : Y 6 Z are two continuous functions thenthe map g 7 f : X 6 Z is continous. (HWS02E04)

Proof: Observe that f ) 1#g) 1 (V )

$= ( g 7 f )) 1 (V ).

DomainRestriction If f : X 6 Y is continous and if A is a subspace of Xthen the restricted function f |A : A 6 Y is continous.

Proof: (HWS03E01) f |A = f 7j , where j is the inclusion map. Apply the above.

DomainRestrictionOrRangeExpansion Let f : X 6 Y be continuous. Iff (X ) . Z . Y then the function g : X 6 Z obtained by restricting therange of f is continuous. If Z - Y then the function h : X 6 Z obtainedby expanding the range of f , is continous.

Proof: Take some B " Open(Z ). Thus &U " Open(Y ) such that B = Z / U.Since Z - f (X ), observe that g) 1 (B ) = f ) 1 (U) " Open(X ).

Next, observe that h = j 7 f where j : Y 6 Z is the inclusion map.

LocalFormulationOfContinuity f : X 6 Y is continous if & {U" " Open(X )}such that X =

+" U" and f |U! is continuous * $ .

Proof: Take some V " Open(Y ).By assumption, & {U" " Open(X )} such that X =

+" U" and f |U! is con-

tinuous * $ .Observe that f ) 1 (V ) =

+"

#f ) 1 (V ) / U"

$=

+"

9#f |U!

$) 1(V )

:

But since f |U! are continuous,#

f |U!$) 1

(V ) " Open(U" ). Thus &V" "Open(X ) such that

#f |U!

$) 1(V ) = U" / V" . Thus

#f |U!

$) 1(V ) " Open(X ).

So that+

"

9#f |U!

$) 1(V )

:" Open(X ).

1.5.21 Theorem 18.3The Pasting Lemma

(HWS03E01 (b) the gluing lemma)Let X = A + B where A, B " Closed(X ). Let f : A 6 Y and g : B 6 Y be

continous. If f (x) = g (x) * x " A / B then the function h : X 6 Y defined by

h (x) :=

6f (x) x " Ag (x) x " B

is continuous.

Proof:Take any V " Open(Y ). Observe that h) 1 (V ) = f ) 1 (V ) + g) 1 (V ) "

Open(X ).Note: This theorem wouldve worked with open sets as well, in which case

it wouldve been a special case of the local formulation of continuity in 1.5.20.Note: We needed f (x) = g (x) * x " A / B for h to be well-defined.

21

1.5.22 CounterexampleDiscontinxuous Because the Domains arenot both closed or both open

The map l : R 6 R, both endowed with the standard topology, defined by

l (x) :=

6x ( 2 x " ((, , 0)x + 2 x " [0, , )

is not continuous. (1, 3) " Open(R) but l ) 1 ((1, 3)) =

[0, 1) /" Open(R).

1.5.23 Theorem 18.4Products of Continuous Functions

Let f x : A 6 X , f y : A 6 Y be maps and let there be a map f : A 6 X ! Y

defined by f (a) :=!f x (a)f y (a)

"* a " A.

Then f is continuous 5 both f x and f y are continuous.The maps f x , f y are called coordinate functions of f .Proof:%

Let ! x : X ! Y 6 X and ! y : X ! Y 6 Y be the projection maps:

! x

%!xy

"&= x and ! y

%!xy

"&= y *

!xy

"" X ! Y .

Claim: ! x : X ! Y 6 X is continuousProof: Take U " Open(X ). Then ! ) 1x (U) = U ! Y . Since Y " Open(Y ),

by the definition of the product topology, U ! Y " Open(X ! Y ).Simiarly, ! y is continuous.Observe that f x = ! x 7 f and f y = ! y 7 f .$

Following 1.5.2, take a basis element of X ! Y : Ux ! Uy such that Ux "Open(X ) and Uy " Open(Y ). Observe that f ) 1 (Ux ! Uy ) = f ) 1x (Ux ) /f ) 1y (Uy ) " Open(A).

Note: # useful criterion for when f : A ! B 6 X is continuous.

1.6 16 The Subspace Topology

1.6.1 Denition

Let X be a topological space and let Y " 2X . The subspace topology onY is defined as follows: Open(Y ) := { Y / U | U " Open(X )} . With such atopology, Y is now called a subspace of X .

1.6.2 Characterization with Closed Sets

Let X be a topological space and let Y " 2X have the subspace topology.Claim: If K " Closed(Y ) then K = F / Y where F " Closed(X ).Proof: Let K " Closed(Y ).% Y \ K " Open(Y )% Y \ K = Y / U where U " Open(X ).% K = Y \ (Y / U) = Y / (Y \ U).

22

"

1.6.3 Lemma 16.1Subspace Topology by Basis

If B is a basis that generates Open(X ) then BY := { B / Y | B " B} is a basisthat generates the subspace topology on Y .

Proof:Following 1.2.7: Take any U " Open(X ). Then U / Y is an arbitrary element

in Open(Y ). Take any y " U / Y . Since U " Open(X ), there exists By " Bsuch that y " By . U. But that means y " By / Y . U / Y .

1.6.4 Lemma 16.2Open Subspaces are Cool

Let Y be a subspace of X . If U " Open(Y ) and Y " Open(X ) then U "Open(X ).

Proof:U " Open(Y ), so &Ux " Open(X ) such that U = Ux / Y . So U is the finite

intersection of two open sets. nu! said.

1.6.5 Lemma 16.3The Subspace topology on a Product is theproduct topology of two subspaces

If A is a subspace of X and B is a subspace of Y , then the product topologyon A ! B where A and B are both endowed with the subpsace topology is thesame as the subspace topology on A ! B as a subspace in X ! Y .

Proof:Claim: The basis that generates the subspace topology on A ! B is the same

as the basis that generates the product topology on A ! B . X ! Y .Proof: The general basis element for X ! Y is U ! V where U " Open(X )

and V " Open(Y ). Thus (U ! V ) / (A ! B ) is the general basis element of thesubspace topology of A ! B . X ! Y .

Observe that (U ! V ) / (A ! B ) = ( U / A) ! (V / B ).Observe that U / A is the general open set in the subspace topology of

A . X and V / B is the general open set in the subpsace topology of B . Y ,(U / A) ! (V / B ) is the general basis element in the product topology on A ! B .

Thus since the two basis are the same they generate the same topology.

1.6.6 The Order Topology and the Subspace Topology Do Not Al-ways Agree!

1.6.6.1 Example of Agreement Take R with the standard topology andY := [0 , 1] with the subspace topology. Thus a general basis element in the

subpsace topology is of the form: (a, b) / Y :=

;

But this basis generates the order topology, so we have agreement.

1.6.6.2 Example of Disagreement Take R with the standard topologyand Y := [ 0 , 1 ) + { 2} . In the subpsace topology on Y , { 2} is open, because{ 2} = Y / (1.999, 2.0001). However in the order topology on Y , { 2} is not open:any basis element of the order topology which would contain 2 is of the form(a, 2] / Y where a " Y , which necessarily cannot be contained in { 2} , becauseit necessarily contains points below it.

1.6.6.3 Example of Disagreement in the Plane Let I = [0 , 1]. Thedictionary order on I ! I gives rise to a topology di!erent than the subspacetopology on I ! I from R ! R with the order topology of the dictionary order.

For example, the set.

12

/! ( 12 , 1] is open in I ! I in the subspace topology

(it is, for example, the intersection of.

12

/!

#12 , 1.111

$" Open

#R2

$and I 2,

which is open in the subspace topology), but not in the order topology, as!1/ 21

"

is not the largest element in I ! I with the dictionary order topology, and sothe set cannot be open.

The set I ! I in the dictionary order topology is called the ordered squareand is denoted by I 20 .

1.6.7 Dention of Convex Sets

Let X be an ordered set. Y " 2X is called convex in X if for each a, b " Y suchthat a < b, { x " X | a < x < b } . Y .

Note that intervals and rays in X are convex in X .

1.6.8 Lemma 16.4Convex Subspaces have the same order topology

Let X be an ordered set in the order topology. Let Y " 2X such that Y convexin X . Then the order topology on Y is the same as the topology Y inherits asa subspace of X .

Proof:Claim: The order Topology on Y contains the subspace topology on YProof: Take any open element in X , it will look like this:

+" $ A (a" , , ) ++

$ $ B ((, , a$ ) where a" " X * $ " A and a$ " X * * " B .Thus any open element in the subspace topology on Y will look like this:9+" $ A (a" , , ) +

+$ $ B ((, , a$ )

:/ Y =

+" $ A ((a" , , ) / Y )+

+$ $ B (( (, , a$ ) / Y ).

But (a" , , ) / Y and ((, , a$ ) / Y are both open in the order topology onY , because they are either open rays in Y (in case a" or a$ are in Y ) or theyare the empty set of they are the whole of Y , in either one of these cases, theywould be open sets.

Claim: The subspace topology on Y contains the order topology on YProof: Take a general open set in the order topology on Y :

+" $ A { x " Y | x > a " }++

$ $ B { x " Y | x > a $ } where a" " Y * $ " A and a$ " Y * * " B .

24

We can write this as an open set in the subspace topology on Y :+

" $ A { x " X | x > a " }/Y +

+$ $ B { x " X | x > a $ } / Y .

1.7 19 The Product Topology

Let7

" $ A X " be the Cartesian product (where A may be finite, countable, oruncountable) of the topological spaces X " .

1.7.1 Denition of the Box Topology

The box topology is generated by the basis Bbox :=. 7

" $ A U" | U" " Open(X " )/

.

1.7.2 Denition of the Product Topology

In analogy with 1.8.5, the collection

S :=2

" $ A

.! ) 1" (U" ) | U" " Open(X " )

/

is a subbasis for the product topology on7

" $ A X " .From this it follows that the basis for the product topology is such that there

are only nitely many open sets in the product which are not the entire space,and the rest of the product has to be the entire space.

Thus a typical basis element has the form:#! ) 1" 1 (U" 1 )

$/

#! ) 1" 2 (U" 2 )

$/ /

#! ) 1" n (U" n )

$=

?

" $ A

U"

where U" = X " whenever $ /" { $ i | i " { 1, . . . , n}} .

1.7.3 Denition of tuple notation

Let J be an index set (finite, countable, or uncountable), and let X be any set.Let x : J 6 X a map, thus (x" )" $ J is the set of points in X indexed by J . AllJ -tuples of elements of X are denoted by X J .

1.7.4 Theorem 19.1Comparison of the box and product topologies

Observe: for finite sets the box and product topology are identical. Secondly,the box topology is finer than the product topology in general.

1.7.5 Theorem 19.2Basis generating product spaces

Let7

" $ A X " be the Cartesian product (where A may be finite, countable, oruncountable) of the topological spaces X " , each of which is generated by thebasis B" .

The basis. 7

" $ A B " | B " " B "/

generates the box topology.The basis

. 7" $ A B " | B " " B " for finitely many$ " A, otherwise B " = X "

/

generates the product topology.

25

Proof:TODO

1.7.6 Theorem 19.3Subspace product topology

Let7

" $ A X " be the Cartesian product (where A may be finite, countable, oruncountable) of the topological spaces X " , each of which has a subspace A " .

Thus7

" $ A A " is a subspace of7

" $ A X " if both products are given the boxtopology, or if both products are given the product topology.

Proof:TODO

1.7.7 Theorem 19.4Hausdor! Products7

" $ A X " is Hausdor! in both the box and product topologies.Proof:TODO

1.7.8 Theorem 19.5Product of Closures

Let A " . X " for all $ " A. Then in both the product and box topologies,

?

" $ A

closureX ! (A " ) = closure!

! # A X !

@?

" $ A

A "

A

Proof:TODO

1.7.9 Theorem 19.6Continuous functions in the product topology

Let f : A 67

" $ J X " be given by the equation f (a) := ( f " (a))a$ J wheref " : A 6 X " for each $ " J . Let

7" $ J X " have the product topology. Then

the function f is continuous 5 each function f " is continuous.Proof:%

Take some * " J .! $ is continuous because if U$ " Open(X $ ), then ! ) 1$ (U$ ) is a subbasis

element in the product topology and thus is open.f b = ! $ 7 f .Since both these functions are continuous, their composition is also contin-

uous.$

Take any subbasis element from the subbasis of the product topology:! ) 1$ (U$ ) where * " J and U$ " Open(X $ ).

Observe that because f b = ! $ 7 f , f ) 19

! ) 1$ (U$ ):

= f ) 1$ (U$ ) " Open(A)as f $ is continuous.

So by 1.5.3 f is continuous.

26

1.7.10 ExampleDiscontinuous in Product Topology

RN =7

n $ N R

Define f : R 6 RN by the equation f (t) :=

(tt

. . .

)

* .

Thus f n (t) := t.Each f n : R 6 R is continuous.By the above theorem, RN with the product topology is continuous then.Claim: f is discontinuous if RN is given the box topology.Proof: Take the basis element B := ( ( 1, 1) !

#( 12 ,

12

$!

#( 13 ,

13

$! . . .

in the box topology basis. Assume f ) 1 (B ) " Open(R). Thus &( > 0 suchthat (( (, ( ) . f ) 1 (B ). Thus f (( ( (, ( )) . B . Apply ! n to both sides of theinclusion to get:

f n (( ( (, ( )) = ( ( (, ( ) .#( 1n ,

1n

$for any n " N

Which is clearly a contradiction.Thus f ) 1 (B ) /" Open(R), so that f is discontinuous.

1.8 15 The Product Topology on X ! Y

1.8.1 Denition of the Product Topology on X ! Y

Let X and Ybe topological spaces. The product topology on X ! Y is the topol-ogy generated by the basis Bproduct := { U ! V | U " Open(X ) 1 V " Open(Y )} .

Claim: Bproduct is a basis for a topology.Proof: Following 1.2.1:

Take any!xy

"" X ! Y . Since X ! Y " B product (as X and Y are open in

their respective topologies), X ! Y would be the basis element containing!xy

".

This satisfies the first condition for a basis.Next, take any two basis elements from Bproduct : U1 ! V1, and U2 ! V2.

Observe that (U1 ! V1) / (U2 ! V2) = ( U1 / U2) ! (V1 ! V2) is a basis elementas the intersection of two open sets is again open. Thus this element fulfils therequirement for the second condition.

1.8.2 Theorem 15.1Going to the basis of the product topologyfrom the constituent basis

If B is the basis for the topology of X and C is the basis for the topology of Ythen the collection D := { B ! C | B " B 1 C " C} is a basis for the topologyof X ! Y .

Proof:D . Open(X ! Y ) because any basis element in the constituent spaces is

open. Thus take any open set in X ! Y ,+

" $ A U" ! V" (that is, U" " Open(X )

and V" " Open(Y )). Next take any point!xy

""

+" $ A U" ! V" (that is x " U$

27

and y " V$ for some * " A). Since B is the basis for Open(X ) &B " B suchthat x " B . U$ . Similarly, &C " C such that y " C . V$ . Thus &B ! C such

that!xy

"" B ! C . U$ ! V$ .

+" $ A U" ! V" . But B ! C " D , so according

to 1.2.7 we are done.

1.8.3 Example 1The basis for Open#R2

$

Thanks to 1.8.2, we can ascertain that Open#R2

$is generated by sets of open

rectangles, rather than products of arbitrary open sets.

1.8.4 Projections

Let ! 1 : X ! Y 6 X be defined by ! 1%!

xy

"&:= x *

!xy

"" X ! Y .

Let ! 2 : X ! Y 6 X be defined by ! 2%!

xy

"&:= y *

!xy

"" X ! Y .

The maps ! 1 and ! 2 are called projections of X ! Y onto its first and secondfactors, respectively.

These maps are surjective.

1.8.5 Theorem 15.2Subbasis for the Product Topology

The collection

S :=22

i =1

.! ) 1i (U) | U " Open(X i )

/

is a subbasis for the product topology on X 1 ! X 2.Proof:Let T denote the product topology on X ! Y . Let T ! denote the topology

generated by S.T ! . T

Note that every element of S belongs to T , and so every arbitrary unions offinite intersections of elements of S (which represent arbitrary members of T !).

T ! - TOn the other hand, every basis element U ! V for the topology T is a finite

intersection of elements of S: U ! V = ! ) 11 (U) / !) 12 (V ). Therefore, U ! V " T ! .

But an arbitrary element of T is an arbitrary union of basis elements, whichT !is closed under, being a topology. Thus T . T ! .

1.9 22 The Quotient Topology

1.9.1 Denition of a Quotient Map

Let X and Y be topological spaces. Let p : X 6 Y be a surjective map. Aquotient map is a map p such that U " Open(Y ) 5 p) 1 (U) " Open(X ).

Note: This condition is stronger than continuity.

28

Equivalently we could define that U " Closed(Y ) 5 p) 1 (U) " Closed(X ),because f ) 1 (Y \ B ) = X \ f ) 1 (B ).

1.9.2 Denition of a Saturated Subset

A subset C " 2X is saturated (with respect to the surjective map p : X 6 Y )i! C - p) 1 ({ y} ) * y " Y such that C / p) 1 ({ y} ) 0= ! .

1.9.3 Another Characterization of Saturated Set

Claim: C is saturated 5 C = p) 1 (B ) for some B " 2Y .Proof:$

Let C = p) 1 (B ) for some B " 2Y .Take some y " Y such that C / p) 1 ({ y} ) 0= ! 5 p) 1 (B ) / p) 1 ({ y} ) 0= ! .

Thus y " B .Take some x " p) 1 ({ y} ). Thus p (x) = y. Thus p (x) " B . Thus x "

p) 1 (B ) = C. So C is saturated.%

Now assume that C is saturated.Define B :=

.y " Y | p) 1 ({ y} ) / C 0= !

/.

Claim: C = p) 1 (B ).)

Take any x " C.Then p (x) " B , because p) 1 ({ p (x)} ) / C - { x} 0= ! .Thus x " p) 1 (B ).8

Take any x " p) 1 (B ).Thus p (x) " B . Thus p) 1 ({ p (x)} ) / C 0= ! .But { x} . p) 1 ({ p (x)} ) / C. So x " C.

1.9.4 Another Characterization of Saturated Set

Claim: C is saturated 5 C = p) 1 (p (C)) .This follows from 1.9.3 together with the fact that for any set, p) 1 (B ) =

p) 1#p

#p) 1 (B )

$$.

1.9.5 Quotient Maps of Saturated Sets

To say that p is a quotient map is equivalent to saying that p is surjective,continuous and p maps saturated open sets of X to open sets of Y (or saturatedclosed sets of X to closed sets of Y ).

Proof:%

Assume that p is a quotient map. Thus p is continuous by definition.

29

Take any saturated set C " Open(X ). We need to show that p (C) "Open(Y ).

But C is saturated, so that C = p) 1 (B ) for some B " Y . But sinceC " Open(X ), p) 1 (B ) " Open(X ). But p is a quotient map, so B " Open(Y ).But p (C) = p

#p) 1 (B )

$. Observe that p

#p) 1 (B )

$= B since p is surjective.

Thus p (C) " Open(Y ).$

Assume that p is continuous and p maps saturated open sets of X to opensets of Y .

Take any U " Open(Y ). Since p is continuous, p) 1 (U) " Open(X ).Next, let U " 2Y such that p) 1 (U) " Open(X ).But p) 1 (U) is exactly an open saturated set.Thus p

#p) 1 (U)

$= U (because p is surjective) and so by our assumption

U " Open(Y ).

1.9.6 Open Maps and Closed Maps

1.9.6.1 Closed maps and Open maps A map is open if * U " Open(X ) , f (U) "Open(Y ).

A map is closed if * U " Closed(X ) , f (U) " Closed(Y ).

(HWS04)

The product of two closed maps is not necessarily closed: for example,let x0 " R, f " { x0} R then f ! idR : R ! R 6 R ! { x0} is not closed.

1.9.6.2 Connection to Quotient Maps If p : X 6 Y is a surjectivecontinuous map that is either open or closed then p is a quotient map.

However, there are quotient maps that are neither open nor closed:

(17:25) (+zeno) The identification map R -> R/Q (where all pointsof Q are identified together) is a quotient map that sends the openset (0,1) to a nonempty proper subset of R/Q, and also sends theclosed set {0} to a nonempty proper subset of R/Q. But R/Q has noopen nonempty proper subsets, as you can check.

(HWS04) (Munkres 22.3) Define A :=0!

xy

"" R2 : x 4 0 2 y = 0

1.

Then define q : A 6 R by q := ! 1|A . Observe that q is a quotientmap that is neither open nor closed.

1.9.7 Example

Let X := [0 , 1] + [2, 3] . R. Define Y := [0 , 2] . R. Define p : X 6 Y as:

p (x) :=

6x x " [0, 1]x ( 1 x " [2, 3]

Observe that p is surjective, continuous (pasting lemma).

30

Claim: p is closed.

Proof 1:p is continuous, so that it takes compact sets to compact sets.But any closed subset of X is going to be compact, so that p (K ) willalso be compact and so closed in Y .

Proof 2:Take any U " Closed(X ).Thus by 1.6.2 U = X / K where K " Closed(R).% U = ([0 , 1] + [2, 3]) / K = ([0 , 1] / K ) + ([2, 3] / K ).Since ([0, 1] / K ) , ([2, 3] / K ) are two disjoint sets,p (U) = p ([0, 1] / K ) + p ([2, 3] / K )By definition, p ([0, 1] / K ) = [0 , 1]/ K = Y / ([0, 1] / K ) " Closed(Y ).andp ([2, 3] / K ) = [1 , 2] / K ( 1 = Y / ([1, 2] / K ( 1) " Closed(Y ).So that the union of two closed sets in Y is again closed and thusp (U) " Closed(Y )."

Thus p is a quotient map. However, p is not open, because [0, 1] " Open(X )gets mapped to [0, 1] /" Open(Y ).

1.9.7.1 Counterexample Let A := [0 , 1)+ [2, 3] . R. Define Y := [0 , 2] .R. Define p : A 6 Y as:

q(x) :=

6x x " [0, 1)x ( 1 x " [2, 3]

Observe that q is surjective, continuous (pasting lemma), but not a quotientmap: [2, 3] " Open(A) is saturated (q) 1 ([1, 2]) = [2 , 3]) and gets mapped to[1, 2] /" Open(Y ).

1.9.8 Example

Let ! 1 : R2 6 R. Then ! 1 (projection onto the first coordinate) is continuousand surjective. Furthermore, ! 1 is an open map. Thus it is a quotient map.

But ! 1 is not a closed map.

Define C :=0!

xy

"" R2 | xy = 1

1

Claim: C " Closed#R2

$

Take any!ab

"" R2\ C.

We could always pick an open rectangle around!ab

"which doesnt

intersect C. Thus R2\ C " Open#R2

$.

31

However, ! 1 (C) = R\ { 0} " Open(R).

1.9.8.1 Counterexample Let A := C +!00

", then the map q : A 6 R ob-

tained by restricting ! 1 is continuous and surjective, but it is not a quotient map:0!00

"1" Open(A) is saturated in A with respect to q (

0!00

"1= ! ) 11 ({ 0} )),

but its image is not open in R.TODO: Example of non-quotient map.

1.9.9 Denition of Quotient Topology

If X is a topological space and A is a set and if p : X 6 A is a surjective map,then there exists exactly one topology T on A relative to which p is a quotientmap. It is called the quotient topology induced by p.

Claim: T =.

U " 2A | p) 1 (U) " Open(X )/

Proof:Define R :=

.U " 2A | p) 1 (U) " Open(X )

/.

Step 1: R is a topology.! Since p is surjective, p1) (! ) = ! and p) 1 (A) = X and so

! , A " R .! Let U" " R* $ " J . Then

#+" $ J U"

$" R because p) 1

##+" $ J U"

$$=+

" $ J p) 1 (U" ) " Open(X ).

! Let Ui " R* i " { 1, . . . , n} , n " N. Then9,

i ${ 1, ..., n } Ui:

" R

because p) 19,

i ${ 1, ..., n } Ui:

=,

i ${ 1, ..., n } p) 1 (Ui ) " Open(X ).

Step 2: If A is endowed with the topology defined by R then p is aquotient map.

! p is continuous:Take any U " R . Thus p) 1 (U) " Open(X ).

! Take some U " Open(X ) such that U is saturated in X withrespect to p. That is, U = p) 1 (B ) for some B " A. Then B " Rbecause p) 1 (B ) = U " Open(X ).

Step 3: If p is a quotient map then Open(A) = R .

! )Take any U " Open(A). Since p is continuous, p) 1 (U) "Open(X ). Thus U " R .

! 8Take any U " R . Thus p) 1 (U) " Open(X ). But since p is aquotient map that means that U " Open(A).

"

32

1.9.10 Example

Let p : R 6 { a, b, c} be a surjective map defined by p (x) :=

; 0b x < 0c x = 0

.

The quotient topology on { a, b, c} is thus { ! , { a, b, c} , { a} , { b} , { a, b}} .

1.9.11 Denition of Quotient Space

Let X be a topological space, and let X * be a partition of X into disjoint subsetswhose union is X . Let p : X 6 X * be a surjective map that carries each pointof X to an element of X * containing it. In the quotient topology induced by p,the space X * is called a quotient space of X . (Thus a quotient space is alwaystaken with respect to a certain partition of the a given topological space).

X * is also called an identification space, or a decomposition space.Thus U " 2X

$is a collection of chunks of X , and p) 1 (U) is the union of

all the elements belonging to any one of those chunks, all together. So theopen sets in X * would be sets of classes whose union is an open set in X .

1.9.12 Example

Let X be the closed unit ball0!

xy

"" R2 | x2 + y2 3 1

1and let X * :=

.S1

/+

00!xy

"1" 2R

2| x2 + y2 < 1

1.

X * # S2

by the following map:f : S2 6 X *

f

B

C

(xyz

)

*

D

E :=

;

1.9.14.1 a A " Open(X ) 2 A " Closed(X ) =% q is a quotient map.

1.9.14.2 b If p is either an open map or a closed map then q is a quotientmap. HOW??

1.9.15 Some Claims about Quotient Maps and Spaces

The composite map of two quotient maps is again a quotient map

The cartesian product of two quotient maps need not be a quotient map.(see example 7)

If p and q are both open maps then p ! q is an open map, thus a quotientmap.

The quotient space of a Hausdor! space need not be Hausdor!. (HWS04E01 Zariski topology on C2)

S/R is Hausdor! $% * [x]R , [y]R " S/R : [x]R 0= [ y]R , &Ax , Ay "Open(S) : Ax / Ay = ! , Ax and Ay are saturated, and [x]R . Axand [y]R . Ay .

If # R : S! S 6 { 0, 1} , S/R is Hausdor! =% (# R )) 1 ({ 1} ) " Closed(S ! S)

If (# R )) 1 ({ 1} ) " Closed(S ! S) 1 q : S 6 S/R is an open map =%S/R is Hausdor!.

If each elemen of a quotient space X * is a closed subset of X then X * willsatisfy the T1 axiom.

1.9.16 Theorem 22.2Continuous maps out of Quotient Spaces

Let p : X 6 Y be a quotient map.Let Z be a space and let g : X 6 Z be a map that is constant on each set

p) 1 ({ y} ) for all y " Y .

Then:

1. g induces a map, f : Y 6 Z such that f 7 p = g.2. f is continuous 5 g is continuous.3. f is a quotient map 5 g is a quotient map.

Proof:

1. Since * y " Y , g#p) 1 ({ y} )

$is a single-point set in Z , define f (y) to

be that point inside of g#p) 1 ({ y} )

$.

2. TODO3. TODO

34

1.9.17 Universal Property of the Quotient Topology

(HWS04)Let S be a topological space, R be an equivalence relation on S and q : S 6

S/R be the quotient map.* topological space T and * f " TS : f (x) = f (y) $% x # R y &!# "

C(S/R, T ) : f = # 7 q.The map is # ([x]R ) := f (x).

1.9.18 Connectedness

(HWS06E01) Any quotient of a connected topological space is connected.

1.9.19 Homeomorphisms

Every homeomorphism is also a quotient map, and the quotient topology isunique.

1.10 20, 21 The Metric Topology

1.10.1 Denition of a Metric

A metric on a set X is a function

d : X ! X 6 R

having the following properties:

1. d (x, y) 4 0* x, y " X

2. d (x, y) = 0 5 x = y* x, y " X

3. d (x, y) = d (y, x) * x, y " X

4. d (x, y) + d (y, z) 4 d (x, z) * x, y, z " X

1.10.2 Denition of the +-ball centered at x

For any + > 0 and any x " X , define:

Bd (x, +) := { y | d (x, y) < +}

1.10.3 Denition of the Metric Topology

If d is a metric on a set X , then the topology induced by the basis { Bd (x, +) | x " X 1 + > 0}is called the metric topology induced by d.

Proof that this is a basis for a topology:

The first point of 1.2.1 is fulfilled because the +-ball centered at anyx " X , for any + > 0 will be a basis element containing x.

35

For the second point, let Bd (x1, +1) , Bd (x2, +2) be two basis ele-ments and take some x3 " Bd (x1, +1) / Bd (x2, +2).Define +3 := min ( { +1 ( d (x1, x3) , +2 ( d (x2, x3)} ).First, observe that +3 > 0 because d (x i , x3) < +i * i " { 1, 2} , whichis true as x3 " Bd (x1, +1) / Bd (x2, +2).

! Claim: x3 " Bd (x3, +3) . B (x1, +1) / Bd (x2, +2)! Proof: First by definition x3 " Bd (x3, +3).

Pick some i " { 1, 2} .Next, take any x " Bd (x3, +3). Thus d (x, x 3) < +3 < +i (d (x i , x3).So that d (x, x 3) + d (x i , x3) < +i .Invoke the triangle inequality to get: d (x, x i ) < +i % x "Bd (x i , +i ).

(HWS05E02) the topology induced by a metric is the coarsest such thatthe distance is a continuous function.

1.10.4 Another characterization of Metric Topology

The metric topology is T :=.

U " 2X | * x " U&+ > 0 : Bd (x, +) . U/

.

1.10.5 Examples

1.10.5.1 The Discrete Topology The metric d (x, y) :=

61 x 0= y0 x = y

in-

duces the discrete topology.Bd (x, 1) = { x} .

1.10.5.2 The Order Topology The metric d (x, y) := |x ( y| on R inducesthe order topology.

1.10.6 Metrizable Spaces

If X is a topological space, X is said to be metrizable if & a metric d on the setX of which the metric topology is equal to the original topology on X .

A metric space is a metrizable space X together with a specific metric d thatinduces the topology of X .

1.10.7 The Standard Bounded Metric

Let X be a metric space with the metric d. Define d : X ! X 6 R by theequation

d (x, y) := min ( { d (x, y) , 1} )

Then d is a metric that induces the same topology as d.Under this metric, every subset is bounded.

36

1.10.8 The Square Metric (The Product Metric with d( )

The metric , (x , y ) := max ( {| x i ( yi | | i " { 1, . . . , n}} ) on Rn is the squaremetric.

Basis elements are squares in R2 for example.It induces the sametopology as the standard topology on Rn .

1.10.9 Lemma 20.2Comparing topologies based on comparing met-rics

Let d and d! be two metrics on the set X , and let T and T ! be the two topologiesinduced by d and d! respectively. Then T ! is finer than T if and only if * x "X, > 0&( > 0 such that Bd! (x, ( ) . Bd (x, ).

Proof: TODO (by Theorem 13.3 of comparing bases for a topology)

1.10.10 Theorem 20.3Metric topologies and Product Topologies

The topologies on Rn induced by the Euclidean metric d and the square metric, are the same as the product topology on Rn .

1.10.11 The Uniform Metric

Let J be some index set, and let x := ( x" )" $ J and y := ( y" )" $ J be two pointsof RJ .

Define the metric , (x , y ) := sup ( { min ( {| x" ( y" | , 1} ) | $ " J } ) on RJ ,called the uniform metric .

The topology it induces is called the uniform topology.

(HWS03E05) The uniform topology on RN is generated by the basis Bu :=.B r ({ xn } ) : r > 0, { xn } " RN

/where B r ({ xn } ) :=

.{ yn } " RN : sup ({| xn ( yn | : n " N} ) < r

/.

Observe that Tbox - T uniform - T product .

1.10.12 Theorem 20.4Relation between the uniform, box and prod-uct topologies

1. The uniform topology on RJ is finer than the product topology and coarserthan the box topology.Proof: TODO (in handwriting)

2. These three topologies are all di!erent if J is infinite.Proof: TODO

1.10.13 RR is not metrizable

(16:50) (+zeno) I think one way is: K uncountable, if R^K is metrizable thenso is {0,1}^K is also, where {0,1} is discrete, but then show this space does nothave a countable local base at each point (the way a metrizable space should).

37

(16:52) (+zeno) any subspace of a metric space is metrized by the induced("same") metric

1.10.14 Theorem 20.5 RN is metrizable

Let x := ( xn )n $ N and y := ( yn )n $ N be two points of RN.

Define the metric D (x , y ) := sup9K

min( {| x n ) yn |, 1} )n | n " N

L:.

Claim: D is a metric that induces the product topology on RN.

Proof: TODO (in handwriting)

1.10.15 Metric Topology vs. Subspace Topology

If A is a subspace of a topological space X and d is a metric for X , then therestriction of d to A ! A is a metric for the topology of A.

Proof: TODO

1.10.16 Metric Topology vs. Order Topology

Some order topologies are metrizable (N and R for instance) and others are not.

1.10.17 Metric Topology vs. The Hausdor! Axiom

The Hausdor! axiom is satisfied by every metric topology.

1.10.18 Metric on Finite Product of Metric Spaces

Let (Si , di ) be a metric space * i " { 1, . . . , n} . Let p 4 1.Define Dp (x, y) := {

M ni =1 [di (x i , yi )]

p}1p for all x, y "

7 ni =1 Si .

Then Dp is a metric on7 n

i =1 Si which induces the product topology on7 ni =1 Si .

1.10.19 Metric Topology on a Countable Product Topology

The countable product topology is metrizable.

1.10.20 Theorem 21.1 ( ( continuity equivalent to continuous func-tions

Let f : X 6 Y be a map from two metrizable spaces X and Y with metrics dXand dY .

Claim: f is continuous $% [ * x " X, * > 0&( > 0 such that dX (x, y) 0&m%" N such that d (f n (x) , f (x)) < * n > m %, x " X

1.10.27 Theorem 21.6Uniform Limit Theorem

Let f n : X 6 Y be a sequence of continuous functions from the set X to themetric space Y . If { f n } converges uniformly to f then f is continuous.

Proof: TODO (in handwriting)

1.10.28 Uniform Metric and Uniform Convergence

Let RX be the space of all functions f : X 6 R, with the uniform metric , .Claim: The sequence of functions f n : X 6 R converges uniformly to f

$% { f n } 6 f when these two are considered as elements of the metric space#RX , ,

$.

Proof: TODO

1.10.29 Example RN with the box topology is not metrizable

Proof: (HWS05) If it were then it would have had a countable basis and then by21.2 for any set, and any point in the sets closure, there would be a sequenceconverging to that point in the set, but if we take the set of all points withpositive coordinates, zero is in the closure of the set, but no sequence convergesto it.

1.10.30 ExampleAn uncountable product of R with itself is notmetrizable

Proof: (HWS05) Similar to above, defined A =.

x" " RJ : x" = 1 for all but finitely many $ " J/

.Again, zero is in the closure, but no sequence converges to zero.

1.10.31 Ultrametrics

(HWS05)A metric space (S, d) is said to be ultrametric $% * x, y, z " S, d (x, z) 3

max ({ d (x, y) , d (y, z)} ).

Example: d (x, y) :=

61 x 0= y0 x = y

is ultrametric.

In an ultrametric space, every triangle is an isosceles (* x, y, z " S, d(x, y) =d (y, z) 2 d (x, y) = d (x, z) 2 d (y, z) = d (x, z)).

In an ultrametric space, * x " S, > 0, Bd (x, ) = Bd (y, ) * y "Bd (x, ).

40

2 Chapter 3 Connectedness and Compactness

2.1 23 Connected Spaces

2.1.1 Separation Denition

Let X be a topological space. A separation of X is two subsets, U, V "Open(X ) \ { ! } such that:

1. U / V = !

2. U + V = X

2.1.2 Connected Space Denition

A space X is connected if # a separation of X .

2.1.3 TheoremHomeomorphism Preserves Connectedness

Proof:Let X be a connected topological space, let Y be any topological space, and

let f : X 6 Y be a homeomorphism. TODO (in handwriting)(HWS06E02)

2.1.4 TheoremConnected Spaces Characterized by Clopen Subsets

Claim: A space X is connected $%.

U " 2X | U " Open(X ) 1 U " Closed(X )/

={ X, ! }

Proof: TODO (in handwriting)

2.1.5 Lemma 23.1Separation (another denition)

If Y is a subspace of X , a separation of Y is two sets A, B " 2Y \ { ! } such that:

1. closureX (A) / B = A / closureX (B ) = !

2. A + B = Y

Claim: # a separation of Y $% Y is connected.

Proof: TODO (in handwritingunsolved!)

2.1.6 Lemma 23.2

If { C, D } is a separation of X , and if Y is a connected subspace of X , thenY . C $ Y . D .

Proof: TODO (in handwriting)

41

2.1.7 [I] Lemma 23.3Union of Not-Disjoint Connected Subspaces isConnected

Let { U" } " $ J be a collection of connected subspaces of X such that,

" $ J U" 0=

! .

Claim:+

" $ J U" is a connected subspace.Proof: TODO (in handwriting)

2.1.8 [I] Theorem 23.4

Let A be a connected subspace of X . If A . B . closureX (A) then B is aconnected subspace of X .

(A+ some (or all) of its limit points is connected)(Thus closure of connected space is connected!)Proof: TODO (in handwriting)

2.1.9 [I] Theorem 23.5The image of a connected space under a con-tinuous map is connected

Let X be a connected space and let Y be any space. Let f : X 6 Y be acontinuous map.

Claim: f (X ) is a connected subspace of Y .Proof: TODO (in handwriting)

2.1.10 Theorem 23.6A nite cartesian product of connected spacesis connected

Let n " N and let { X i } i ${ 1,..., n } be a finite collection of connected spaces.Claim:

7i ${ 1,..., n } X i is a connected space.

2.1.11 Example 6 RN is not connected in the box topology

Separate RN to two open sets: all bounded sequences in RN and all unboundedsequences in RN.

2.1.12 Example 7 RN is connected in the product topology

What is R# ??? TODO unsolved(19:16) (+Polytope) Munkres says that R^omega is the set of all sequences

of real numbers, and R^oo is the subspace consisting of all sequences that areeventually zero.

(19:22) (+Psycho_pr) Hrmmm (19:24) (+Psycho_pr) R^oo = U_{n\in N}R^n (19:24) (+Psycho_pr) R^omega = \prod_{n\in N} R

42

2.1.13 Totally Disconnected Denition

A space is called totally disconnected if its only connected subspaces are one-point sets.

Claim: If X has the discrete topology, then X is totally disconnected.Claim: & totally disconnected spaces with a topology di!erent from the

discrete topology: Q (HWS06EC).

2.1.14 Example

(HWS06)R and R2 are not homeomorphic because if they were, then R\ { x0} and

R2\ { f (x0)} would be homeomorphic (where f is the homeomorphism). How-ever, the former is not connected whereas the second is.

2.1.15 Product of Connected Spaces

(HWS06E03) The (countable, uncountable) product of connected spaces in theproduct topology is connected.

2.1.16 R\ { 0} is not connected

Thus it is not path-connected. R\ { 0} = ( (, , 0) + (0, , ) are its connectedcomponents.

2.1.17 Q is not connected

Thus it is not path-connected. Q =+

a$ Q { a} are its connected components.

2.2 24 Connected Subspaces of the Real Line (and pathconnectedness)

2.2.1 Linear Continuum Denition

A simply ordered set (1.3) L having more than one element, such that:

1. L has the least upper bound property.

2. * x, y " L such that x < y &z " L such thatx < z < y .

is called a linear continuum.

2.2.2 Theorem 24.1Linear Continuums in the Order Topology areConnected

Let L be a linear continuum with the order topology.

1. Claim: L is connected.

2. Claim: (a, b) , [a, b] , [a, b), (a, b] are connected * a, b " L such that a < b.

43

3. Claim: (a, , ) and ((, , a) are connected * a " L .

Proof: TODO (in handwriting)Thus R, as well as rays and intervals in R are connected.

2.2.3 [I] Theorem 24.3Intermediate Value Theorem

Let X be a connected space and Y be an ordered set with the order topology.Let f : X 6 Y be a continuous map. If a, b " X such that a 0= b, and ifr " (f (a) , f (b)) then &c " X such that f (c) = r .

Proof: TODO (in handwriting)

2.2.4 Example 1The Ordered Square is a Linear Continuum

(HWS06E03) The order square is not locally path-connected. It is howeverconnected.

2.2.5 Example 2The product of a well-ordered with [0, 1) is a linearcontinuum

2.2.6 Denition of a Path

Let X be a topological space and let there be two points x, y " X . A path in Xfrom x to y is a continuous map f : [a, b] 6 X where a, b " R and a < b, suchthat f (a) = x and f (b) = y.

2.2.7 Denition of Path Connectedness

A space X is said to be path connected if * x, y " X & a path in X from x to y.

2.2.8 [I] Path Connectedness means Connectedness

Proof: TODO (in handwriting)

2.2.9 The continuous image of a path connected space is path con-nected

2.2.10 Example 3The unit ball in Rn is connected

2.2.11 Example 4Punctured Euclidean Space

Punctured Euclidean space Rn \ { 0} is path connected * n " N\ { 1} .

2.2.12 Example 5The Unit Sphere Sn ) 1 . Rn

Sn ) 1 := { x | ||x || = 1 }

If n > 1, it is path connected: g : Rn \ { 0} 6 Sn ) 1 defined by g (x ) := x|| x ||is continuous and surjective.

44

2.2.13 A connected space need not be path connected

2.2.13.1 Example 6 The ordered square is connected (because it is a linearcontinuum).

Claim: It is not path connected.Proof: TODO (in handwriting)

2.2.13.2 [I] Example 7The Topologists Sine Curve S :=0!

xsin

#1x

$"

" R2 | x " (0, 1]1

Because S is the image of a connected set (0, 1] under a continuous map, Sis connected. Therefore its closure closureR2 (S) is also connected by Theorem23.4 is also connected.

The topologists Sine Curve is:

closureR2 (S) = S +0!

0x

"" R2 | x " [( 1, 1]

1

Claim: closureR2 (S) is not path connected. (HWS06E03)Proof: TODO (in handwriting). Its two path-connected components are S

and0!

0x

"" R2 | x " [( 1, 1]

1

2.2.14 ExampleOrthogonal Group

(HWS06E04) On (R) is not connected, because there is a continuous functionfrom it to { 0, 1} which is not constant, namely, the function (1 7 det. SO (n) isconnected because SO (n) (SO (2)) ,

n2 - and SO (2) S1. Its two connected

components are SOn (R) and On (R) SOn (R).

2.2.15 Example SU (2)

(HWS06E05) SU (2) is isomorphic to S3 which is connected.

2.2.16 A riddle

(HWS06)There cannot be an f " C (R, R) such that f (R\ Q) . f (Q) and f (Q) .

f (R\ Q). If there were, then |f (R\ Q)| 3 9 0 =% | f (R)| 3 9 0. But R isconnected, so f (R) is connected. So f (R) is connected countable subset of R,so it is a point, so f (x) = x0, which is a contradiction.

2.3 25 Components and Local Connectedness

2.3.1 Components

Let X be a topological space. Define an equivalence relation on X be settingx # y $% & A " 2X such that A is connected and x, y " A. The equivalenceclasses of X are called components or connected components of X .

Proof that # is an equivalence Relation: TODO (in handwriting)

45

2.3.2 Theorem 25.1Description of Components

The components of X are connected disjoint subspaces of X whose union is X ,such that each nonempty connected subspace of X intersects only one of them.

Proof: TODO

Each component is a closed subset of X . (Since closures are connectedand each connected subspace intersects only one component.

If there are only finitely many components then each component is alsoopen.

2.3.3 Path Components

Let X be a topological space. Define an equivalence relation on X be settingx # y $% & path in X from x to y. The equivalence classes of X are calledpath components of X .

Proof that # is an equivalence Relation: TODO (in handwriting)

2.3.4 Theorem 25.2

The path components of X are path-connected disjoint subspaces of X whoseunion is X such that each nonempty path connected subspace of X intersectsonly one of these components.

Proof: TOOD (in handwriting)

Path components need not be closed or open.

2.3.5 Example 1Components of Q

The components of Q are all of its single points. None of the components areopen.

2.3.6 Example 2The Topologists Sine Curve

The Topologists sine curve has one component (since it is connected) and twopath components. One path component is S and another is 0 ! [( 1, 1].

Claim: S " Open(closureR2 (S)) but S /" Closed(closureR2 (S))

Claim: 0! [( 1, 1] /" Open(closureR2 (S)) yet 0! [( 1, 1] " Closed(closureR2 (S)) .

Claim: closureR2 (S) \0!

0x

"" R2 | x " [( 1, 1] / Q

1has only one com-

ponent but uncountably many path components.

2.3.7 Local Connectedness at x

Let X be a space and let x " X .A space X is locally connected at x if * U " Open(X ) : x " U, &V "

Open(X ) : x " V . U such that V is connected.

46

2.3.8 Locally Connected Space

If X is locally connected at each of its points, it is locally connected.

2.3.9 Local Path Connectedness at x

Let X be a space and let x " X .A space X is locally path connected at x if * U " Open(X ) : x " U,

&V " Open(X ) : x " V . U such that V is path connected.

2.3.10 Locally Path Connected Space

If X is locally connected at each of its points, it is locally path connected.

2.3.11 Example 3

Each interval and each ray in R is locally connected (and as stated previ-ously also connected).

[( 1, 0) + (0, 1] is not connected but it is locally connected.

The topologists sine curve is connected but not locally connected.

Q is neither connected nor locally connected.

2.3.12 Theorem 25.3A Criterion for Local Connectedness

A space X is locally connected $% * U " Open(X ), each component of U isopen in X .

Proof: TODO (in handwriting)

2.3.13 Theorem 25.4A criterion for Local path-connectedness

A space X is locally path connected $% * U " Open(X ), each path componentof U is open in X .

Proof: TODO

2.3.14 Theorem 25.5A relation between path components and comp-nents

Let X be a topological space.

1. Each path component of X lies in a component of X .Proof: Each path component is path connected and is thus connected andso lies in exactly one component.

2. If X is locally path connected, then the components and the path compo-nents of X are the same.Proof: TODO (in handwriting)

47

2.4 26 Compact Spaces

Every metric space (for instance, Y) is Hausdor!. Then, you can use the factthat every continuous and bijective map f : X 6 Y is a homeomorphism (thatis, f" 1 is continuous), if X is compact and Y Hausdor!.

To prove that claim, you can do as follows: in order to see that a mapg : X 6 Y is continuous, you can prove that, for every closed subset C . X ,g) 1 (C) . Y is closed. In our case, g=f" 1, so it su"ces to show that, for everyclosed subset C . X ,

#f ) 1

$) 1(C) = f (C . Y ) is closed. Right?

But, if C is a closed subset of a compact space X, then its itself compact andthe image of a compact set by a continuous map, f, is compact. Hence f(C) is acompact subset of a Hausdor! space, Y. Every compact subset of a Hausdor!space is closed. Thus, f(C) is closed. qed.

2.4.1 Cover

A collection A . 2X covers X , or is a covering of X , if+

A $A A = X .

2.4.2 Open Covering

A is an open covering if it is a covering such that A " Open(X ) * A " A .

2.4.3 Compact Space

A space X is said to be compact if every open covering A of X contains a finitesubcollection that also covers X .

2.4.4 Example 1 R is not compact

A = { (n, n + 2) | n " Z} is an open cover of R which has no finite subcollectionthat covers R.

2.4.5 Example 2

X = { 0} +.

1n | n " N

/is compact.

Each cover of X will contain 0, and so will contain the infinite number ofpoints that are near it. All the rest of the points are finite.

2.4.6 Example 3Finite Spaces are Compact

2.4.7 Example 4

(0, 1] is not compact, since the open cover A :=.

( 1n , 1] | n " N/

contains nofinite subcollection covering (0, 1].

(0, 1) is not compact (same reason as above) with A :=.

( 1n , 1 (1n ) | n " N

/.

2.4.8 A cover for a subspace

If Y is a subspace of X , a collection A . 2X is said to cover Y if+

A $A A - Y .

48

2.4.9 Lemma 26.1Compact Subspaces

Let Y be a subspace of X .

Claim: Y is compact $% * covering of Y by sets in Open(X ), & a finitesubcollection which covers Y .

Proof: TODO (in handwriting)

2.4.10 [I] Lemma 26.2Every closed subspace of a compact space iscompact

Proof: TODO (Proven in Rudin PMA chapter 2.)

2.4.11 Lemma 26.3Every compact subspace of a Hausdor! space isclosed

Proof: TODO (in handwriting)

2.4.12 Lemma 26.4

If Y is a compact subspace of the Hausdor! space X , and x0 " X \ Y , then&U, V " Open(X ) such that x0 " U, Y . V and U / V = ! .

2.4.13 Example 6Theorem 26.3 breaks down without Hausdor!Condition

Take X = R with the co-finite topology. This space is not Hausdor!.Closed(X ) =

.U " 2X | |U| < ,

/+ { X }

Claim: Every subset of R is compact in X .Proof: TODO (in handwriting)

2.4.14 Theorem 26.5The image of a compact space under a contin-uous map is compact

Proof: TODO (Proven in Rudin PMA chapter 4)

2.4.15 [I] Theorem 26.6Verifying that a map is a homeomorphism

Let f : X 6 Y be a bijective continuous function. If X is compact and Y isHausdor! then f is a homeomorphism.

Proof: TODO (in handwriting)

2.4.16 Theorem 26.7The product of nitely many compact spacesis compact

Proof: TODO (in handwriting)

49

2.4.17 Lemma 26.8The Tube Lemma

Assumptions:

Let X be a space and Y be a compact space.Let x0 " X .Let N " Open(X ! Y ) such that x0 ! Y . N .

Claim: &W " Open(X ) such that:

x0 " W

W ! Y . N .

Proof: Proven as part of Theorem 26.7.

2.4.18 Example 7The Tube Lemma does not hold if Y is not com-pact

Let Y be the y-axis of R2. Define N :=K

x ! y : |x| < 1y2 +1

L.

Then 0 ! R " N " Open#R2

$, but it contains no tube about 0 ! R.

2.4.19 Tychono! TheoremThe Innite Product of Compact Spacesis Compact

Proof: TODO(HWS09E06)

2.4.20 Finite Intersection Property

A collection C of subsets of X is said to have the finite intersection property iffor every B " 2C such that |B| < , ,

3

C $B

C 0= !

2.4.21 Theorem 26.9Criterion for Compactness Formulated in termsof Closed Sets

Let X be a topological space.X is compact $% *C " 2C losed (X ) such that C has the finite intersection

property,,

C $C C 0= ! .Proof: TODO (Rudin PMA chapter 2, theorem 2.36anyway, in handwrit-

ing)Note: If we have a nested sequence, Ci - Ci +1 * i " N and Ci 0= ! * i " N,

then { Ci } has the finite intersection property. Thus,

i $ N Ci 0= ! .

50

2.5 27 Compact Subspaces of the Real Line

2.5.1 Theorme 27.1Every closed interval is compact

Let X be a simply ordered set having the least upper bound property.Claim: X in the order topology has the property that each closed interval in Xis compact.Proof: TODO (in handwriting)Corollary 27.2: Every closed interval in R is compact.

2.5.2 [I] Theorem 27.3Characterization of compact subspaces of Rn

(Heine-Borel Theorem)

Let d be the Euclidean metric on Rn and , be the square metric on Rn .Claim: A subspace A of Rn is compact $% A " Closed(Rn )1 (diamd (A) < , 2 diam& (A) < , )Proof: (Rudin Chapter 2 Theorem 2.34?) Consider only one metric (why?). For=% , build a covering by boxes of length in N. For $ = , A is contained insidesome compact box of some finite size, thus, being closed, it is compact also.

2.5.3 Example 1The unit sphere and the closed unit ball are com-pact in Rn

Because they are closed and bounded.

2.5.4 Example

A =.

x !#

1x

$|x " (0, 1]

/is closed but not bounded, thus it is not com-

pact.

S =.

x ! sin#

1x

$|x " (0, 1]

/is bounded but not closed, thus it is not

compact.

2.5.5 Theorem 27.4Extreme Value Theorem

Let f : X 6 Y be continuous, where Y is an ordered set in the order topology.Claim: If X is compact then & points xmin , xmax " X such that f (xmin ) 3f (x) 3 f (xmax ) for any x " X .Proof: TODO (in handwriting)

2.5.6 Distance from a point to a set

Let (X, d ) be a metric space and let A " 2X \ { ! } . * x " X define the distancefrom x to A by the equation:d (x, A ) := inf ( { d (x, a) |a " A} )Observe that for a fixed A, d : X 6 R is a continuous function, becaused (x, A ) ( d (y, A) 3 d (x, y).

51

2.5.7 Lemma 27.5The Lebesgue Number

Let A be an open covering of the metric space (X, d ).Claim: If X is compact, &(A > 0 such that * S " 2X such that diam (S) < ( A ,&AS " A such that S . AS .The number (A is called the Lebesgue number of A .Proof: TODO (in handwriting)

2.5.8 Uniform Continuity

Let (X, d X ) and (Y, dY ) be two metric spaces. A function f : X 6 Y is said tobe uniformly continuous if * > 0&( > 0 : * (x0, x1) " X 2, dX (x0, x1) < ( =%dY (f (x0) , f (x1)) < .

2.5.9 Theorem 27.6Uniform Continuity Theorem

Let (X, d X ) be a compact metric space and (Y, dY ) a metric space, and letf : X 6 Y be a continuous function.Claim: f is uniformly continuous.Proof: TODO (Rudins PMA chapter 4also in handwriting)

2.5.10 Isolated Point

If X is a space, x " X is said to be an isolated point of X if { x} " Open(X ).

2.5.11 Theorem 27.7A set without isolated points is uncountable

Let X be a nonempty compact Hausdor! space. If X has no isolated points,then X is uncountable.Proof: TODO (in handwriting)Corollary 27.8: Every closed interval in R (thus compact with no isolated points)is uncountable.But every superset of au uncountable set is uncountable, thus R is uncountable.

2.5.12 Examples

(HWS07)

Q / [0, 1] is not compact. If it were, then Q / [0, 1] would have been inClosed(R) by 26.3, however, it is not closed, since it does not contain itslimit points: (R\ Q) / [0, 1].

O (n) is closed, and bounded, and thus, it is compact.

A finite space is always compact.

The Cantor set is an inifinite intersection of closed intervals, hence closed.It is also bounded, hence, compact.

52

2.6 28 Limit Point Compactness

2.6.1 Limit Point Compactness

A space X is said to be limit point compact if every infinite subset of X has alimit point.

2.6.2 Theorme 28.1Compactness implies limit point compactness

Proof: TODO (in handwriting)

2.6.3 Example 1

Y := { a, b} , Open(Y ) := { Y, ! }X := N ! Y

Claim: X is limit point compact.

Proof: every nonempty subset of X has a limit point.

Claim:X is not compact.

Proof: the covering of X by open sets Un = { n} ! Y has no finitesubcollection covering X .

2.6.4 Example 2

Let S! be the minimal uncountable well-order set, in the order topology.TODO!!!

Claim: S! is not compact.

Proof: It has no largest element.

2.6.5 Example

(HWS09E02) In [0, 1]N with the uniform topology, The setK

a " [0, 1]N : &i 0 " N : ! i 0 (a) = 1 1 ! i (a) = 0 * i " N\ { i 0}L

is infinite but has no limit point.

(HWS09E03) [0, 1] is not limit point compact in the lower limit topology.Take the set

.1 ( 1n : n " N

/, it has no limit points. The neighborhood

[1, 2) of 1 does not intersect the set at all.

2.6.6 Limit Point Compactness and Continuous Functions

(HWS09E04) If f " C (X, Y ) and X is limit point compact, f (X ) is notnecessarily limit point compact

Take X :=#N, 2N

$! ({ a, b} , { ! , { a, b}} ) and let the continuous

function be the projection onto the first coordinate. X is limit pointcompact, but

#N, 2N

$is not.

53

2.6.7 Limit Point Compactness and Closedness

(HWS09E04) If X is limit point compact and A " Closed(X ) then A is limitpoint compact.

2.6.8 Sequential Compactness

Let X be a topological space. If every sequence of points of X has a convergentsubsequence then X is sequentially compact