Notes on Topology

These are links to PostScript files containing notes for various topics in topology.

I learned topology at M.I.T. from Topology: A First Course by James Munkres. At the time, the first edition was just coming out; I still have the photocopies we were given before the printed version was ready!

Hence, I'm a bit biased: I still think Munkres' book is the best book to learn from. The writing is clear and lively, the choice of topics is still pretty good, and the exercises are wonderful. Munkres also has a gift for naming things in useful ways (The Pasting Lemma, the Sequence Lemma, the Tube Lemma).

(I like Paul Halmos's suggestion that things be named in descriptive ways. On the other hand, some names in topology are terrible --- "first countable" and "second countable" come to mind. They are almost as bad as "regions of type I" and "regions of type II" which you still sometimes encounter in books on multivariable calculus.)

I've used Munkres both of the times I've taught topology, the most recent occasion being this summer (1999). The lecture notes below follow the order of the topics in the book, with a few minor variations.

Here are some areas in which I decided to do things differently from Munkres:

● I prefer to motivate continuity by recalling the epsilon-delta definition which students see in analysis (or calculus); therefore, I took the pointwise definition as a starting point, and derived the inverse image version later. ● I stated the results on quotient maps so as to emphasize the universal property. ● I prefer Zorn's lemma to the Maximal Principle; for example, in constructing connected components, I use Zorn's lemma to construct maximal connected sets. ● I did Tychonoff's theorem at the same time as the other stuff on compactness. The proof for a finite product is long enough that I decided to save some time by omitting it and just doing the general case. ● It seems simpler to me to do Urysohn's lemma by indexing the level sets with dyadic rationals in [0,1] rather than rationals.

● How do I view/print the notes?

● How may I use/distribute the notes? How did you write them?

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● funct.ps (86,912 bytes; 5 pages): Review of topics in set theory.

● top.ps (168,720 bytes; 6 pages): Topological spaces, bases.

● closed.ps (55,509 bytes; 4 pages): Closed sets and limit points.

● haus.ps (132,025 bytes; 2 pages): Hausdorff spaces.

● contin.ps (100,021 bytes; 3 pages): Continuous functions.

● homeo.ps (83,891 bytes; 3 pages): Homeomorphisms.

● prod.ps (51,962 bytes; 3 pages): The product topology.

● metric.ps (140,993 bytes; 8 pages): Metric spaces.

● quot.ps (116,395 bytes; 5 pages): Quotient spaces.

● conn.ps (82,400 bytes; 4 pages): Connectedness.

● realconn.ps (55,934 bytes; 3 pages): Connected subsets of the reals.

● pathco.ps (45,136 bytes; 3 pages): Path connectedness, local connectedness.

● compact.ps (240,878 bytes; 8 pages): Compactness (including Tychonoff's theorem).

● compreal.ps (69,020 bytes; 3 pages): Compact subsets of the reals.

● limcomp.ps (107,691 bytes; 4 pages): Limit point compactness and sequential compactness.

● loccpt.ps (70,399 bytes; 3 pages): Local compactness; the one-point compactification.

● countab.ps (40,734 bytes; 2 pages): The countability axioms.

● sep.ps (139,569 bytes; 5 pages): The separation axioms: regularity and normality.

● urysohn.ps (69,043 bytes; 2 pages): Urysohn's lemma.

● tietze.ps (45,448 bytes; 3 pages): The Tietze Extension Theorem.

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Last updated: August 16, 2000

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Review Set TheoryFunctions and Relations

This is a brief review of material on set theory functions and relations Ill concentrate on things which

are often not covered or given cursory treatment in courses on mathematical pro of

Denition Let A and B b e sets and let f A B b e a function from A to B

f is injective if for all a a A f a f a implies a a

f is surjective if for all b B there is an element a A such that f ab

f is bijective if f is injective and surjective

Denition Let f A B b e a function Let B B

The image of f is the set

f Aff a j a Ag

The inverse image of B is the set

f B fa A j f a B g

Remark f is surjective if and only if f AB

Example a Let f A B and g h B C b e functions Supp ose that f is surjectiveandg f h f

Prove that g h

Let b B Imust showthatg bhb

Since f is surjectiveImay nd an element a A suchthatf abTheng f h f implies that

g f a hf a so g bhb

Therefore g h

b Let f A B Supp ose that for all g h B C if g f h f then g hProvethat f is surjective

First Ill take care of the cases where A or A contains a single element

If A and B then f is vacuously surjective Every elementofB there are none is an image of

an elementofA

If A and B then B contains at least one element b Dene g B f g by g band

b It is vacuously true that g f a hf a for all a A since there are no as h B f g by h

in A but g h Therefore the hyp othesis isnt satised and this case is ruled out

Supp ose A consists of a single element aIfB thenf is vacuously surjective If B consists of a

single element then f is obviously surjective

Finally supp ose that B contains more than one element f a is an elementofB supp ose that b B

and b f a Consider the functions g h B B given by

g bb for all b B

b if b b

f a if b b

Then g f a f ahf a so g f h f However g h Therefore the hyp othesis isnt

satised and this case is ruled out

This concludes the pro of for the trivial cases

Assume that A contains more than one element

Supp ose on the contrary that f A B Ill show that the hyp othesis is violated

If b f A choose an element a A suchthatf a b Since A contains more than one element I

b b

may nd elements a a A such that a a Dene g h B A by

if b f A a if b f A a

b b

g b and g b

a if b f A a if b f A

If a A then f a f A so

g f a a hf a

f a

Thus g f h f

However since f A B there is an element b B f A For this element g b a but

hb a Therefore g h

This proves that the hyp othesis is violated This contradiction in turn establishes that f AB sof

is surjective

Remark This result is often expressed bysaying that surjective functions are rightcancellable The dual

result is also true Injective functions are leftcancellable

Example Let f A B and supp ose that B B B Prove that f B B f B f B

Supp ose a f B B Then f a B B so f a B or f a B

If f a B then a f B so a f B f B If f a B thena f B so

a f B f B

Therefore a f B f B so f B B f B f B

B f B so a f B ora f B Conversely supp ose that a f

If a f B then f a B sof a B B and hence a f B B If a f B then

f a B so f a B B and hence a f B B

Therefore a f B B so f B f B f B B

This proves that f B B f B f B

Denition Let A b e a set A relation on A is a simple order or linear order if

ComparabilityFor all a b A exactly one of the following is true a b baor a b

Nonreexivity There is no element a A suchthataa

TransitivityFor all a b c Aifa b and bcthen ac

Example Let b e a simple order on a set AThedictionary order or lexicographic orderon A A

is given by a b a b if and only if

a a or

a a and b b

The words in a dictionary are ordered by their rst letter Among words with the same rst letter they

are ordered by their second letter and so on This explains the name of this order relation

To give a sp ecic example in R under the dictionary order and

Denition If A is an ordered set a b Aanda bthenthe op en interval from a to b is the set

a b fx j axbg

It may happ en that this set is empty In that case since there are no elements b etween a and b its

natural to say that a is the predecessor of b and b is the successor of a

Unfortunatelya b can mean a point in a Cartesian pro duct or an open interval in a simple order

Toavoid confusion Ill sometimes write a b for the p oint when b oth concepts o ccur in the same discussion

Example Consider the dictionary order on R The op en interval from the p oint to the p oint is

the union of

The op en ray f y j yg

The lines fx y j xg

The op en ray f y j yg

2 x 4

1 x 2

Denition Supp ose A and B are sets with simple orders and resp ectivelyLetf A B be a

A B

function f is orderpreserving if for all a a A

a a implies f a f a

A B

Example With the usual order relation on R dene f R R by f xx Then f is orderpreserving

In fact a function R R is orderpreserving if it is an increasing function

Denition Let A and B b e ordered sets A and B havethesame order typ e if theres an orderpreserving

bijection from A to B

Denition A set is wellordered if every nonempty subset has a smallest element

The WellOrdering Axiom says that every set can b e wellordered This is an axiom of set theory

its equivalent to other results which can b e taken as axioms suchasthe Axiom of Choice and Zorns

lemma

Wellordering means that for a given set there is some order relation on the set relative to which the

set is wellordered That wellordering will usually have little to do with familiar order relations

Denition Let A b e an ordered set and let B A

c B is the largest element of B if b c for all b B b c

Likewise c B is the smallest element of B if cb for all b B b c

An element a A is an upp er b ound for B if ba for all b B b aIfB has an upp er b ound B

is b ounded ab ove

hasalower Likewise an element a A is a lower b ound for B if ab for all b B b aIfB

b ound B is b ounded b elow

If the set of upp er b ounds of B is nonempty then the smallest elementofthatsetisthe least upp er

b ound for B It is denoted lub B or sup B

Likewise if the set of lower b ounds of B is nonempty the largest element of that set is the greatest

lower b ound for B It is denoted glb B or inf B

A set has the least upp er b ound prop erty if every nonempty subset that is b ounded ab ovehasa

least upp er b ound

A set has the greatest lower b ound prop erty if every nonempty subset that is b ounded b elowhas

a greatest lower b ound

Example The real numb ers with the usual ordering satisfy the least upp er b ound prop erty In fact it is

one of the axioms for the real numb ers

Mathematics is based on a collection of axioms for set theory They are called the ZermeloFraenkel

Axioms with the Axiom of Choice often abbreviated to ZFC

While the Axiom of Choice is accepted and used by most mathematicians the situation was dierent

earlier in this century Some mathematicians felt that the Choice Axiom was problematic and refused to use

it others to ok care to note when they used the Choice Axiom in a pro of Many of the ob jections stemmed

from the often startling results which can b e proved using the axiom

Axiom of Choice Let fX g b e a collection of disjoint nonempty sets There exists a set X containing

exactly one element from each X

The Axiom of Choice is equivalent to the WellOrdering Axiom and also to an imp ortant result known

as Zorns lemma The statement of Zorns lemma requires some preliminaries

Denition A relation on a set X is a partial order if

Reexivity x x for all x X

AntisymmetryIf x y and y x then x y

TransitivityIf x y and y z then x z

Example The set of subsets of a set is partially ordered by inclusion

Example The usual less than or equal to relation on R is a partial order

Denition Let X b e a partially ordered set

A subset Y X is totally ordered or linearly orderedifforallx y Y either x y or y x

A totally ordered subset is also said to b e a chain

If Y X then Y has an upp er b ound if there is an element x X suchthaty x for all y Y

An element x X is maximal if y X and x y implies x y

Zorns Lemma If X is a nonempty partially ordered set in whichevery chain has an upp er b ound then

X has maximal elements

Zorns lemma is an immensely useful form of the Axiom of Choice For example you mayhave seen a

pro of that every nitedimensional vector space has a basis Zorns lemma can b e used to show that every

vector space has a basis

Example Let G b e a group and let g G g There is a subgroup HGsuchthatg H and H is

maximal among subgroups with this prop erty

Let S b e the collection of subgroups of G whichdonotcontain g S since fg S S is partially

ordered by inclusion

Let T be a chain in S Let

H H

H T

Thus H is just the union of all the elements of the chain

I claim that H is a subgroup of G which do es not contain g

First H for all H T so certainly H

If h h H then h H and h H for some H H TButT is a chain so either H H or

H H

Supp ose H H Then h h H so since H is a subgroup hh H Therefore hh H and

hence H is a subgroup

If g H then g H for some H TButH Sso g H by denition of S Therefore g H

Ive shown that H is a subgroup of G which do es not contain g soH SSinceH H for all

H Titfollows that H is an upp er b ound for T in S

Since every chain has an upp er b ound Zorns lemma implies that S has a maximal element H Then

H is a subgroup of G which is maximal among subgroups which do not contain g

by Bruce Ikenaga

Top ological Spaces

A function f R R is continuous at x a if and only if

For every there is a such that if jx aj then jf x f aj

In words

You can make f x arbitrarily close to f abymakingx suciently close to a

Continuity is imp ortant in analysis and calculus The idea of closeness is an imp ortant part of the

denition of continuityWhatwould closeness mean in an abstract setting one in which theres no way

to measure the distance between two p oints

A top ology for a space is roughly sp eaking a sp ecication of all closeness relations in the space

Denition Let X b e a set A top ology on X is a collection T of subsets of X satisfying the following

conditions

T and X T

The union of elements of T is an elementofT

The nite intersection of elements of T is an elementof T

The elements of T are called op en sets The set X together with the top ology T is a top ological

space

Example If X is a set take D to b e P X the p ower set of X Recall that P X is the set of all subsets

of X

D is clearly a top ology on X it is called the discrete top ology In the discrete top ology all subsets

are op en

The other extreme is the indiscrete top ology on X In this case I fXg Again it is clear that

this is a top ology on X

Denition If T and T are top ologies on a set X then T is ner than T or T is coarser than T if

T T In other words ner top ologies contain more op en sets

Thus if D is the discrete top ology on X thenD is ner than any other top ology on X IfI is the

indiscrete top ology on X then I is coarser than any other top ology on X

It is of course p ossible for two top ologies on X to b e incomparable ie neither may b e ner than the

other

Analysts often refer to stronger and weaker top ologies but their usage is often at variance with the

usage of top ologists

Example Let X f g and consider the following top ologies on X

T f fg fg f gXg

T f fg f gXg

T f f gXg

1 1 2 3

2 1 2 3

3 1 2 3

4 1 2 3

Then T is ner than T and T is ner than T On the other hand T is not comparable with the

other three

As this example shows top ology do es not necessarily concern itself with the concepts of measurement

or distance They might not b e applicable to the space under consideration

Example The standard top ology on R is the top ology whose op en sets consist of arbitrary unions of op en

intervals

The empty set is considered to b e a degenerate op en interval and R can b e written as a union of op en

intervals

A union of unions of op en intervals is a union of op en intervals

It remains to show that this top ology is closed under nite intersections Its fairly obvious that two

op en intervals either do not intersect or intersectinanopeninterval I wont write out the details The

general case for nite intersections reduces to this case by distributing intersections over unions

Example The standard top ology on R is obtained by starting with op en balls

B x r fy R jjx y j rg

B x r is called the op en r ball centered at x The op en sets are arbitrary unions of op en balls

Ill defer the verication that this gives a top ology since it will follow easily once I develop the notion

of a basis

The examples ab ove giveatypical way of dening a top ology Start with a collection of subsets and

then extend the collection to a top ology by taking arbitrary unions The next denition makes precise the

conditions I need to b e able to do this

Denition Let X b e a set A collection B of subsets of X is a basis if

Every elementofX is contained in some elementofB

If B B B and x B B there is an element B B such that

x B B B

B1 x

The elements of B are called basic subsetsor basis elements

Remark By induction it follows that if B is a basis B B Bandx B B then there is

n n

an element B B such that

x B B B

Example The op en intervals in R form a basis

If x R then x x x for instance

If x is contained in the intersection of twoopenintervals the intersection is itself an op en interval which

contains x andiscontained in the intersection

n n

The op en balls B x r inR form a basis Obviouslyevery p ointinR is contained in some op en ball

TakeopenballsB x r andB x r Assume that they intersect and let x B x r B x r

Ihave to nd an op en ball centered at x whichiscontained in the intersection

Let s r jx x jThenB x s B x r

r11 - | x - x |

| x - x1 |

B(x11 ; r )

Likewise let s r jx x jThenB x s B x r

Therefore if I let r mins s then

B x r B x r B x r

This shows that the op en balls in R form a basis

Prop osition Let X b e a set and let B b e a basis Dene T to b e the collection of subsets U X satisfying

the following prop erty

U T if and only if for all x U there is an element B B such that x B U

Then T is a top ology on X

T is the top ology generated by the basis B

Pro of The empty set is in T it satises the dening condition vacuously since it has no elements

X TIfx X the rst axiom for a basis implies that there is an element B B such that x B X

Let fU g b e a collection of elements of T I need to show that the union U U is in T

a a

U T there is an element B B such that x B U Let x U Then x U for some aSince

a a a

Then

x B U U

Hence U T

V is in T Letx V Thenx V for all i Since Let V V TImust show that V

i i n

V T for each i Imay nd an element B B suchthatx B V By the remark ab ove I may nd an

i i i i

B Then element B B such that x B

n n

x B B V V

i i

Hence V T

Therefore T is a top ology

Notice that elements of a basis are automatically elements of the top ology they generate Basic sets are

op en

Denition If T is a top ology and B is a basis then B is a basis for T if the top ology generated by B is

the same as T

Lemma Let X b e a set B a basis and T the top ology generated by B ThenT is equal to the collection

of all unions of elements of B

Pro of Since basis elements are op en unions of basis elements are op en Therefore the collection of all

unions of basis elements is contained in T

Conversely let U TImust express U as a union of basis elements For each x U nd a basis

element B such that x B U Then U B

x x x

Here are two more facts ab out bases The pro ofs are straightforward

Let T and T b e top ologies on X generated by bases B and B resp ectively Then T is ner than T

if and only if for every U B and x U there is an element V B such that x V U

Given a top ology T on X a sub collection B of op en sets is a basis for T if for all U T and x U

there is an element B B such that x B U

Its p ossible to go to a lower level and generate a top ology by taking unions and nite intersections

Denition If X is a set a subbasis is a collection of subsets of X whose union is X

tersections of elements of S is a Prop osition If S is a subbasis in X the collection of all unions of nite in

top ology on X

Pro of Let T denote the collection of all unions of nite intersections of elements of S

The union of the elements of the subbasis is X so X TTaking the empty union gives the empty set

so T

The union of unions of nite intersections of elements of S are unions of nite intersections of elements

of S sosuch a union is an elementofT

The verication that T is closed under nite intersections is routine using the fact that intersections

distribute over unions Ill omit the details

Example Order top ologies Let X b e a set linearly ordered by a relation Consider the following

collection of subsets of X

All intervals s b where bs and s is the smallest elementofX if there is one

All intervals a l where al and l is the largest elementofX if there is one

All intervals a b where ab

I claim that this collection is a basis

First if x X Imust nd an element of the basis containing xIfx is the largest element takean

interval of the form a x if x is the smallest element takeaninterval of the form x b If x is neither

largest nor smallest I may nd a b suchthataxbThenx a b

Toverify the second axiom for a basis just observe that the intersection of twointervals is either empty

or an interval

The top ology generated by this basis is called the order top ologyOnR for instance this top ology

is the standard top ology

Example Partition top ologies Let X b e a set and let P b e a partition of X ThenP is a basis the

union of the sets in the collection is X and since the sets do not intersect the second basis axiom is satised

vacuously

The top ology generated by P is called the partition top ology on X The op en sets are unions of

partition elements

Note that if the partition consists of X itself the partition top ology is the indiscrete top ology on X

And if the partition elements are the p oints of X the partition top ology is the discrete top ology

Example Pro duct Top ology Supp ose X S and Y T are top ological spaces That is S is the

top ology on X and T is the top ology on Y

Dene a basis on the pro duct X Y by

B fU V j U SV Tg

I need to verify that this is a basis

First if x y X Y nd op en sets U X V Y such that x U X and y V Y Then

x y U V B

Now let U and U b e op en in X and let V and V be open in Y Supp ose that

U V U V x y

Note that

U V U V U U V V

Find op en sets U X and V Y suchthatx U U U and y V V V Then U V B

and

V x y U V U U V

Therefore B is a basis

The top ology generated by B is the pro duct top ology on X Y

In fact if C is a basis for the top ology on X and D is a basis for the top ology on Y then

E fU V j U CV Dg

is a basis for the pro duct top ology on X Y

For instance give R the standard top ology generated by the op en intervals Then the pro duct top ology

on R is generated by pro ducts of op en intervals ie op en rectangles In fact this top ology is the same

as the top ology generated by op en balls This will follow from the next lemma whichgives an easy wayof

telling when two top ologies are the same

Lemma Let T and T b e top ologies on X ThenT T if and only if for every U T and x U there

is an element V T such that x V U

Pro of Supp ose T T IfU T and x U then U T sox U U and the condition holds

Conversely supp ose that for every U T and x U there is an element V T suchthatx V U

Let U b e op en in T Iwanttoshowthat U T For each x U nd an element V T suchthat

S S

Hence T T x V U Then V T but clearly U Therefore U T

x x

xU xU

Corollary The pro duct top ology on R is the same as the top ology generated by the op en balls

Pro of The picture b elow indicates the idea

Example Subspace top ology Let X b e a top ological space and let A X Dene a top ology on A by

stipulating that the op en sets b e the intersections of op en sets of X with A

It is easy to check that this is a top ology It is called the subspace top ology on A When a subset A

of X is given the subspace top ology A is said to b e a subspace of X

It is also easy to check that if B is a basis for the top ology on X then

fU A j U Bg

is a basis for the subspace top ology on A

Remark If A X a subset of A that is op en in the subspace top ology need not b e op en in X

For example consider the subset Z R where R has the standard top ology Then fg is op en in the

subspace top ology on Z since is op en in R and fg Z

However fg isnotopeninRTake fg There is no op en interval a binR suchthat a b

In fact the subspace top ology on Zis just the discrete top ology

Theorem Let X and Y b e top ological spaces Let A b e a subspace of X and let B b e a subspace of Y

The pro duct top ology on A B agrees with top ology it inherits as a subset of X Y

Pro of Let U b e op en in A and let V b e op en in B Thus

U A U and V B V

where U is op en in X and V is op en in Y U V is a basic op en set in the pro duct top ology

Now

U V A U B V A B U V

A B Thus the pro duct top ology is contained Therefore U V is op en in the subspace top ology on

in the subspace top ology

Going the other way consider a basic op en set A B U V in the subspace top ology on A B

Then

A B U V A U B V

This is a pro duct of an op en set in A and an op en set in B so its op en in the pro duct top ology on

A B

Since an arbitrary op en set in the subspace top ology on A B is a union of basic op en sets it follows

that the subspace top ology is contained in the pro duct top ology

by Bruce Ikenaga

Closed Sets and Limit Points

Denition If X is a top ological space a subset A X is closed if X A is op en

Example In the standard top ology on R closed intervals are closed and p oints are closed

Note however than p oints need not b e closed in an arbitrary top ological space

In R the closed r ball centered at x

B x r fy R jjx y jr g

is closed

Closed sets satisfy complementary prop erties relative to op en sets

Prop osition Let X b e a top ological space

X and are closed

Finite unions of closed sets are closed

Arbitrary intersections of closed sets are closed

Pro of X is closed since X X is op en is closed since X X is op en

Let C C b e closed For each i write C X U where U is op en Then

n i i i

n n n n

X C X X U X X U U

i i i i

T S

n n

tersection of op en sets Therefore U is op en b ecause its a nite in C is closed

i i

Let fC g b e a family of closed sets For each a A write C X U where U is op en Then

a aA a a a

X C X X U X X U U

a a a a

aA aA aA aA

T S

C is closed U is op en b ecause its a union of op en sets Therefore

a a

aA aA

Its p ossible to construct a top ology by sp ecifying the closed sets rather than the op en sets Then you

can dene the op en sets to b e the complements of closed sets

for n are op en sets However Example Give R the standard top ology The intervals

n n

their intersection is

n n

which is a closed subset of R

Likewise the intervals are closed sets However their union is

n n

whichisopeninR

Op en sets in the subspace top ology are intersections of the subspace with op en sets in the big space

The next result says that the same is true of closed sets

Prop osition Let X b e a top ological space let Y b e a subspace of X and let C Y C is closed in Y if

and only if C Y D where D is closed in X

Note that closed in Y means closed in the subspacetopology on Y

Pro of Let D b e closed in X Write D X U where U is op en in X Then

D Y X U Y Y Y U

If x X U Y thenx Y and x U sox Y U Thismeansx Y Y U Converselyif

x Y Y U then x Y and x U since x Y U Therefore x X U so x X U Y

Y U is op en in Y so Y Y U isclosedinY

Conversely supp ose C is closed in Y Then Y C is op en in Y soY C Y U where U is op en in

X Hence

C Y Y C Y Y U X U Y

X U is closed in X soC is the intersection of a closed set in X with Y

Prop osition Let X b e a top ological space let Y b e a subspace of X and let Z Y

If Z is op en in Y and Y is op en in X then Z is op en in X

If Z is closed in Y and Y is closed in X then Z is closed in X

Pro of Ill prove the rst assertion the pro of of the second is similar

Supp ose Z is op en in Y and Y is op en in X Z Y U where U is op en in X ButY is op en in X

so Z is an intersection of two op en sets of X hence it is op en in X

Denition Let X b e a top ological space and let Y X

The closure of Y denoted Y or clY is the intersection of all closed sets containing Y

The interior of Y denoted Y or intY is the union of all op en sets contained in Y

Notice that the closed set X contains every subset Y so the intersection formed in taking the closure

is not taken over an empty collection

Remarks

The closure of a set is closed and the interior of a set is op en

The closure of a set is the smallest closed set containing the set likewise the interior of a set is the

largest op en set contained in the set

Y a set Y is op en if and only if Y Y A set Y is closed if and only if Y

Example In the standard top ology on R the closure of a bisa b and the interior of a bisa b

More interestingly Q R A set whose closure is the whole space is said to b e densethus the

rationals are dense in the reals

Example The closure of even a single p oint can b e quite large For example give Zthe top ology whose

op en sets are the op en intervals n nforn Z together with and Z

What is the closure of fg The only closed sets whichcontain are Zand Z Now

f g so the intersection of these two sets is ZfgThus the closure of a single p oint turns out to b e

everything in Zbut

Lemma Let X b e a top ological space let Y b e a subspace and let Z Y Then

cl Z Y Z

Here cl Z is the closure of Z in Y with the subspace top ology whereas Z is the closure of Z in X

Pro of cl Z is closed in Y socl Z C Y where C is closed in X Now

Y Y

Z cl Z C Y C

so C is a closed set in X containing Z

Therefore Z C so

Z Y C Y cl Z

Conversely Z is closed in X and contains Z soZ Y is closed set in Y containing Z Therefore

cl Z Z Y

Hence cl Z Y Z

Terminology

If X is a top ological space and x X aneighborhood of x is anyopensetcontaining x

Some authors use neighborhood to mean a set containing x in its interior

If X and Y are sets Ill saythatX intersects Y or X meets Y if X Y

Lemma Let X b e a top ological space and let Y X x Y if and only if every neighb orho o d of x

intersects Y

Y LetU b e a neighb orho o d of x Supp ose on the contrary that U Y Then Pro of Supp ose x

U X Y so X U X X Y Y

Now X U is a closed set containing Y soX U Y But then x Y X U implies x U

contrary to the assumption that U was a neighborhood of x

Conversely supp ose every neighborhood of x intersects y Let C b e a closed set containing Y Imust

show that x C since C is arbitrary this will provethatx Y

Supp ose on the contrary that x C Y C implies X Y X C that is X C do es not intersect

Y Further x X C and X C is op en Thus X C is a neighborhood of x which do es not meet Y

contrary to assumption Therefore x C whichiswhatIwanted to show

Remark If B is a basis for the top ology on X then x Y ifandonlyifevery B B intersects Y

In a sense the closure of a set consists of the original set together with other p oints that are close

to the original set you might picture these additional p oints as b oundary p oints I wanttomake the

notion of a b oundary p oint precise

Denition Let X b e a top ological space Y X Apoint x X is a limit p oint of Y is every neighb orho o d

of x intersects Y in some point other than x

Example In the standard top ology on Revery p ointofa b is a limit p ointofa b

On the other hand is not a limit p ointof For example the op en set contains but

do es not intersect

n Z has as its only limit p oint The set

Zhas no limit p oints in RAboutevery p ointinR Zyou can construct an op en interval whichdoes

not contain anyintegers And if n Z the op en interval n n intersects Zin the p oint n

Every real numb er is a limit p ointofQ since every op en interval ab out a real numb er contains rational

numb ers

Denition If Y is a subset of a top ological space X the derived set Y is the set of limit p oints of Y

Prop osition If Y is a subset of a top ological space X then

Y A A

Pro of Supp ose x Y Y

If x Y thenx Y

If x Y then x Y soevery neighb orho o d of x meets Y inapoint other than x but this fact

isnt relevant Since every neighb orho o d of x meets Y an earlier result implies that x Y

Conversely supp ose that x Y Ifx Y thenx Y Y and Im done

Supp ose then that x Y Sincex Y every neighborhood of x intersects Y But the intersection cant

consist of x since x Y Therefore every neighb orho o d of x intersects Y in a p oint other than xHence

x Y and certainly x Y Y

Corollary Let X b e a top ological space and let Y X ThenY is closed if and only if Y Y

Pro of If Y is closed then Y Y Y Y soy Y

ConverselyifY Y then Y Y Y so Y Y andY is closed

The last result is often expressed bysaying that a closed set contains al l its limit pointsThus you

often show a set is closed by taking a limit p oint and showing that it lies in the set

by Bruce Ikenaga

Hausdor Spaces

If you try to study top ological spaces without any conditions the variety b ecomes overwhelming

there is little you can say ab out top ological spaces in general In order to say more you have to assume

more The Hausdor condition is a common assumption mathematicians make ab out top ological spaces

Denition A top ological space X is Hausdor or T if for all x y X x y there are neighb orho o ds

U of x and V of y suchthatU V

y V

Remark An easy induction shows that if X is Hausdor and x x X then there are neighb orho o ds

U of x U of x such that the U s are pairwise disjoint

n n

Example R is Hausdor in the standard top ology

Let x and y b e distinct p oints in RLet jx y jThenx x and y y are disjoint

op en sets containing x and y resp ectively

ε ε ε ε

Example Consider the top ology on X f g in which the op en sets are f gand f g

123

X is not Hausdor since it isnt p ossible to nd disjoint neighb orho o ds containing and

Prop osition The pro duct of Hausdor spaces is Hausdor

Pro of Let X and Y b e Hausdor spaces Let x y x y X Y and assume that x y x y

Iwant to nd disjointneighborhoods in X Y containing the p oints

Since the p oints arent the same they must dier in at least one co ordinate Assume x x the

argument is similar if instead y y

Since X is Hausdor I can nd disjoint neighborhoods U of x and U of x Then U Y and U Y

are disjointneighborhoods in X Y x y U Y andx y U Y

(x1 , y1 ) y1

(x2 , y2 ) y2

U U x X

1 x1 2 2

Therefore X Y is Hausdor

Prop osition A subspace of a Hausdor space is Hausdor

Pro of Let X b e Hausdor and let Y b e a subspace of X Lety y Y where y y

Since X is Hausdor there are disjoint neighb orho o ds U of y and U of y in X ThenU Y is a

neighb orho o d of y in Y and U Y is a neighborhood of y in Y andU Y and U Y are disjoint

Therefore Y is Hausdor

Prop osition If X is Hausdor and x x X thenfx x g is closed

n n

Pro of First Ill show that a single p oint is closed Let x X I claim that X fxg is op en

Take y X fxg Since x y Imay nd disjointneighborhoods U of x and V of y Since V do esnt

intersect U and x U itfollows that V X fxgThus V is a neighb orho o d of y contained in X fxg

Since y was arbitrary X fxg is op en

X then fx x g is closed Since a nite union of closed sets is closed if x x

n n

If y is a limit p ointofaset Y then every neighborhood of y meets Y in a p oint other than y Ina

Hausdor space every neighb orho o d of a limit p oint meets the set in innitely many points

Prop osition Let X b e Hausdor and let Y X x is a limit p ointofY if and only if every neighb orho o d

of x meets Y in innitely many p oints

Pro of If every neighb orho o d of x meets Y in innitely many p oints then surely every neighb orho o d meets

Y in a p oint other than Y Therefore x is a limit p ointof Y

Conversely supp ose x is a limit p ointofY LetU b e a neighborhood of xImust show that U meets

Y in innitely manypoints

Supp ose U meets Y in only nitely manypoints Then U meets Y fxg in only nitely many p oints

x x Note that U must meet Y fxg in at least one point since x is a limit p ointof Y

Now X fx x g is op en and contains xsoU X fx x g is op en and contains x But

n n

in forming this set Ivethrown out the p oints x x where U meets Y so it follows that U X

fx x g do es not meet Y at all This contradicts the fact that every neighb orho o d of x must intersect

Hence U intersects Y in innitely manypoints

by Bruce Ikenaga

Continuous Functions

For a function f R R f is continuous at x a if for every there is a such that if

jx aj then jf x f aj

Interpreted in terms of intervals ie basic op en sets in R thissays that for anyinterval I containing

f a there is an interval J containing asuch that f maps J into I

This motivates the following denition

Denition Let f X Y b e a function where X and Y are top ological spaces f is continuous at x X

if for every neighborhood V of f x there is a neighb orho o d U of x suchthatf U V

V U x f

f(x) f(U)

f is continuous on X ifitiscontinuous at every p ointofX

The denition Ivegiven is often called the pointwise denition of continuity Ill give the equivalent

global form shortly

n m

Example Any function f R R that is continuous in the sense is continuous according to the

denition ab ove

Example Let X b e a top ological space The identity function id X X is continuous

To prove this let x X and let V be a neighb orho o d of idxxThenV is a neighb orho o d of x

which satises idV V

Example Let X and Y b e top ological spaces and let y Y The constant function c X Y dened by

c x y for all x X is continuous

To see this supp ose x X and V is a neighb orho o d of c xy X is a neighb orho o d of xand

c X fy gV Therefore c is continuous at x

y y

Example Let Y b e a subspace of the top ological space X The inclusion map i Y X given by iy y

for all y Y is continuous

To prove this let y Y and let V beaneighborhood of iy y in X Since V is op en in X V Y is

op en in Y andy V Y Clearly iV Y V so V Y is a neighborhood of y in Y which i maps into

V Therefore is continuous at y

Example The comp osite of continuous functions is continuous

To b e sp ecic supp ose that X Y and Z are top ological spaces f X Y is continuous at xand

g Y Z is continuous at f x Then g f X Z is continuous at x

Thus if f is continuous on X and g is continuous on Y then g f is continuous on X

To prove the result at a p oint assume that f X Y is continuous at x and g Y Z is continuous

at f x Let W b e a neighb orho o d of g f x in Z Bycontinuityofg at f x there is a neighborhood V

of f x suchthatg V W

By continuityoff at x there is a neighborhood U of x suchthatf U V

x f V

f(x)

f(U) g W

f(V) g(f(U))

g(f(x))

Then g f U g V W sog f is continuous at x

Example The restriction of a continuous function is continuous

Sp ecicallylet f X Y be a continuous function b etween top ological spaces Let U b e a subspace

of X Then f jU U Y dened by f jU xf x for all x U is continuous

The result is immediate from the last two examples since f jU f i where i U X is the inclusion

map and the comp osite of continuous functions is continuous

Example If f X Y is a continuous function b etween top ological spaces and Y is a subspace of Z then

the function f X Z dened by f xf xforallx X is continuous f is said to b e obtained by

expanding the range

This result is also easy since f i f where i Y Z is the inclusion map and the comp osite of

continuous functions is continuous

Next Ill show that the p ointwise denition of continuity is equivalent to the following global denition

Prop osition Let f X Y b e a function b etween top ological spaces f is continuous if and only if for

every op en set V in Y f V isopeninX

This is expressed more concisely bysaying that theinverseimageofanopen set is open

Pro of Supp ose f is continuous and let V b e op en in Y Iwanttoshow that f V isopeninX

Let x f V Iwant to nd a neighborhood of x contained in f V Now f x V soby

I can nd a neighb orho o d U of x suchthatf U V Then U f V and U is the desired continuity

neighb orho o d of xThisproves that f V isopen

Conversely supp ose that the inverse image of an op en set is op en I want to show that f is continuous

Let x X and let V beaneighborhood of f x I want to nd a neighborhood U of x such that f U V

By assumption f V is op en Since f x V it follows that x f V Moreover f f V V

Thus f V is a neighborhood of x that f maps into V Hence f is continuous

I b elieve that the name of the next result is due to James R Munkres its an enormously useful to ol

for constructing continuous functions

Lemma The Pasting LemmaLet X and Y b e top ological spaces and let U and V b e op en sets in X

such that X U V Supp ose that f U Y and g V Y are continuous and

f xg x for all x U V

Then the function h X Y dened by

f x if x U

g x if x V

is continuous

Pro of Let W b e op en in Y Then

h W f W g W

By continuity f W isopeninU since U is op en in X f W isopeninX

By continuity g W isopeninV since V is op en in X g W isopeninX

Therefore f W g W isopeninX soh W isopeninX By the preceding result h is

continuous

An analogous result holds if U and V are closed the pro of follows from the next result whichsays that

you can also express continuity in terms of closed sets

Prop osition Let f X Y b e a function b etween top ological spaces The following statements are

equivalent

f is continuous

For all U X f U f U

The inverse image of a closed set is closed

U U U I have Pro of Supp ose f is continuous and let U X Since

f U f U U f U f U

Clearly f U f U I need to showthatf U f U

Let x b e a limit p ointof U andlet V be a neighborhood of f x I must showthatV intersects f U

Since f V is a neighborhood of x and x is a limit p ointofU there is a y f V U suchthaty x

Then

f y f f V U f f V f U V f U

tersects f U so f x f U Thus f U f U and so f U f U f U This proves that V in

Therefore f U f U

Supp ose that for all U X f U f U I wanttoshowthattheinverse image of a closed set is

closed

Let C b e closed in Y Iwanttoshowthat f C is closed in X Now

f f C f f C C C

where the rst inclusion follows from statement the second inclusion follows from f f C C and

the last equality follows from the fact that C is closed

Thus f C f C But clearly f C f C so f C f C and f C is closed

Supp ose the inverse image of a closed set is closed I wanttoshow that f is continuous

Let V b e an op en subset of Y ThenY V is closed so by assumption f Y V is closed But

f Y V f Y f V X f V

Therefore X f V is closed so f V is op en Therefore f is continuous

by Bruce Ikenaga

Homeomorphisms

Denition Let X and Y b e top ological spaces and let f X Y f is a homeomorphism if f is bijective

and b oth f and f are continuous

If there is a homeomorphism b etween top ological spaces X and Y then X and Y are homeomorphic

Homeomorphic spaces are the same as top ological spaces in the same way that isomorphic groups

are the same as groups The big problem in top ology whichismuch to o dicult to deal with in this

generality is to classify all spaces up to homeomorphism

The following prop erties are obvious

The identity map of a top ological space is a homeomorphism

The comp osite of homeomorphisms is a homeomorphism

The inverse of a homeomorphism is a homeomorphism

Example Dene f R R by

f xx k

where k R Then f is a homeomorphism

Its clear that f is continuous since the inverse image of an op en interval is an op en interval its just

a translate of the original interval Then the inverse function f R R given by f x x k is also

continuous since it can b e written as f x x k

Dene g R R by

g xrx

where r R and r Then g is a homeomorphism

Again its clear that g is contin uous since the inverse image of an op en interval is an op en interval

shows that the inverse g R R is continuous Replacing r with

From these facts it follows that anytwononemptyopenintervals in R with the subspace top ologies

are homeomorphic For if a b is a nonemptyopeninterval then

f xx a maps a btob a

g x maps b ato

b a

f and g are homeomorphisms by observations ab ove This proves that every op en interval is homeo

morphic to But homeomorphism is transitive since the comp osite of homeomorphisms is a homeo

morphism so anytwononemptyopenintervals are homeomorphic

The function tan R is continuous and its inverse arctan R is continuous

Therefore is homeomorphic to RButevery nonemptyopeninterval is homeomorphic to

so it follows that every nonemptyopeninterval is homeomorphic to R

Denition Let f X Y b e a function b etween top ological spaces f is op en if f takes op en sets to op en

sets Likewise f is closed if f takes closed sets to closed sets

Lemma Homeomorphisms are op en maps and closed maps

Pro of Ill do the op en map case the closed map case is similar

Let f X Y b e a homeomorphism and let U b e op en in X Iwanttoshowthatf U isopeninY

Since f X Y is a homeomorphism it has a continuous inverse f By continuityf U is

op en in Y Butf U f U so f U isopeninY

Example You can tell that two spaces are not homeomorphic if you can nd a top ological prop erty that one

has that the other do es not In a sense this is a circular observation A topological property is a prop erty

preserved by homeomorphisms So this observation b ecomes useful as you develop a rep ertoire of prop erties

that are preserved by homeomorphisms

For example the circle

o n

x y S x y R

with the subspace top ology is not homeomorphic to the interval in R with the subspace top ologyOne

reason Removing a p oint from leaves a disjoint union of twosetsopenin but removing a p oint

from S do es not leave a disjoint union of two sets op en in S

Take for granted that the assertion ab out S is true Toshow that this dierence implies that the

spaces arent homeomorphic supp ose f is a homeomorphismTake x so fxg S

x x a disjoint union of op en subsets of

Since f is a homemorphism f x and f x are disjoint op en subsets of S and S ff xg

f x f x This contradicts my claim ab out S so S and arent homeomorphic

The assertion ab out S will follow from results Ill prove later ab out connected set

Example Removing a p ointfromR leaves a disjoint union of two op en sets The same is not true when a

n n

p oint is removed from R for n Therefore R is not homeomorphic to R for n

m n

The fact that R and R arent homeomorphic for m n is a fact called invariance of domain the

b est pro ofs use metho ds from algebraic top ology

The standard informal example of homeomorphic spaces is that a coee cup is homeomorphic to a

doughnut Popular accounts often say that two spaces are homeomorphic if you can imagine deforming one

into the other with cutting tearing or breaking

One of these things is not elik the others.

In the picture ab ove the three solid ob jects are homeomorphic the ob ject shap ed like a ring is not

homeomorphic to the others This kind of assertion like the one ab out the coee cup and the doughnut

must b e taken in the right spirit No one would set out to prove the assertion by writing down sp ecic

homeomorphisms

The notion of homeomorphism do es not include the idea of deforming one thing into another In the

rst place a deformation requires a big space inside of which the deformation takes place In the second

place a deformation is something whichtakes places in time Neither of these notions is part of the denition

of a homeomorphism The notion of isotopy comes closer to the idea of a continuous deformation of one

space into another

Homeomorphism is also indep endentoftheemb edding of a space in a larger space For example the

trefoil knot on the left and the circle on the right are homeomorphic as top ological spaces

In order to get a hold of the obvious dierence that exists b etween the two you need to consider the

complements of these sets in R In this case the ambient space R is resp onsible for the knottedness of

the trefoil

by Bruce Ikenaga

The Pro duct Top ology

If X and Y are top ological spaces the pro duct top ology on X Y is the top ology generated by the

basis consisting of all sets U V where U is op en in X and V is op en in Y Thismay b e extended to

a pro duct of nitely many spaces in the obvious fashion Howdoyou dene a top ology on a pro duct of

innitely many spaces

X is the top ology Denition Let fX g b e a family of top ological spaces The box top ology on

a a aA

generated by the basis consisting of sets U where U is op en in X

a a a

Let B b e the collection of sets U where U is op en in X Iwanttoshow that B is a basis

a a a

Q Q

X Since X is op en in X Let x X is an elementofB and it obviously contains

a a a a aA a

aA aA

a aA

Q Q

Let x U V where U and V are op en in X for each aThenU V is

a aA a a a a a a a

aA aA

U V is an elementofB Moreover op en in X so

a a a

Y Y Y

U V U V x

a a a a a aA

aA aA aA

This proves that B is a basis

There is another way to generate a top ology on a pro duct Ill need some preliminary results

Lemma If fT g is a family of top ologies on a set X thenT T is a top ology on X

a aA a

Pro of Since X T for all a it follows that X T

If fU g is a collection of op en sets in T then for each aIhave U T for all i Since T is a

i iI i a a

S S

U T for all a Hence U TThus T is closed under arbitrary unions top ology

i a i

iI iI

If fU U g is a collection of op en sets in T then for each aIhave U T for all i n Since

n i a

T T

n n

U TThus T is closed under nite intersections U T for all a Hence T is a top ology

i i a a

i i

Therefore T is a top ology

Denition Let fX g b e a family of top ological spaces let X beasetandletff X X j a Ag

a aA a a

b e a family of functions The top ology induced on X by the family ff g is the smallest top ology

a aA

on X whichmakes all the f s continuous

There is at least one top ology on X whichmakes all the f s continuous namely the discrete top ology

If I intersect al l the top ologies on X whichmakethe f s continuous I get a top ology by the last lemma

which makes all the f s continuous It is clearly the smallest such top ology in the sense that its contained

in every top ology whichmakes all the f s continuous

While this construction accomplishes the goal of making the f s continuous it would b e nice to have

a more concrete description of the op en sets

Lemma Let fX g b e a family of top ological spaces let X b e a set and let ff X X j a Ag be

a aA a a

a family of functions The top ology induced on X bythe f s is the same as the top ology generated by the

subbasis

o n

op en

U j U X a A f

Pro of First this collection is a subbasis If x X xa A and consider f x X Then X is op en in

a a a

X is an element of the collection which contains x X and f

a a

Now let S b e the top ology induced on X b ythe f s and let T b e the top ology generated by the subbasis

If U is op en in X and f U is an element of the subbasis then f U isopeninT Since U and

a a

f were arbitrary this shows that T makes all the f s continuous But S is the smallest top ology which

a a

makes the f s continuous Hence ST

ConverselyletU b e op en in X and let f U b e an element of the subbasis I claim that f U is

a a

an op en set in S For if not then f is not continuous since the inverse image under f of the op en set U

a a

isnt op en in X

Now I know that all the elements of the subbasis are contained in S ButS is a top ology so arbitrary

unions of nite intersections of elements of the subbasis are also contained in S Therefore TS

Hence the two top ologies are the same

Heres an imp ortantproperty of this top ology

Prop osition Let fX g b e a family of top ological spaces let X b e a set and let ff X X j a Ag

a aA a a

b e a family of functions Give X the top ology induced on X bythef s and let W b e a top ological space

A function g W X is continuous if and only if f g W X is continuous for all a

a a

Pro of If g is continuous then f g W X is continuous for all a since f g is a comp osite of continuous

a a a

maps

Conversely supp ose f g W X is continuous for all aIwanttoshow that g is continuous Let

a a

U b e op en in X Imust showthatg U isopeninW

To b egin with take U to b e a subbasic op en set Thus supp ose U f V where V is op en in X

Then

g U g f V f g V

Since f g is continuous f g V isopeninW

a a

For a general op en set U simply write U as a union of a nite intersections of subbasic op en sets then

use the fact that g commutes with unions and intersections

Now sp ecialize to the case where X X and give X the top ology induced by the family of

pro jection maps

X X j b A

b a b

The top ology generated in this way is called the pro duct top ology it is the smal lest topology which

makes al l the projection maps continuous

Notice that Ive already used the name pro duct top ology in the case of a pro duct of nitely many

spaces to mean the top ology generated by the basis consisting of pro ducts of op en sets from the factors In

fact if there are nitely many factors these two top ologies are the same

To see this its helpful to get an explicit description for the basic op en sets in the pro duct top ologyA

typical basis element has the form

U U U

a a a

where U is op en in X Note that U V U V so I can combine terms with the same

i a

i a a a

index Thus I mayaswell assume that a a a are distinct

In this case the set ab oveis

X if a a a a

a n

V where V

a a

U if a a i n

i i

Thus a typical basis element for the product topology is a product of sets open in the factors whereall

y factors but nitely many of these sets arethewholespaces In the case where there are only nitely man

this is simply pro ducts of sets op en in the factors the same as the pro duct top ology I dened earlier

The next result says that a function into a pro duct is continuous if and only if its factors are continuous

Corollary Let fX g b e a family of top ological spaces and give X the pro duct top ology Let W

a aA a

b e a top ological space and let g W X b e a function g is continuous if and only if g is

a a

continuous for all a

Example Dene f R R by

f t cos t sin t t

In this case the pro jections are

f tcost f t sin t f tt

Since these are continuous as functions R R it follows from the corollary that f is continuous as well

A nal note A basis element in the b ox top ology is a pro duct of sets op en in the factors Since this is

more general than the condition for basis elements of the pro duct top ology it follows that the box topology

is ner than the product topology

by Bruce Ikenaga

Metric Spaces

Denition If X is a set a metric on X is a function d X X R such that

dx y for all x y X dx y if and only if x y

dx y dy x for all x y X

Triangle InequalityFor all x y z X

dx y dy z dx z

Lemma If X is a set with a metric the collection of op en balls

B x fy X j dx y g

is a basis

Pro of If x X then x B x

Supp ose B x and B x are op en balls Let x B x B x

Let

min dx x dx x

Then x B x B x B x Therefore the collection of op en balls forms a basis

Denition If X is a set with a metric the metric top ology on X is the top ology generated by the basis

consisting of op en balls B x where x X and A metric space consists of a set X together with

a metric dwhere X is given the metric top ology induced by d

Remark In generating a metric top ology it suces to consider balls of rational radius

Example The usual metric on R is dened by

dx y x y

i i

where x x x and y y y

n n

It is clear that dx y and that dx x for all x y R Ifx x x andy y y

n n

and dx y then

n n

X X

x y so x y

i i i i

i i

This is only p ossible if x y forall i and this in turn implies that x y for all i Therefore

i i i i

x y

It is obvious that dx y dy x for all x y R

Note that dx y jx y j where jj is the standard norm whichgives the length of a vector Now

juj u u where denotes the dot pro duct in R By standard prop erties of the dot pro duct

u v v v juj u v jv j juj jujjv j jv j juj jv j ju v j u v u v u u

The inequalityfollows from the Schwarz inequality ju v jjujjv j Then

ju v jjuj jv j

Now let u x y and v y z Then

jx z j jx y j jy z j or dx z dx y dy z

Here is a pro of of the Schwarz inequalityincaseyou ahvent seen it Given u u u v

v v R Iwanttoshow that ju v jjujjv j Ill show that ju v j juj jv j and the result will

followby taking square ro ots

Set A juj C jv j and B ju v jIwanttoshowthat B AC

If A then u and the result is obvious Assume then that A For all x R

u x v

i i

n n n

X X X

x u x u v v

i i

i i i

x A xB C

Take x The last inequality yields

B B

A B C

A A

AC C

This completes the pro of of the Schwarz inequality

Thus the standard metric on R satises the axioms for a metric Obviously the metric top ology is

just the standard top ology

Lemma Comparison Lemma for Metric Top ologiesLet d and d b e metrics on X inducing top ologies

T and T T is ner than T if and only if for all x X and there is a suchthatB x

B x

Pro of Supp ose rst that T TLetx X and let B x isopenin T soitsopeninT Since

the op en dballs form a basis for T there is an op en ball B x such that

x B x B x

d d

Conversely supp ose that for all x X and there is a such that B x B x I want

d d

to show that T T

Let U b e op en in T Iwanttoshow that its op en in T Let x U Since the d balls form a basis for

T there is an suchthat

x U x B

By assumption there is a suchthat

x B x B x

d d

B x U Therefore x

Now B x isa T op en set containing x and contained in U Since x U was arbitrary U is op en in

T Therefore T T

The standard metric on R is unb ounded in the sense that you can nd pairs of p oints whichare

arbitrarily far apart However you can always replace a metric with a bounded metric whichgives the same

top ology

Denition If X d is a metric space and Y X then Y is b ounded if there is an M R such that

dx y M for all x y Y

Lemma Let X b e a metric space with metric d Dene

dx y mindx y

d is a metric

d and d induce the same top ology on X

Pro of Let x y X Since dx y dx y mindx y and

dx x mindx x min

If dx y mindx y then dx y so x y This shows that the rst metric axiom holds

Since dx y mindx y mindy x dy x the second metric axiom holds

Toverify the third axiom take x y z X Begin by noting that if either dx y ordy z

then dx y or dy z Therefore

dx y dy z dx z

Assume that dx y anddy z Then

dx y dy z dx y dy z dx z dx z

This veries the third axiom so d is a metric

x The idea is to apply the Comparison Lemma shrinking Observe that for B x B

balls if necessary to make their radii less than

Let x X and let

If then x B x B x

If then

x B x x B x B

d d

Therefore the dtop ology is ner than the d top ology The other inclusion follows bysimplyswapping

the ds and d s

It follows that boundedness is not a top ological notion since every subset is b ounded in the standard

b ounded metric

Example The square metric on R is given by

x y maxjx y jjx y j

n n

Relative to this metric B x isan ncub e centered at x with sides of length

First Ill showthat is a metric Let x x x y y y R

n n

Clearly x y andx x If x y then jx y j for all isox y

i i

Its also obvious that x y y x

If x y z R then for each j

x y y z maxfjx y jg maxfjy z jg jx y j jy z j jx z j

i i i i j j j j j j

i i

Therefore

x y y z maxfjx z jg x z

j j

Thus is a metric

Lemma induces the same top ology on R as the standard metric

Pro of The idea of the pro of is depicted b elow

Note that

x y maxfjx y jg jx y j x y x y dx y

i i j j j j j j

p p

n x y nx y dx y x y n maxfx y g

j j i i

These inequalities maybeusedtogetballs contained in dballs and dballs contained in balls by

the Comparison Lemma this shows that the top ologies are the same

Lemma The square metric induces the pro duct top ology on R

Pro of If x x x R then

B x x x x x x x

n n

The set on the right is op en in the pro duct top ology Since the square metricbasic sets are op en in the

pro duct top ologyany square metricop en set is op en in the pro duct top ology

Conversely let

U a b a b a b

n n

It is easy to check that sets of this form comprise a basis for the pro duct top ology

Let x x x U soa x b for all i Dene

n i i i

minfx a b x g

i i i i

Then

x x a b for all i

i i i i

Y Y

B x x x a b U

i i i i

i i

It follows that U is op en in the square metric top ology Since the pro duct top ology basic sets are op en

in the square metric top ologyany pro duct top ology op en set is op en in the square metric top ology

Lemma Metric top ologies are Hausdor

Pro of Let X d b e a metric space and let x and y b e distinct p oints of X Let dx y Then B x

and B y are disjoint op en sets in the metric top ology whichcontain x and y resp ectively

Lemma If X d Y d are metric spaces the denition of continuityisvalid That is a map f X Y

is continuous at x X if and only if for every there is a such that if dx y implies that

df xfy

Pro of First supp ose that f X Y is continuous at x X Let and consider the ball B f x

Since this is an op en set containing f x continuity implies that there is a suchthat

f B x B f x

Now consider the conclusion to b e established Supp ose y X satises dx y Then y B x

so f y f B x Therefore f y B f x so df xfy

Conversely supp ose that for every there is a suchthatifdx y implies that

df xfy I want to show that f is continuous

Let x X and let V b e an op en set in Y containing f x I want to nd a neighb orho o d U of x such

that f U V

Since the balls form a basis for the metric top ologyI mayndan suchthatf x B f x V

By assumption there is a such that if dx y implies that df xfy

Now consider the ball B x This is an op en set containing xIfy B x then dx y

Therefore df xfy so f y B f x This shows that f B x B f x V sof is

continuous

Denition If X is a set a sequence in X is a function x Z X

Its customary to write x for xn in this situation and to abuse terminology by referring to the

collection fx g as the sequence

Denition Let X b e a top ological space A sequence fx g converges to a p oint x X if for every

neighb orho o d U of x there is an integer N such that x U for all n N

x x means that fx g converges to x

n n

Lemma Let X b e a Hausdor space Convergent sequences converge to unique points

x y Iwanttoshowthatx y Pro of Let x x and

n n

Supp ose x y Since X is Hausdor I can nd disjointneighborhoods U of x and V of y Since

x xIcanndaninteger M such that n M implies x U Since x y I can nd an integer N

n n n

such that n N implies x V Therefore for n maxM N I have x U V This is nonsense

n n

so x y

In particular limits of sequences are unique in metric spaces

Lemma The Sequence Lemma Let X b e a top ological space let Y X and let x X If there is a

sequence fx g with x Y for all n and x xthen x Y The converse is true if X is a metric space

n n n

Pro of Supp ose that there is a sequence fx g with x Y for all n and x xLetU b e a neighb orho o d

n n n

Y of x Find an integer N suchthatx U for all n N Obviously U meets Y This proves that x

the ball B x Conversely supp ose that X is a metric space and x Y For each n Z meets Y

Y I claim that x x soImaycho ose x B x

n n

Let U b e a neighborhood of x Since the op en balls form a basis for the metric top ologyI maynd

such that B x U thenImaynd N Z such that so B x B x

n N

Since B x U I have x U for all n N sox B x For all n N Ihave

n n

n N N N

This proves that x x

Theorem Let X b e a metric space let Y b e a top ological space and let f X Y f is continuous if and

only if x x in X implies that f x f xinY

n n

More succinctly continuous functions carry convergent sequences to convergent sequences

Pro of Supp ose f is continuous and supp ose x x in X Let V b e a neighb orho o d of f xinY By

continuity there is a neighborhood U of x suchthatf U V

Since x x there is an integer N such that x U for all n N Then f x f U V for all

n n n

n N This proves that f x f x

Conversely supp ose that x x in X implies that f x f xinY To show f is continuous it will

n n

suce to show that for all A X Ihave f A f A

Thus take x cl AIwanttoshowthatf x f A

Now X is a metric space and x cl Asoby the Sequence Lemma there is a sequence of p oints fx g A

with x xByhyp othesis this implies that f x f x Since ff x g is a sequence in f A the

n n n

Sequence Lemma implies that f x f A Therefore f is continuous

Denition Let ff X Y g b e a sequence of functions from X to Y where Y is a metric space ff g

n n

converges uniformly to a function f X Y if for every there is an integer N such that

df xfx for n N and x X

Theorem Let ff X Y g b e a sequence of continuous functions from X to Y where Y is a metric space

If ff g converges uniformly to f X Y then f is continuous

This is often expressed bysaying that a uniform limit of continuous functions is continuous

Pro of Let f a Y and let B f a b e a neighb orho o d of f a I want to nd a neighb orho o d U of a

such that f U B f a

First uniform continuity implies that there is an integer N such that

for n N x X df xfx

In particular

df afa

f is continuous so there is a neighborhood U of a suchthatf U B f a Thus

N N N

d f xf a for all x U

N N

Moreover restricting to n N and x U I have

df xfx for all x U

Therefore the triangle inequality implies that

df xfa df xfx d f xf a df afa

N N N N

for all x U

This proves that f is continuous

Example For n Z let f R R b e dened by

f x

For xed x lim Thus this sequence of functions converges p ointwise to the constant

function

The picture b elow shows the graphs of f for n ontheinterval x n

0.5

0.4

0.3

0.2

0.1

0.2 0.4 0.6 0.8 1

I will show that the convergence is uniform on the interval x Thus cho ose Imust nd

an integer N such that if n N then

jf xj

f is an increasing function f so it follows that Since f x

n n

nx n

jf xj for x

Then if n N Nowcho ose N such that

jf xj

n N

This proves that f converges uniformly to on x

Example For n Z let f R R b e dened by

f x

Thus this sequence of functions converges p ointwise to the constant For xed x lim

function for x Itconverges p ointwise to for x

The picture b elow shows the graphs of f for n ontheinterval x n

0.2 0.4 0.6 0.8 1

0.8

0.6

0.4

0.2

I will show that the convergence is not uniform on the interval x In fact I will show that there

is no integer N such that if n N then

jf xj

To see this it suces to note that f so there will always b e a p ointinx where the

function exceeds

Therefore ff g converges p ointwise but not uniformly to the zero function

by Bruce Ikenaga

Quotient Spaces

You mayhave encountered quotient groups or quotient rings in an algebra course In those situa

tions the idea was to turn a set of cosets of a subgroup or an ideal into a group or a ring

A set of cosets is a partition of a group or a ring In the top ological setting a quotientspace arises as

a top ology on a set of partition elements You can use this to construct new top ological spaces by gluing

and pasting

If X and Y are top ological spaces a function f X Y is continuous if and only if when V is op en in

Y f is op en in X

Denition Let f X Y where X and Y are top ological spaces f is a quotientmapif

f is surjective

V is op en in Y if and only if f V isopeninX

Remarks

Quotient maps are continuous

If f X Y is a homeomorphismthen f is a quotien tmap

Since f is a homeomorphism its continuous hence if V is op en in Y then f V isopeninX

Conversely supp ose V Y and f V isopeninX Sincef is continuous

V f f V f f V is op en

The comp osite of two quotient maps is a quotientmap

In the denition of a quotient map there are top ologies on b oth X and Y More often you start with

a top ological space and a partition of the space then try to top ologize the set of partition elements so that

the asso ciated function carrying an element of the space to the partition elementcontaining it is a quotient

map

b e a surjection Dene a collection Lemma Let X b e a top ological space let Y b e a set and let f X Y

of subsets of Y by requiring that V T if and only if f V isopeninX ThenT is a top ology on Y

T is called the quotient top ology on Y

Pro of f is op en in X so is op en in Y f Y X is op en in X soY is op en in Y

Supp ose fV g is a collection of subsets of Y suchthatf V isopeninX for all i I Now

i iI i

f V f V

i i

iI iI

S S

and f V isopeninX Therefore V is op en in Y

i i

iI iI

Finally supp ose V V is a collection of subsets of Y suchthatf V isopeninX for i n

n i

n n

f V f V

i i

T T

n n

and f V isopeninX Therefore V is op en in Y

i i

is a top ology Therefore T

Example Partition Top ologiesLet X b e a top ological space and let X fX g b e a partition of

a aA

X Dene f X X by

f xX if x X

a a

This makes sense b ecause every elementof X lies in exactly one X

f is clearly surjective the quotient top ology on X induced by f is called the partition top ology

In fact all quotient top ologies are partition top ologies For if X is a top ological space and f X Y is

surjective the collection of inverse images ff y j y Y g forms a partition of X The partition top ology

induced by this partition is the same as the quotient top ology induced by f

t to glue parts of a space together to make a new space In this case In applications you often wan

the partition elements are as follow A p oint that isnt b eing glued to anything else is in a partition element

by itself Otherwise a group of p oints that are b eing glued together form a partition element

For example take the interval Partition by dening f g to b e one partition elementand

dening fxg to b e a partition elementforx This has the eect of gluing the endp oints together

the quotient space is homeomorphic to a circle

glue the endpoints

glue the

sides

Similarlyifyou take a rectangle and glue two opp osite sides together you get a cylinder In this case

the partition elements consist of individual p oints not on the glued sides and pairs of corresp onding p oints

on the sides b eing glued

If you twist the rectangle b efore gluing the opp osite sides the resulting quotient space is a Mobius strip

If you glue b oth pairs of opp osite sides together with no twisting you get a torus

Lemma Let X and Y b e top ological spaces If f X Y is continuous op en and surjective then f is a

quotientmap

Pro of If V is op en in Y then f V isopeninX bycontinuity

Supp ose V Y and f V isopeninX Sincef is op en f f V is op en since f is surjective

V f f V Therefore V is op en

Hence f is a quotient map

In general quotient maps do not need to b e op en

Covering spaces provide imp ortant examples of quotient maps they are fundamental ob jects in

algebraic top ology

Denition Let p E B b e a continuous surjection p is a covering map if every p oint x B has

U U is a a neighb orho o d U such that p U is a disjointunion U of op en sets and pj

a a U

homeomorphism for all a A

A neighborhood U which satises this condition is evenly covered

Thus p U lo oks like a stack of pancakes where each pancake is carried homeomorphically onto U

by p

Lemma Covering maps are op en

Pro of Let p E B be a covering map and let U E be open I wanttoshow that pU isopeninB

Take x U sopx pU I need to nd a neighborhood V of pxcontained in pU Let W be an

evenly covered neighborhood of px Since p W is a disjoint union W of op en sets homeomorphic

to W and since x p W x must lie in one of the W s

Thus let W be a neighborhood of x mapp ed homeomorphicallyonto W by pThenW U is op en

a a

p maps it homeomorphically onto pW U and px pW U I claim that pW U isopeninB

a a a

To see this observethatpj W pW W is a homeomorphismsotheinverse pj W W

W a a a

W U pW U isopen is continuous Now W U is op en in E so its op en in W Hence pj

a a a a

in pW But pW W is op en in B sopW U isopeninB

a a a

Ive found a neighb orho o d pW U ofx whichiscontained in pU Therefore pU is op en and p

is an op en map

Corollary Covering maps are quotient maps

Example The map e R S given by

etcost sint

is a covering map it wraps the real line around the circle Eachopeninterval of length in R is mapp ed

homeomorphically to the circle minus a p oint

Example If X Y is a pro duct of top ological spaces the pro jection maps X Y X and

X Y Y are op en continuous and surjective Therefore they are quotient maps

The next result tells you how to construct a map out of a quotient space it is analogous to the results

in algebra whichtellyou how to construct group maps out of quotient groups or ring maps out of quotient

rings

Denition Let p X Y b e a quotientmapAb er of p is a set p y for y Y

Theorem Universal Prop erty for QuotientsLet p X Y b e a quotientmapThenanycontinuous

map f X Z which is constant on eachberofp factors uniquely through p that is there is a unique

continuous map f Y Z suchthatfp f

Z Y

Pro of Let y Y Since f is constantonp y I may takeany x p y and dene

f y f x

By construction fp f Ihavetoshowthatf is continuous

Let U b e op en in Z I need to show f U isopeninY By denition of the quotient top ology f U

is op en in Y if and only if p f U is op en in X But since fp f

p f U f U

whichisopeninX by continuityof f

This proves that f is continuous

To prove uniqueness supp ose that f and f are continuous maps Y Z such that f p f and

f p f Thenf p f f p since p is surjective it is rightcancellable so f f

If you think of the quotient top ology as gluing or identifying the elements of each b er the universal

prop ertysays that the quotient top ology is the smallest way of accomplishing this gluing in the sense

that any other map that accomplishes this gluing receives a map from the quotient space

Corollary Let f X Z b e a continuous surjective map and supp ose p X X is the quotient map

obtained from the partition of X by the b ers ff z j z Z g

There is a continuous bijection f X Z suchthatfp f

f is a homeomorphism if and only if f is a quotient map

f x f is Pro of p is the function which sends x X to the partition element f z X where z

trivially constant on each b er so the Universal Prop erty yields a continuous map f X Z suchthat

fp f

To b e explicit consider a b er f z X Let x f z Then px f z so

f f z f px f xz

The function g Z X given by g z f z isob viously the inverse of f as a set mapso f is

bijective Thus f is a continuous bijection

If f is a homeomorphism then f fp is a comp osite of quotient maps so its a quotient map

Conversely supp ose f is a quotientmap p is constant on eachberoff so the Universal Prop erty

applied to the quotientmapf yields a continuous map h Z X such that hf p

Now fhf f Thatisfh makes the outer triangle b elow commute

f f

Z X Z

But the outer triangle clearly commutes if I replace fh with id By the uniqueness part of the Universal

Prop erty applied to the quotientmapf it follows that fh id

Likewise hfp p In other words hf makes the outer triangle b elow commute

p p

Z X X

By the uniqueness part of the Universal The outer triangle will commute if I replace hf with id

Prop erty applied to the quotientmappitfollows that hf id

Therefore h f In fact h is the map g dened in the rst part of the pro of

Since f is a continuous bijection with a continuous inverse its a homeomorphism

Corollary Let f X Z b e a continuous surjective map and supp ose p X X is the quotient map

obtained from the partition of X by the b ers ff z j z Z gIfZ is Hausdor then so is X

Pro of Supp ose f z and f z are distinct elements of X Then z z soImay nd disjoint

neighb orho o ds U of z and U of z in Z

U These are op en in X since f is continuous Since U and U are Lo ok at f U andf

disjoint f U andf U are disjoint

Finally f f z z U shows that f z f U and likewise f z f U

Thus I have disjoint neighb orho o ds of f z and f z inX so X is Hausdor

by Bruce Ikenaga

Connected Spaces

Denition A space is connected if it cannot b e written as the disjoint union of twononemptyopensets

A separation of a top ological space X consists of two disjoint nonempty op en sets U and V suchthat

X U V

Thus a connected space is one that cannot b e separated

Example If X is connected and Y is homeomorphic to X then Y is connected

Example A p oint is connected considered as a subspace of an arbitrary top ological space

Example R is connected in the standard top ology

On the other hand Zis not connected when considered as a subspace of R The subspace top ology is

the discrete top ology in this case

More generally a set with more than one p oint is not connected in the discrete top ology

Example If X U V is a separation of X thenU and V are b oth op en and closed

ConverselyifA is a subset of a top ological space X A XandA is b oth op en and closed then

X A X A is a separation of X

To put it another way a space is connected if and only if there is no prop er nontrivial subset whichis

b oth op en and closed

For example any nontrivial partition top ology is not connected

Prop osition Let X U V b e a separation of X IfY X is connected then Y U or Y V

Pro of Supp ose on the contrary that Y U and Y V Then Y U and Y V are nonemptyopen

subsets of V whose union is V contradicting the fact that Y is connected

If a space X is not connected what are the biggest connected subspaces of X

Lemma Let fU g b e a collection of connected subsets of a top ological space X Supp ose that U

a aA a

Then U U is connected

U Thenx V or Pro of Supp ose on the contrary that U V W is a separation of U Let x

x W without loss of generality supp ose x V

For each a the connected set U is contained in either V or W by the preceding result But x U for

a a

all a so all the U s meet V Therefore U V for all a and hence U V

a a

This means that W which contradicts the fact that V and W separated U

Therefore U is connected

Theorem Let X b e a top ological space

Every p oint x X is contained in a unique maximal connected subset C of X

C C or C C

x y x y

TosaythatC is maximal means that C is not prop erly contained in any other connected subset of

x x

Pro of Let x X Consider the collection C of connected subsets of X whichcontain x C is nonempty

since fxg C OrderC by inclusion Ill use Zorns Lemma to showthatC has maximal elements

Let D be a chain in C Then every U D contains xso V U is a union of connected sets with

U D

a p oint in common By the preceding result V is connected since it contains x and contains every U D

it is an upp er b ound for D in C

By Zorns Lemma C has maximalelements

Next Ill showthatC has a unique maximal element

If U and V are maximal elements of C thenU V is a connected set since U and V are connected

and have x in common Thus U V is a connected set containing xsoitsanelementofC Since U and

V are maximal and are contained in U V this is only p ossible if U V

Note that if C is the unique maximal elementinC any connected set containing C would b e a

x x

connected set containing x henceanelementof C containing the maximal element C Therefore there

is no connected set which properly contains C

Supp ose C C Letz C C ThenC C is connected since its the union of connected

x y x y x y

sets having a p oint in common Moreover C C contains the maximalsets C and C By part this is

x y x y

imp ossible unless C C

x y

The maximal connected sets describ ed by the theorem form a partition of X They are called the

connected comp onents of X Here is another sense in which they are maximal

Lemma Let X b e a top ological space Every connected subset of X is contained in a comp onentof X

Pro of Let U b e a connected subset of X U must intersect some comp onentof X supp ose U C where

U is a comp onent Then U C is a union of connected sets with nonemptyintersection so its connected

Since C U C and since C is maximal I must have C U C Therefore U C

Theorem The continuous image of a connected set is connected

Pro of Let f X Y b e continuous and surjective and supp ose X is connected I wanttoshow that Y is

connected

Supp ose on the contrary that Y U V is a separation of Y Since U and V are op en f U

and f V are op en since U and V are disjoint f U and f V are disjoint Therefore X

f U f V is a separation contradicting the fact that X is connected

The theorem will b e useful once I have some connected spaces to play with

Im working toward showing that the pro duct of connected spaces is connected the preliminaries are

interesting in their own right

Lemma Let X b e a top ological space let Y X and let Y A B b e a partition of Y A and B separate

Y if and only if neither A nor B contains an X limit p oint of the other

Pro of Supp ose A and B separate Y Then A and B are closed in Y so

A Y cl A A

Supp ose x is an X limit p ointofAso x AIfx B thenx Y sox A Y A But then

x A B contradicting the fact that A and B are disjoint

Therefore B contains no X limit p oints of A

A similar argumentshows that A contains to X limit p oints of B

Conversely supp ose neither A nor B contains an X limit p oint of the other Let A denote the X limit

p oints of A Then A Y is the set of X limit p oints of A contained in Y by assumption none of these are

in B soany suchpoints must b e in A That is A Y AThen

A Y A A Y A Y A Y A A A A Y

Therefore A A Y so A is closed in Y

A similar argumentshows that B is closed in Y

Since A and B partition Y they form a separation of Y

Lemma If A is connected and

A B A

then B is connected

Heuristicallyifyou add some limit p oints of a connected set to the set you get a connected set

Pro of Supp ose B U V is a separation of B Thus U and V are disjoint nonempty op en subsets of B

Since A is connected A lies entirely in U or V Without loss of generalitysay A U Taking closures in

X yields A U

U and V are disjoint and by the preceding lemma V contains no limit p oints of U Thus A V

But B AsoB do esnt intersect V either This means that V is empty which is a contradiction

Therefore B is connected

Corollary The closure of a connected set is connected

Corollary Connected comp onents of a space are closed

Pro of If C is a connected comp onentofX thenC is a connected set containing C by the previous corollary

By maximality C C so C is closed

Lemma The pro duct of nitely many connected spaces is connected

Pro of It suces to prove this for two spaces since the general result will followby induction

Thus supp ose X and Y are connected I wanttoshow that X Y is connected The idea is to use the

fact that a union of connected sets with a p oint in common is connected

To do this picture X Y as a plane Ill decomp ose it as the union of crosses each cross consists

of the union of a xed horizontal line with an arbitrary vertical line

{x} x Y

X x {y0 } y0

Fix y Y For each x X consider the set

C X fy g fxgY

X fy g is connected since its a homeomorphic copyof X fxg Y is connected since its a homeo

morphic copyof Y The sets have the p ointx y in common Therefore C is connected

Next the union of the C s is connected b ecause its a union of connected sets having X fy g in

common But the union of the C s is X Y so X Y is connected

Theorem An arbitrary pro duct of connected spaces is connected

Pro of First Ill cho ose a p oint in the pro duct Ill construct a union of connected sets having this p ointin

common the earlier result on unions shows that the union is connected Finally Ill show that the closure

of the union is the whole pro duct this will prove that the pro duct is connected by the earlier result on

closures

Let X X b e a pro duct of connected spaces IwanttoshowthatX is connected Let

y y X

a aA

For each nite collection fa a g of indices from A dene

X a a x j x y if a a a g

n a aA a a n

Thus a p ointofX a a agrees with y except p ossibly in the a a p ositions

n n

Note that this do esnt mean that suchanx doesnt agree with y in the a a p ositions it means

only that suchanx does agree with y in the other p ositions

I claim that X a a is connected Ill do this byshowing that its homeomorphic to a nite

pro duct of X s which is connected by the preceding lemma

Dene X X X a a by

a a n

x if a a for some i

a i

x x

a a a

y if a a for any i

a i

Thus plugs x a a comp onents and uses y for the other comp onents x into the

a n a

Dene X a a X X by

n a a

x x x

a a a

and are easily seen to b e inverses so is bijective

A basis element for the pro duct top ology on X X is a pro duct of op en sets from the factors

a a

A basis element for the subspace top ology on X a a consists of the intersection of X a a with

n n

Q Q

a basis element for the pro duct X X A basis element for X X consists of a pro duct of

a a

aA aA

op en sets from the factors where at most nitely many are not the whole space

With this description in mind it is clear that carries a basis elementfor X X onto a

a a

basis elementforX a a and carries a basis elementforX a a onto a basis element for

n n

X X Therefore is a homeomorphism

a a

Since X X is a nite pro duct of connected spaces its connected Therefore X a a

a a n

is connected

Since y X a a for any nite subset fa a g A the union

n n

o n

a gA X X a a fa

n n

is a union of connected sets with a p oint in common hence connected

Finally I claim that X X Let x X and let U b e a neighb orho o d of xwhere U is op en

a a

in X and U X for all but nitely many as Let a a b e the indices for which U X Dene

a a a n a a

z z by

a aA

x if a a for some i

a i

y if a a for any i

a i

X Then z U and z X a a

a n

X X Thus every basic neighb orho o d of x intersects X Since x X was arbitrary it follows that

Since X is the closure of a connected set its connected

by Bruce Ikenaga

Connected Subsets of the Real Line

Prop osition is connected in the standard top ology

Pro of Supp ose U V is a separation so U and V are nonempty disjoint op en sets

The idea of the pro of is to nd a p oint on the b oundary b etween U and V then show that sucha

p oint cant b e in either U or V

Without loss of generality supp ose that U Let

S fs j s U g

Since S S is nonempty Moreover since U and U is op en there is an suchthat U

i h

U so S Thus S fg Then

S is a nonemptysetwhich is b ounded ab ove so it has a least upp er b ound Let

s supS

Since S fg s

s is the sort of b oundary p oint I alluded to ab ove Before using it to obtain a contradiction Ill

provetwo preliminary results

Claim If s S and rs then r S

r s U since s S

Therefore r S

Claim If ss then s S and s U

Supp ose ss but s S Ift s then t S else s S by Claim Since no elementofS is

greater than s it follows that s is an upp er b ound for S Butss and s is the least upp er b ound for S

This contradiction implies that s S thus s U so s U

Finally Ill obtain a contradiction byshowing that s cantbeinS or its complement

Supp ose s S sos U Note that s for if s then s U contradicting

the fact that V is nonempty

U is op en so there is an suchthats s U Hence Nows s U and

h i h i

s s s s U

S This contradicts the fact that s is the least upp er b ound for S so s

It follows that s S Therefore s U By Claim s U sos U Therefore s V

V and Now s and V is op en so there is an suchthats s V In particular s

this contradicts Claim

It follows that there is no such separation and hence is connected

Corollary Any closed interval in R is connected

Pro of Any closed interval in R is homeomorphic to

Corollary R is connected

Pro of

R n n

represents R as a union of connected sets having p oints in common sp ecically the p oints in are

common to all of the intervals By an earlier result this proves that R is connected

Corollary R is connected

Pro of R is a pro duct of connected spaces so its connected

Corollary Anyopeninterval in R is connected

Pro of Anyopeninterval in R is homeomorphic to R

In fact more is true Let L b e an ordered set with at least twoelements L is a linear continuum if

L has the least upp er b ound prop erty every subset b ounded ab ove has a least upp er b ound and if xy

implies there exists z suchthatx z yFor example R is a linear continuum One mayshow that every

linear continuum is connected

You can show that manysetsin R are connected by using the results ab ove and general stu ab out

connected sets

Example S is a connected subset of R

To see this note that the continuous function f R S given by f t cos t sin t maps the

connected set R onto S

More generallyif f R R is continuous then the image is connected For example the graph of a

function y f x dened for all x R is a connected set

Example The top ologists sine curve consists of the graph of y sin for x The closed

top ologists sine curve consists of the union of the top ologists sine curve with the segment fy j

y g

The top ologists sine curve is connected b ecause its a continuous image of a connected set The closed

top ologists sine curve is the closure of the top ologists sine curveinR so its connected as well

Heres a familiar result from calculus

Theorem Intermediate Value Theorem Let f X Y be a continuous function where X is connected

and Y is an ordered set with the order top ologyLeta b X and let d b e an elementof Y between f a

and f b so f a d f borf b d f a Then d f c for some c X

Pro of If d f aord f b then Im done So assume d f afb Then f a dfbor

f b dfa

f X is connected since X is connected Consider the sets f X dandf X d These

are disjoint op en subsets of f X they are nonempty since one contains f a and the other contains f b

Therefore their union isnt all of f X Hence f X fdg is nonempty

An element f X fdg is an element f cforc X suchthatf cd

by Bruce Ikenaga

PathConnectedness and Lo cal Connectedness

Denition Let X b e a top ological space A path in X is a continuous function X and

are the endp oints of the path is apathfrom to

Denition Let X b e a top ological space X is path connected if for all x y X there is a path in X

from x to y

Example Points are path connected

n n

Example R is path connected If x y R then

t tx ty

is a path in R from x to y

Lemma A path connected space is connected

Pro of Supp ose X is path connected and supp ose that X U V is a separation of X Letx U and let

y V By assumption there is a path X suchthat x and y

U and V are disjointopensetsin and U V This contradicts the

fact that is connected

Therefore X is connected

Example The top ologists sine curve S consists of the graph of y sin for x

The top ologists sine curve is connected since its a continuous image of the connected set The

space T S f g is also connected since its the union of S with one of its limit p oints in R However

T is not path connected

To prove this Ill show that the p oint cannot b e connected by a path to the p oint sin

Supp ose on the contrary that S is a path from to sin Notice that is closed

in T so is a closed subset of I will show that isalsoopenin this will

prove that

Let x Let V b e the op en rectangle The intersection of V with S

is an op en set in S containing that is a neighb orho o d of in S BycontinuityI mayndan

op en interval a binR containing x such that

a b V S

The intersection a bisanopeninterval or a halfop en interval so its connected Therefore

a b is connected V S is a disjoint union of op en arcs in S with the p oint Hence

a b must map entirely into one of these comp onents

However I know the p oint x a bmapsto Therefore

a b

Thus a b is a neighborhood of x contained in This proves that is op en

Since is a nonempty op en and closed and since its a subset of the connected set it

must b e all of Thus contradicting the fact that is a path from to sin

It follows that there is no such path and therefore T isnt path connected

Connected comp onents were dened as the maximal connected subsets of a space their existence followed

from Zorns Lemma The path comp onents of a space are dened in a similar fashion I need a lemma

which is analogous to one I prove for connected sets

Lemma Let fU g b e a collection of path connected subsets of a top ological space X Supp ose that

a aA

T S

U Then U U is path connected

a a

aA aA

Pro of Let y z U Imust showthaty and z canbejoinedby a path in U

Supp ose y U and z U Since x U and U is path connected x and y canbejoinedby a path

a b a a

in U Since x U and U is path connected x and z can b e joined by a path in U

a b b b

Concatenating the two paths pro duces a path in U U U joining y and z Therefore U is path

a b

connected

Theorem Let X b e a top ological space

Every p oint x X is contained in a unique maximal path connected subset C of X

C C or C C

x y x y

Tosay that C is maximal means that C is not prop erly contained in any other path connected subset

x x

of X

The pro of will b e omitted it is essentially the same as the result for connected comp onents You

can verify that the connected comp onents pro of required only the lemma on unions of connected sets with

nonemptyintersection I just proved the analogous lemma for path connected sets ab ove

Denition Let X b e a top ological space The path comp onents of X are the maximal path connected

subsets of X

Denition Let X b e a top ological space

X is lo cally connected if for every x X and every neighb orho o d U of x there is a connected

neighborhood V of x such that V U

X is lo cally path connected if for every x X and every neighb orho o d U of x thereisapath

connected neighborhood V of x suchthatV U

Example Since a path connected neighb orho o d is a connected neighb orho o d every lo cally path connected

space is lo cally connected

Example R is lo cally path connected since a ball B x is path connected

Example Let S denote the top ologists sine curve

x x sin S

Let T S f g T is not lo cally connected a neighb orho o d of consists of together with

a disjoint union of op en arcs which is not a connected set

Theorem A space is lo cally connected if and only if the comp onents of any op en subset are op en

In particular the comp onents of a lo cally connected space are op en

Pro of Supp ose X is connected and U X is op en Let C b e a comp onentofU Iwanttoshow C is op en

Let x C By lo cal connectedness there is a connected neighb orho o d V of x such that x V U

Now V C is a union of connected sets having the p oint x in common since C is maximal V C C so

V C

Since every p ointof C has a neighb orho o d contained in C C is op en

Conversely supp ose that comp onents of op en subsets are op en Let x X and let U b e a neighb orho o d

of xImust nd a connected neighborhood V of x such that x V U Take V to b e the connected

comp onentofU whichcontains x By assumption V is op en and x V U

Hence X is lo cally connected

An essentially identical argumentproves the following

Theorem A space is lo cally path connected if and only if the path comp onents of any op en set are op en

In particular the path comp onents of a lo cally path connected space are op en

The relationship b etween comp onents and path comp onents is describ ed by the following result

Prop osition Let X b e a top ological space

Every path comp onent is a subset of a comp onent

If X is lo cally path connected then the path comp onents and comp onents coincide

Pro of A path comp onent is path connected and path connected sets are connected Since every

connected set is contained in a comp onent ie a maximal connected set every path comp onent is contained

in a comp onent

Supp ose X is lo cally path connected Let C be a component I havetoshowthatC is a path comp onent

Let P be any path comp onent contained in C For instance let x C and take P to b e the path

comp onent containing x Let P b e the union of the path comp onents of C other than P

Since X is lo cally path connected P is op en X is lo cally connected so C is op en again since X is

lo cally path connected the path comp onents of the op en set C are op en Hence P is op en since its a

union of op en sets

P is a disjoint union of op en sets Since C is connected P and P cant b e nonempty Now C P

since P Ihave P Thus C has no path comp onents other than P which means that C P

by Bruce Ikenaga

Compact Spaces

In the real line compactness is equivalent to b eing closed and b ounded The HeineBorel Theorem says

that a closed and b ounded subset B of R has the prop ertythatiffU g is a collection of op en intervals

i iI

whose union contains B ie the collection covers B then some nite sub collection of the U s also covers

B This is the appropriate way to generalize closed and b ounded to arbitary spaces

Denition

A top ological space X is covered by a collection fU g of subsets if X U

i iI i

If X is a top ological space Y X and fU g is a collection of subsets of X thenfU g covers Y

i iI i iI

if Y U

If fU g covers X and each U is op en then fU g is an op en cover of X

i iI i i iI

If a sub collection of a covering also covers the space the sub collection is a sub covering

Example The collection

fn n j n Zg n n n Z

is an op en cover of R

Every p ointofR Zis contained in an interval of the form n n for some n ZEachinteger n

is contained in n n Thus the collection covers R since each element of the collection is op en

it is an op en cover

The collection of op en intervals with rational endp oints is also an op en cover of R

More generally the collection of op en balls with rational centers and rational radii forms an op en cover

of R

Denition A space X is compact if for every op en cover fU g of X some nite sub collection fU U g

i iI n

also covers X

Tosay it another way a space is compact if every covering has a nite sub covering

It is clear that compactness is preserved by homeomorphisms If X is compact and X is homeomorphic

to Y then Y is compact

Example Any nite set of p oints with any top ology is compact since anyopencover can contain at most

nitely many distinct op en sets

Example R is not compact

Consider the following op en cover of R

n Z n fn n j n Zg n

Eachinteger n is contained in exactly one elementofthecover namely n n Therefore

any nite sub collection contains at most nitely manyintegers so no nite sub collection can cover R

Since every op en interval a b is homeomorphic to Ropenintervals arent compact

On the other hand Ill show later that a closed interval a b is compact

The following condition is equivalent to compactness it is extremely imp ortant in analysis

Denition Let X b e a top ological space A collection of subsets C of X satises the nite intersection

condition if the intersection of any nite sub collection of C is nonempty

In other words if C C C then C

n i

Theorem Let X b e a top ological space X is compact if and only if every collection of closed sets satisfying

the nite intersection condition has nonemptyintersection

Pro of Supp ose X is compact and let fC g b e a collection of closed subsets of X such that every nite

i iI

sub collection has nonemptyintersection I want to show that C

C Then Supp ose on the contrary that

X X C X C

i i

iI iI

Therefore fX C g is an op en cover of X By compactness some nite sub ollection fX C X

i iI

C g covers X Thus

n n

X X C X C

i i

Hence C contrary to assumption

Therefore C

Conversely supp ose that every collection of closed subsets of X satisfying the nite intersection condition

has nonemptyintersection Let fU g b e an op en cover of X IwanttoshowthatfU g has a nite

i iI i iI

sub cover

Supp ose on the contrary that no nite sub collection of fU g covers X Consider the complements

i iI

fX U g This is a collection of closed sets If fX U X U g is a nite sub collection then

i iI n

n n

X U X U

i i

For X U since no nite sub collection of fU g covers X

i i iI

Thus every nite sub collection of fX U g has nonemptyintersection By assumption fX U g

i iI i iI

has nonemptyintersection Therefore

X U X U

i i

iI iI

This contradicts the fact that fU g covers X

i iI

Hence some nite sub collection of fU g covers X andX is compact

i iI

Thus the intersection condition is really just a translation of the denition of compactness from op en

to closed sets In any situation you can cho ose the version that is easier to apply

The results that follow describ e how compactness b ehaves in connection with other top ological concepts

Lemma Let X b e a top ological space and let Y X Y is compact if and only if every op en cover of Y

has a nite sub cover

Reminder to saythatfU g is an op en cover of Y as a subset of X means that each U is op en in X

i iI i

U That is the U s are subsets of X and not necessarily subsets of Y and Y

i i

Pro of Supp ose Y is compact Let fU g b e an op en cover of Y bysetsopeninX ThenfU Y g

i iI i iI

is an op en cover of Y bysetsopenin Y By compactness some nite sub collection fU YU Y g

covers Y Then fU U g covers Y

versely supp ose that every op en cover of Y has a nite sub cover Let fU g b e an op en cover of Con

i iI

Y bysetsopeninY For each i U Y V where V is op en in X

i i i

Now fV g is an op en cover of Y by sets op en in X soby assumption a nite sub collection fV V g

i iI n

covers Y Then fY V Y V g fU U g covers Y Therefore Y is compact

n n

One way of putting this is Compactness is intrinsic to the space That is to say Y is compact in the

sense of the op en cover prop erty has the same meaning whether Y is considered as a subspace of another

space or as a space in its own right In contrast whether a set is closed or not dep ends on the top ology of

the ambient space

Theorem A closed subset of a compact space is compact

Pro of Let X b e a compact top ological space and let C b e a closed subset of X LetfU g be an open

i iI

cover of C and consider the collection fU g fX C g Thisisanopencover of X soby compactness

i iI

it has a nite sub cover Ui

X - C

If X C o ccurs in this sub cover throw it out The remaining sets fU U g cover C Thus Ive

found a nite sub cover of C so C is compact

Theorem A compact subset of a Hausdor space is closed

Pro of Let X b e a Hausdor space and let C b e a compact subset of X Ill show that the complementof

C is op en

Let x X C Iwant to nd a neighborhood of x that is contained in X C

Since X is Hausdor for each y C there are disjoint neighb orho o ds U of x and V of y Now

y y

fV g is an op en cover of C soby compactness there is a nite sub cover fV V gLet

y y Y y y

U U

Uy 1 Uy 2

Uy Vy 3 2

y 2 y 1

y Vy 3 1 C Vy

If z U thenz U for i n Hence z V for i nButfV V g cover C so

y y y y

i i n

z C ie z X C Thus U is a neighborhood of x contained in X C Hence X C is op en so C is

closed

Remark In the course of the pro of I showed that in a Hausdor space a compact set C and a p ointin

the complementxmay b e separated by disjoint op en sets

Theorem The continuous image of a compact space is compact

Pro of Let X b e a compact top ological space let Y b e a top ological space and let f X Y b e a surjective

continuous function I wanttoshowthat Y is compact

Let fU g b e an op en cover of Y Thenff U g is an op en cover of X soby compactness I may

i iI i iI

nd a nite sub cover ff U f U g

Nowify Y theny f x for some x X Findi fng suchthat x f U Then

y f x U This shows that fU U g covers Y

i n

Hence Y is compact

The pro of of the following theorem makes nice use of the last three results

Theorem Let X and Y b e top ological spaces where X is compact and Y is Hausdor Let f X Y be

a continuous bijection Then f is a homeomorphism

Pro of I need to show that f is continuous I will showthattheinverse image of a closed set is closed

Let C b e closed in X Iwanttoshowthatf C f C is closed in Y Since X is compact and

C is closed C is compact Since f is continuous and C is compact f C is compact Since Y is Hausdor

and f C is compact f C is closed

Therefore f is continuous and f is a homeomorphism

Example Dene f S by

f t cos t sint

0 1

f isacontinuous bijection and S is surely Hausdor However f is not a homeomorphismIntuitively

f unwraps the circle onto the interval it fails to b e continuous at the p oint where the circle needs

to b e cut in order to do the unwrapping This is not continuous b ecause p oints on either side of the cut

p oint which start o close together wind up far apart at opp osite ends of

Let X b e the union of countably many copies of and countably many copies of S Construct a

continuous bijection of X to itself by dening the map to b e the wrapping map f from one of the s

to one of the S s and the identitymapsid and id S S on all the other pieces

0 1

0 1 0 1

The inverse fails to b e continuous for the same reason that f is not continuous

This gives an example of a continuous bijection from a Hausdor space to itself which is not a homeo

morphism

The following result can b e used to showthatanite pro duct of compact spaces is compact its of

interest in its own right

Theorem The Tub e LemmaLet X and Y b e top ological spaces and supp ose Y is compact Let x X

and let W X Y b e an op en set containing fxg Y Then there is a neighborhood U of x suchthat

U Y W

Y w

{x} x Y

U x Y

U X

Pro of For each y Y there is a neighborhood U V of x y such that U is a neighb orho o d of x in X

y y y

V is a neighborhood of y in Y andU V W Itfollows that fU V g is an op en cover of fxgY

y y y y y y Y

Now fxgY is compact since its homeomorphic to Y Therefore there is a nite sub collection

fU V U V g whichcovers fxg Y

u y y y

n n

Let U U U is an op en set containing xIfz y U Y then y V for some i fng

y y

i i

So z y U V W and hence U Y W

Example

W fy j y Rg x y j y x

is an op en subset of R which contains the y axis ie fg R

However there is no op en subset U of R suchthat

fy j y RgU R W

The Tub e Lemma do es not apply b ecause R is not compact

Lemma Let X b e a top ological space Let D b e a collection of subsets of X which is maximal with resp ect

to satisfying the nite intersection condition Then

D is closed under nite intersections

If C X and C meets every elementofD thenC D

Pro of Supp ose D D DImust showthatD D D

n i

Consider the set D DfD g I claim that D satises the nite intersection condition

Take E E DIfE E Dthen E since D satises the nite intersection

m m i

condition

D and E E DThen Otherwise one of the E s is D without loss of generalitysay E

m n

E D E E D E E

i m i m

i i

since this is a nite intersection of elements of D

Now D satises the nite intersection condition and it contains D Bymaximalityof D D D which

means that D D

Supp ose C X and C meets every elementofD Iwanttoshowthat C D

Consider the set D DfC g I claim that D satises the nite intersection condition

Take E E DIfE E Dthen E since D satises the nite intersection

m m i

condition

Otherwise one of the E s is C without loss of generalitysay E C and E E DNow

E E D bypartso

C E E

since C meets every elementofD

Now D satises the nite intersection condition and it contains D Bymaximalityof D D D which

means that C D

All of the pro ofs of Tychono s theorem are sophisticated its a dicult and imp ortant result Lang

credits the following pro of to Nicholas Bourbaki the nomdeplume of a group of mathematicians who

wrote an extremely inuential series of exp ository monographs

Theorem TychonoLet fX g b e a family of compact top ological spaces Then X X is

a aA a

compact

Pro of Let C b e a collection of closed subsets of X satisfying the nite intersection condition I need to

C show that

C C

Step Consider the set of collections of subsets of X closed or otherwise which satisfy the nite

intersection condition and contain C I claim that has maximal elements

Order by inclusion

Let be a chain in Let

F G

Since each G contains C F also contains C I claim that F satises the nite intersection condition

Let F F FSupposeF G F G where G G Since is a chain there is

n n n n

an index k fng such that F G for i n Since G satises the nite intersection condition

i k k

F F This proves that F satises the nite intersection condition

Since F satises the nite intersection condition and contains C itsanelement of and its clearly an

upp er b ound for By Zorns lemma has maximal elements

Step Let D b e a maximal element of Ill construct an element x D

D D

Let X X b e the a pro jection map Fix a A and consider the family of sets f D g

a a a D D

or any nite sub collection f D D gtheintersection D D is nonempty b ecause D F

a a n n

satises the nite intersection condition Since

D D D D

a n a a n

it follows that D D

a a n

Therefore f D g satises the nite intersection condition

a D D

Hence f D g satises the nite intersection condition

a D D

Now f D g is a family of closed subsets of the compact space X Therefore there is an element

a D D a

x D

a a

D D

Dene x x

a aA

Step Let U be a neighb orho o d of x in X I claim that U D

a a a a

Since U is a neighb orho o d of x and x D for all D D U meets D for all D D But if

a a a a a a

U D then U D To see this note that if z U D then z y for some

a a a a a a

y D and so y U Thus y U and hence y U D

a a a a

a a

Thus U meets every set D D By the second part of the lemma U D

a a

Step Every basic op en set containing x is contained in D

The sets U fora A and U a neighborhood of x in X form a subbasis for the op en sets

a a a a

containing x A basic op en set containing x is a nite intersection of U sets such a nite intersection

is in D by the rst part of the lemma

C Step Finally Ill showthat x

C C

Let D D and let U b e a basic op en set containing x Since D satises the nite intersection condition

U D Since every basic op en set containing x meets D itfollows that x D

In particular if D is a closed set in D then x D But all the sets in C are closed and CDsox C

for all C C Therefore x C

C C

This completes the pro of that X is compact

Serge Lang Real analysis edition Reading Massachusetts AddisonWesley Publishing Company

ISBN

by Bruce Ikenaga

Compact Sets and the Real Numb ers

Theorem If X is an ordered set satisfying the least upp er b ound prop ertythenany closed interval a bin

X is compact

Pro of The outline of the pro of is as follows Let fU g b e an op en cover of a b Construct the set C

i iI

of p oints y a b suchthata y can b e covered by a nite sub collection of fU g Take the least upp er

i iI

b ound c of C and showthatc C Finallyshow that c b

Step Supp ose a xb I claim that for some yxtheinterval x y canbecovered byatmosttwo

elements of fU g

i iI

If x has an imm ediate successor x thenx x has only two elements so it can b e covered byat

most two U s

If x do es not have an immediate successor nd U containing x Pick y x suchthatx y U this

i i

is p ossible since U is op en Since x do es not have an immediate successor there is an element y suchthat

x y y Thenx y U andx y iscovered by a single elementof fU g

i i iI

Step Nowlet

C fy a b j a y canbecovered by nitely many U g

By Step there is an element yasuchthata y can b e covered byatmosttwo elements of fU g

i iI

Therefore C is nonempty Let c b e the least upp er b ound of C in a b

Step I claim that c C

Find U containing c U is op en and c asoImay nd an interval d c U Since d cant b e an

i i i

upp er b ound for C there is an elementofC larger than dLetc C where d c cThena c can

c c can b e covered by nitely be covered by nitely many U s and c c U Therefore a c a c

many U s Hence c C

Step I claim that c b

Supp ose that c b By Step there is a ycsuchthatc y can b e covered byatmosttwo elements

of fU g Since c C a ccanbecovered by nitely many elements of fU g Soa y a c c y

i iI i iI

can b e covered by nitely manyelements of fU g and therefore y C This contradicts the fact that c

i iI

was the least upp er b ound of C Hence c b

Since b C a b can b e covered by nitely many elements of fU g Therefore a b is compact

i iI

Corollary A closed interval a bin R is compact

Theorem HeineBorel A subset C R is compact if and only if it is closed and b ounded relativeto

the standard metric or the square metric

Remark Recall that the standard metric d and the square metric are related by

x y dx y n x y

Hence a set b ounded relative to one metric is b ounded relative to the other

Pro of Supp ose that C is compact R is Hausdor so C is closed The collection of balls fB n j n Z g

is an op en cover of R soitisanopencover of C By compactness there is a nite sub cover the element

of the sub cover with the largest radius contains C so C is b ounded

Conversely supp ose C is a closed and b ounded subset of R Supp ose that x y s for all x y C

Fix x C and let t x Then for all y C

dy dy xdx s t

n n

It follows that C s ts t However s ts t is a pro duct of compact spaces so its

compact Thus C is a closed subset of the compact set s ts t so it is compact

Theorem Let X b e a compact top ological space let Y b e an ordered set with the order top ology and let

f X Y be continuous There are p oints a b X suchthat

f a f x f b for all x X

Pro of f X is a compact subset of Y Ill show that f X has a largest element and a smallest element

Supp ose that f X doesnothave a largest element Then the collection of op en rays

ffx j x X g

cover f X By compactness I maynd x x X suchthat

ffx fx g

cover f X

Find m fng such that

f x maxff x j i ng

m i

Then f X fx but f x fx This contradiction shows that f X hasa

m m m

largest element a similar argumentshows that f X has a smallest element

In the case where X a b is a closed interval in R and Y R this is the familiar result from calculus

whichsays that a continuous function on a closed interval has a max and a min on the interval

The next result and its corollary are more amusing than imp ortant since youve probably seen the

Cantor diagonalization pro of of the uncountabilityofl But this top ological pro of do es not makeany

reference to representations of real numb ers as innite decimal

Theorem Let X b e a nonempty compact Hausdor space Supp ose every p ointof X is a limit p ointof

X Then X is uncountable

Pro of

Step Let U b e a nonempty op en subset of X andletx X Ill construct an op en set V U suchthat

x V

If U fxg then U is a neighborhood of x whichcontains only xThiscontradicts the assumption that

every p ointofX is a limit p ointofX Now x may not b e in U in the rst place but this shows that in any

event U cant consist of x alone

Thus I maycho ose y U suchthaty x Since X is Hausdor there are disjointneighb orho o ds A of

x and B of y Let V B U Now

V B U B X A

V X A In particular since x A V do es not contain x and X A is closed Therefore

V = U U B y

Step Next Ill show that there is no surjection f Z X

Supp ose that f Z X For the purp oses of numb ering take V X By Step I maynda

V nonemptyopenset V such that f

Supp ose inductively that n and I have found a nonemptyopenset V V suchthat

n n

f n V Use Step to nd a nonemptyopenset V V suchthatf n V

n n n n

Consider the collection fV g The collection has the nite intersection condition since

V V V

By compactness there is a p oint x V Thenx f n for any n Z so f is not surjective

Since there is no surjective function from Z to X X is uncountable

Corollary The interval in R is uncountable

Obviously it follows that R and op en intervals in R are uncountable

by Bruce Ikenaga

Limit Point Compactness and Sequential Compactness

Denition Let X b e a top ological space

X is limit p oint compact if every innite subset has a limit p oint

X is sequentially compact if every sequence in X has a convergent subsequence

If X is sequentially compact then X is limit p oint compact For every innite subset of X contains

an innite sequence the limit of a convergent subsequence is a limit p oint of the original subset A partial

converse is contained in the following theorem

Theorem Let X b e a top ological space

If X is compact then X is limit p oint compact

If X is a metric space and X is limit p oint compact then X is sequentially compact

Pro of Let X b e compact and let S X b e an innite subset I must showthat S has a limit p oint

Supp ose that S do es not have a limit p oint It is vacuously true that S contains all its limit p oints so

S is closed Since X is compact S is compact

Let x S Since x is not a limit p ointofS there is a neighb orho o d U of x which do es not intersect

S in a p oint other than x fU g is an op en cover of S soby compactness there is a nite sub cover

x xS

fU U g

x x

Now S fU U but each U meets S only in x Therefore S fx x g which

x x x i n

n i

contradicts the assumption that S is innite

Therefore S has a limit p oint

Supp ose X is a metric space and X is limit p oint compact Let fx g b e a sequence in X Iwanttoshow

that fx g has a convergent subsequence

First I have to eliminate any rep etitions in the sequence so dene

S fx j n Z g

Thus S is the set of points in fx g

Supp ose S is nite Then for some x X x x for innitely many n Z Supp ose this is true for

n n Then fx g is a convergent subsequence of fx g b ecause all the terms are equal to x

n n

The only other p ossibility is that S is innite Then by limit p oint compactness S has a limit p oint x

k Z Since metric spaces are Hausdor Consider the nested sequence of neighborhoods B x

and x isalimitpointofS each B x contains innitely manypoints of S

Cho ose n Z suchthatx B x Assuming that k and that n has b een chosen cho ose

n k

n n such that x B x

k k n

for k N Then x B x fork N Inowhave x xFor if cho ose N so that

n n

k k

Example The subsets fn n j n Z g form a partition of Z The corresp onding partition top ology

is called the o ddeven top ology Let X denote Z with this top ology

X is limit p oint compact To see this let A b e an arbitrary nonemptysubsetof X and let a AFor

some n Z a n n Notice that n n is the smallest op en set containing n orn

If a n then n is a limit p ointof Asinceevery neighborhood of n contains a

If a nthenn is a limit p ointof Asinceevery neighborhood of n contains a

In either case A has a limit p oint Since A was an arbitrary nonempty subset of X surely every innite

subset of X has a limit p oint Thus X is limit p oint compact

On the other hand X is not sequentially compact For example the sequence f g has no

convergent subsequence

This example shows that limit p oint compactness do es not in general imply sequential compactness

In fact X is not compact The op en cover fn n j n Z g do es not have a nite sub cover Thus

limit p oint compactness do es not in general imply compactness

Recall that if Y is a subset of a metric space X d the diameter of Y is

diamY supfdx y j x y Y g

Theorem Leb esgue Number LemmaLetX b e a sequentially compact metric space and let fU g

i iI

b e an op en cover of X There is an suchthatevery subset of X of diameter less than is contained

in an elementof fU g

i iI

is called a Leb esgue number for the cover

Pro of Supp ose on the contrary that no such exists Then for every n Z there is a set A of

which is not contained in any U Letx A Illshowthatfx g has no convergent diameter less than

i n n n

subsequence which will contradict sequential compactness

Supp ose on the contrary that fx g isaconvergent subsequence converging to x Supp ose that x U

n i

Find such that B x U Then cho ose n suciently large so that

i k

d x x and

n k

1/n k

x B(x;ε ) nk B(x ,1/n ) x nk k x

ε xn

An B(x;ε )

Since diamA and x Supp ose y B x A itfollows that A B x

n n n n n n

k k k k k k

n n n

k k k

Then

dx y dx x dx y

n n

k k

Therefore B x B x Hence

A B x B x U

n n i

k k

is not contained in any U Therefore there is anumber satisfying This contradicts the fact that A

i n

the conclusion of the theorem

Theorem Let X b e a metric space The following are equivalent

X is compact

X is limit p oint compact

X is sequentially compact

Pro of Ive already proven and Ineedtoprove Supp ose then that X is

sequentially compact I need to show that X is compact

Step For every there is a nite covering of X by balls

Let x X and construct B x If X B x then Im done Otherwise cho ose x X B x

x Im done otherwise cho ose x X B x and construct B x Again if X B x B

B x Keep going Notice that each B x contains only one x namely x

n k n

Supp ose that the pro cess do es not terminate Consider the sequence fx g I claim that it has no

g is a subsequence converging to xthen B x convergent subsequence Indeed if fx can contain at

most one x If it contains two x s then they are less than apart Therefore there is an ball ab out

n n

k k

an x containing another x

n n

k k

It follows that the pro cess must terminate so X is covered by a nite number of balls

Step Every op en cover of X contains a nite sub cover

Let fU g b e an op en cover of X Let b e a Leb esgue numb er for the cover Cover X with a nite

i iI

balls fB B g Each B has diameter less than soeach B is contained in some U number of

n j j i

Then fU j j ng is a nite sub collection of fU g whichcovers X Therefore X is compact

i i iI

Denition Let X and Y b e metric spaces and let f X Y f is uniformly continuous if for every

there is a such that d a b implies d f afb for all a b X

X Y

Theorem Let X and Y b e metric spaces let f X Y be a continuous function and supp ose X is

compact Then f is uniformly continuous

n o n o

Pro of Given the op en balls B y y Y cover Y Therefore f B y y Y is

an op en cover of X

By the Leb esgue Numb er Lemma there is a such that every set of diameter less than is

contained in a f B y In particular if d a b then fa bg is a set of diameter less than so

a b f B y for some y This means that f afb B y so d f afb

Example Consider the function f R R given by f xx f is continuous but f is not uniformly

continuous

To show that f is not uniformly continuous Ill show that there is no suchthatifjx y j

and consider the p oints then jf x f y j Supp ose on the contrary that sucha exists Cho ose x

and x x

First

x x

The p oints are less than units apart

However

x x x

f x f x x x x x x x x x

The images are more than unit apart

This shows that f is not uniformly continuous

by Bruce Ikenaga

Lo cal Compactness

Denition Let X b e a top ological space X is lo cally compact if for all x X there is a compact set C

and a neighborhood U of x suchthatx U C

If X is compact then X is lo cally compact For any x X X is a compact set containing the

neighb orho o d X of x

Example If X is a space with the discrete top ologythen X is lo cally compact If x X then fxg is a

compact set containing the neighborhood fxg of x

n n

Example R is lo cally compact If x R the closed ball B x is compact and it contains the op en ball

B x

Theorem Let X b e a Hausdor space X is lo cally compact if and only if for every x X and every

neighb orho o d U of x there is a neighb orho o d V of x suchthatV is compact and V U

Pro of Supp ose for every x X and every neighborhood U of x there is a neighborhood V of x suchthat

V is compact and V U Cho ose any neighb orho o d U of x for example X and nd a neighb orho o d

V of x with compact closure suchthatV U ThenV is a compact set containing a neighborhood V of x

so X is lo cally compact

Conversely supp ose X is lo cally compact Let x X and let U b e a neighb orho o d of x Let C be a

compact set whichcontains a neighb orho o d V of x

C is compact and X is Hausdor therefore C is closed It follows that C U C X U is closed

since its a subset of the compact set C C U is compact Note that x U sox C U

I showed earlier that in a Hausdor space a compact set and a p oint not contained in it maybe

separated by disjoint op en sets Thus I may nd disjoint op en sets W and W such that C U W and

x W U

C x

Next Ill show that W V is compact C is a closed set containing V soV C ButW V V

so W V V C Now W V is a closed subset of the compact set C so W V is compact

Moreover W V W X W X W is closed so W V X W Since W V do es not

W V do es not intersect C U ButW V is a subset meet W and since C U W it follows that

of C so W V U

Thus W V is a neighborhood of x with compact closure whose closure is contained in U

Corollary Let X b e a lo cally compact Hausdor space and let Y b e a subspace of X IfY is op en or closed

in X then Y is lo cally compact

Pro of Supp ose Y is op en in X Let x Y By the preceding result I may nd a neighb orho o d U of x such

that U is compact and U Y Note that U is automatically op en in Y Thus x is contained in a compact

set whichcontains a Y neighb orho o d of x Therefore Y is lo cally compact

Supp ose Y is closed Let y Y and let C b e a compact set containing a neighb orho o d U of y C is

closed since X is Hausdor therefore C Y is a closed subset of C Since C is compact C Y is compact

Moreover C Y contains U Y which is a neighborhood of y Therefore Y is lo cally compact

Example Let S denote the top ologists sine curve

S x sin x

Let X S f g X is not lo cally compact

Take a ball B ab out the origin Consider the intersection of X with the line y The

intersection is an innite sequence of p oints with all but nitely many terms lying inside B The

sequence converges to which is not in X

If there is a compact set C in X containing a neighb orho o d of then that neighborhood contains

B X for suciently small By the argumentabove suchaneighb orho o d contains an innite

subset of X with no limit p oint The same is true for C whichcontradicts the fact that compact sets are

limit p oint compact

It is often useful to emb ed a space in a compact Hausdor space b ecause compact Hausdor spaces

are very nicely b ehaved For example Ill show later that compact Hausdor spaces are normalAnytwo

disjoint closed sets can b e separated by disjoint op en sets If a space is lo cally compact Hausdor it can b e

emb edded in a compact Hausdor space called the onep oint compactication

Lemma Let X b e a lo cally compact Hausdor space Dene Y X fg where is a p oint not in X

Let T b e the collection of subsets of Y consisting of

Any op en subset of X

The complementin Y of a compact subset of X

Then T is a top ology on Y

Y T is called the onep oint compactication of X

Pro of is op en in X so its op en in Y is compact so Y Y is op en in Y

Toverify the axioms for unions and intersections its necessary to take cases Ill show the work for

unions the pro of for intersections is similar

The union of op en sets in X is op en in X so its op en in Y

g is a family of compact subsets of X then If fC

Y C Y C

i i

iI iI

Now X is Hausdor so compact subsets are closed Therefore C is closed Since its contained

in any one of the compact sets C its also compact Therefore Y C is the complementin Y of a

i i

compact set in X so its op en in Y

The remaining p ossibility is a union of sets op en in X with complements of compact subsets of X Using

the rst two cases this reduces to showing that if U is op en in X and C is a compact subset of X then

U Y C isopeninY But

U Y C Y C U

C is closed in X U is op en in X soC U is closed in X ButC U C and C is compact therefore

C U is compact Hence U Y C is a complementin Y of a compact set in X so its op en in Y

Here are some prop erties of the onep oint compactication

Lemma Let X b e a lo cally compact Hausdor space and let Y b e its onep oint compactication

The subspace top ology on X is the same as the original top ology on X That is the inclusion of X

into Y is a homeomorphism onto its image

If X is not compact then Y X

Y is compact

Y is Hausdor

Pro of If U is op en in X then its op en in Y by denition Conversely an op en set in Y is either

an op en set U in X in which case U X U is op en in X orY C whereC is compact in X But

Y C X X C whichisopeninX since C is closed in X

Since the only p ointofY not in X is this amounts to showing that is a limit p ointofX

Neighb orho o ds of are sets Y C where C is compact and since X is not compact C X Thus Y C

must in tersect X and so X

Let fU g b e an op en cover of Y One of the U s say U must contain so U mustbeasetof

i iI

the form Y C where C is compact in X The sets U X for U U form an op en cover of C let

i i

fU XU X g b e a nite sub cover Then fU U U g covers Y

i i i i

n n

Let x and y b e distinct p oints of Y If theyre b oth in X theymay b e separated by disjointneighb orho o ds

is Hausdor and these neighb orho o ds are also op en in Y in X since X

Otherwise Im trying to separate x X from SinceX is lo cally compact I may nd a compact set

C containing a neighb orho o d U of xThenY C and U are disjointopensetsin Y separating x and

Corollary A space X emb eds as an op en subset of a compact Hausdor space if and only if X is lo cally

compact Hausdor

by Bruce Ikenaga

The CountabilityAxioms

Denition Let X b e a top ological space and let x X

X has a countable basis at x if there is a countable collection of neighb orho o ds fB g of x such

i iI

that if U is a neighborhood of x then B U for some i

X is rst countable if X has a countable basis at each p oint

X is second countable if X has a countable basis for its top ology

Obviously a second countable space is rst countable

Example Every metric space is rst countable If X is a metric space and x X then B x n Z

is a countable basis at x

Example R is second countable since the balls with rational centers and radii form a countable basis for

the usual top ology

On the other hand R with the discrete top ology is not second countable

Example Consider R with the nite complement top ologyThus the op en sets are and any set whose

complement is nite

I claim that R do es not have a countable basis at anypoint Supp ose for example that there is a

countable basis fB B g at I claim that B fg Clearly is in the intersection I must show

that no nonzero p oint is in the intersection

Supp ose that x The complementofR fxg is nite so its an op en set containing Hence

B R fxg for some i Then x B sox B

i i i

Thus

R B R fg R B

i i

Eachset B is nite so R B iscountable But R is uncountable so R fg is uncountable

i i

This contradiction shows that there is no countable basis at obviously the argumentworks for any x R

Thus R is not rst countable or second countable in the nite complement top ology

The following results were proved for metric spaces An examination of the pro ofs shows that they only

dep ended on the fact that metric spaces are rst countable

Theorem Let X and Y b e top ological spaces and supp ose X is rst countable

Let A X x A if and only if there is a sequence of p oints of A converging to x

Let f X Y f is continuous if and only if whenever fx g is a convergent sequence in X ff x g is

n n

a convergent sequence in Y

Prop osition

A subspace of a rst countable space is rst countable

A countable pro duct of rst countable spaces is rst countable

A subspace of a second countable space is second countable

A countable pro duct of second countable spaces is second countable

Pro of Supp ose X is rst countable and supp ose Y is a subspace of X Let y Y Let fB B g be

a countable basis at y in X Then fB Y B Yg is a countable basis at y in Y

X Let B be a countable basis for x Supp ose X X are rst countable spaces Let x

n n n n

U where U X for all but nitely in X Consider the collection U of pro duct neighborhoods

n n n n

Q Q

many n X is X so and if U X then U B Then U is a countable basis at x in

n n n n n n n

n n

rst countable

Supp ose X is second countable and supp ose Y is a subspace of X Let fB B g be a countable basis

for the top ology on X ThenfB Y B Y g is a countable basis for the subspace top ology on Y

Supp ose X X are second countable spaces Let B be a countable basis for the top ology on

X Consider the collection U of pro duct neighborhoods y U where U X for all but nitely man

n n n n

X so n and if U X thenU B Then U is a countable basis for the pro duct top ology on

n n n n n

X is second countable

Denition Let X b e a top ological space

X is Lindelof if every op en cover of X has a countable sub cover

X is separable if X has a countable dense subset

Example Any compact space is Lindelof since every op en cover has a nite sub cover

Example R is separable since Q is a countable dense subset of R

Prop osition Let X b e a second countable top ological space

X is Lindelof

X is separable

Pro of Let U b e an op en cover of X Let fB B g be a countable basis for X For each n Z let

U fU U jB U g

n n

Next for each nonempty U cho ose V U

n n n

For each x X there is a U U suchthatx U Moreover there is a basis element B suchthat

B V x B U ThenU is nonemptyso V is dened and x

n n n n n

Now x was an arbitrary p ointofX so the collection fV g covers X fV g is a countable sub cover of U

n n

Let fB B g b e a countable basis for the top ology of X Cho ose x B for each n Z Then

n n

fx g is a countable dense subset of X

by Bruce Ikenaga

The Separation Axioms

Denition Let X b e a space in which singletons are closed

X is regular if a closed set and a p oint outside it can b e separated by disjointopensets

X is normal if disjoint closed sets can b e separated by disjointopensets

regular

normal

Obviously a normal space is regular and a regular space is Hausdor

Terminology Recall that a Hausdor space is also known as a T space

X is a T space if given distinct p oints x y there is a neighborhood U such that either x U and y U

or y U and x U

X is a T space if given distinct p oints x y there are neighborhoods U of x and V of y suchthaty U

and x V

X is a T space if distinct p oints have neighb orho o ds whose closures are disjoint

If you dont require that p oints b e closed then a space in which disjoint p oints and closed sets can b e

separated is T and a space in which disjoint closed sets can b e separated is T

Lemma Let X b e a space in which singletons are closed

X is regular if and only if for all x X and every neighborhood U of x there is a neighborhood V of

V U x such that

X is normal if and only if for every closed subset A of X and every op en set U containing A there is

an op en set V containing A suchthatV U

Pro of Supp ose X is regular x X andU isaneighborhood of xThenX U is a closed set disjoint

from xsoby regularityI may nd disjointopensets V containing x and W containing X U

In particular x V X U since X U is closed V X U

Conversely supp ose that for all x X and every neighborhood U of x there is a neighborhood V of x

such that V U Letx X and let C b e a closed set whichdoesnotcontain x

X C is an op en set containing x so there is a neighb orho o d V of x suchthatV X C Then X V

is an op en set containing C and it is disjoint from the op en set V containing x Therefore X is regular

The pro of for normality is the same as the pro of for regularity with the p oint x replaced with the closed

set A

Corollary If X is a regular space then X is a T space

Pro of Supp ose X is regular Let x and y b e distinct p oints in X Points in X are closed so X fy g is a

neighb orho o d of x By the preceding result there is a neighborhood U of x such that U X fy gAgain

by the preceding result there is a neighb orho o d V of x suchthatV U

U is a neighborhood of y andX U X U SinceX U is closed X U X U Since X

V U and X U X U V and X U are disjoint

Thus V is a neighborhood of x X U is a neighb orho o d of y and their closures are disjoint Therefore

X is T

Corollary If X is a lo cally compact Hausdor space then X is regular

Example Irrational slop e top ology Example Let

X Q Q

the p oints with rational co ordinates in the closed upp er halfplane Fix an irrational number

If x R and dene a subset of the xaxis B x by

B x fy Q jjx y j g

Then for x y X and set

y y

B x N x y fx y gB x

N x y consists of the p ointx y together with twoopen intervals of rationals on the xaxis The

intervals are centered at p oints on the xaxis which determine lines of slop es and with the p ointx y (x,y)

B(x - y/θ ,ε ) B(x + y/θ ,ε )

Note that if y N x y reduces to an op en interval of rationals ab out x in the xaxis Moreover

if x Q then N x isabasicopenset

Dene a top ology by taking as a basis all the sets N x y for all real and all x y X Its

clear that the sets cover X

If a b N x y N x y then either a bx y ora b lies in the intersection of the

intervals in the xaxis

In the rst case if min then N x y is a basis element containing a b and contained

in N x y N x y

y y

In the second case a ba and a B x B x and this is the intersection

of op en intervalsinthe xaxis Cho osing a suciently small I may nd an interval N a contained

in this intersection

If a b N x y N x y where x y x y then again a bmust lie in the

intersection of intervals in the xaxis As b efore I may nd a basic interval containing x contained in the

intersection of the original intervals

Therefore the collections of sets N x y for all real and all x y X forms a basis

Next Ill show that the closures of anytwo basic neigh borhoods must intersect

What is N x y It consists of two diagonal strips of slop es and emanating from the intervals

y y

B x and B x

(x,y)

B(x - y/θ ,ε ) B(x + y/θ ,ε )

To see this observe that a basic neighborhood of a point in these strips will contain intervals in the

y y

xaxis whichintersect B x or B x

It is clear that anytwosuch pairs of strips must intersect Therefore the closures of anytwo basic

neighb orho o ds must intersect By the result ab ove X is not regular

I proved a sp ecial case of the rst part of the following prop osition Im rep eating the pro of of the second

part for the sake of completeness

Prop osition

A pro duct of Hausdor spaces is Hausdor

A subspace of a Hausdor space is Hausdor

Pro of Let fX g b e a family of Hausdor spaces Let x y b e distinct p oints in X For

a aA a a a

some index b x y Cho ose disjointopensets U V X suchthatx U and y V

b b b b b

Dene

n n

X if a b X if a b

a a

U and V

a a

U if a b V if a b

Q Q Q Q

V are U and U V is a neighborhood of y and is a neighborhood of x

a a a a a a

aA aA aA aA

disjoint Therefore fX g is Hausdor

a aA

Let X b e a Hausdor space and let Y X Letx y Y Find disjoint neighborhoods U of x and V of

y in X Then U Y is a Y neighb orho o d of x V Y is a Y neighb orho o d of y and U Y and V y are

disjoint Therefore Y is Hausdor

Prop osition

A pro duct of regular spaces is regular

A subspace of a regular space is regular

Pro of Let fX g b e a family of regular top ological spaces For each a X is Hausdor Therefore

a aA a

Q Q

X is Hausdor so p oints in X are closed

a a

aA aA

Q Q

Let x beapointin X Takeaneighb orho o d of x in X replacing the neighb orho o d

a a a a

aA aA

with a smaller one if necessaryImay assume that it has the form U where U is op en in X and

a a a

U X for all but nitely many as

a a

suchthatV U Ifa is an Since X is regular for each aI mayndaneighborhood V of x in X

a a a a a a

Q Q

index for which U X I will take V X as well Then V is a neighb orho o d of x in X

a a a a a a a

aA aA

and

Y Y Y

V V U

a a a

aA aA aA

It follows by an earlier result that X is regular

Let X b e a regular top ological space and let Y X Apointof Y is closed in X so it is closed in Y

Let x Y and let C b e a closed subset of Y which do es not contain xWrite C Y D where D is

closed in X Find disjointopensets U and V in X suchthatx U and D V ThenY U and Y V are

disjointopensetsin Y x Y U andC Y V Therefore Y is regular

Remark The preceding results are false for normal spaces A subspace of a normal space need not b e

normal and a pro duct of normal spaces need not b e normal

The following results say that the class of normal spaces is large enough to contain manyinteresting

spaces sp ecically metric spaces compact Hausdor spaces and regular spaces having countable bases

Theorem Every metric space is normal

Pro of Let X d b e a metric space Since metric spaces are Hausdor p oints in X are closed

Let C and D b e disjoint closed sets in X SinceC X D and X D is op en for every x C Imay

nd an op en ball B x contained in X D

By a similar argument for every y D there is a ball B y contained in X C

Let

y x

and V B y U B x

y D xC

U and V areopensets C U and D V I claim that U and V are disjoint

x y

Supp ose on the contrary that z U V say z B x for x C and z B y for y D

Then

y x

dx y dx z dy z

If then

x y

x y y y

dx y

This means that x B y so x C contradicting the fact that x C

Likewise implies that dx y which in turns implies that y B x This means that

y x x x

y D contradicting the fact that y D

This proves my claim that U and V are disjoint Since Ive separated the closed sets C and D with

disjointopensets U and V it follows that X is normal

Theorem A compact Hausdor space is normal

Pro of In a Hausdor space p oints are closed

I need to show that disjoint closed sets can b e separated by disjoint op en sets First recall that in a

Hausdor space a p oint and a compact set that do esnt contain it can b e separated by disjointopensets

Let C and D b e disjoint closed subsets of the compact Hausdor space X Since D is closed and X is

compact D is compact By the observation ab ove for each x C Imay nd disjointopensets U and V

x x

such that x U and D V

x x

fU g is an op en cover of C C is compact since its a closed subset of a compact space Let

x xC

fU U g b e a nite sub cover of C Dene

x x

n n

U U and V V

x x

i i

U and V are op en sets C U andD V Ify U V then y U for some i and y V V so

x x

i i

y U V This contradiction shows that U and V are disjoint

x x

i i

Therefore X is normal

Theorem A regular space with a countable basis is normal

Pro of Let X b e a regular space with a coutable basis fB g Since X is regular p oints are closed

Let C and D b e disjoint closed subsets of X Letx C Since X D is an op en set containing x

Imay nd a neighborhood U of x suchthatx U X D Next regularity implies that there is a

V U Finally there is a basis element B containing x such that B V neighb orho o d V of x such that

i i

Now B V U X D so B misses D

i i

Rep eat this pro cedure for each x C I wind up up with a countable sub collection fC g of fB g which

j i

covers C and which satises C D for all j

D C Likewise I may nd a countable sub collection fD g of fB g whichcovers D and which satises

k k i

for all k

What Id liketodoistotake the unions of the two sub collections as my neighb orho o ds of C and D but

these unions may not b e disjoint However since the sub collections are countable I can inductively adjust

the sub collections to pro duce new sub collections whose unions will b e disjoint

Thus for each j and k dene

C C D and D D C

j k k j

j k

Then let

A C and B D

j k

I claim that A and B are disjoint op en sets C A and D B

Each C and each D is op en since eachisanopensetminus a closed set Therefore A and B are

unions of op en sets so they are op en

D s b ecause these sets missed The union of the C s contains C This is not changed by subtracting

k j

C contains C Similarly B contains D C anyway Therefore A

FinallyIllshow that A and B are disjoint Supp ose y A B Theny C D for some j and k

Without loss of generality supp ose that j k Now y C implies y C but y D implies y C for

j j

j k This contradiction proves that A B

Therefore A and B are disjoint neighborhoods of C and D and X is normal

Lynn A Steen and J Arthur Seebach Counterexamples in Topology New York Holt Rinehart and

Winston Inc ISBN

by Bruce Ikenaga

Urysohns Lemma

Urysohns lemma is often expressed bysaying that disjoint closed sets in a normal space can b e

separated by a continuous function that is there is a continuous realvalued function which is on one

of the closed sets and on the other

Note that if A and B are disjoint closed sets in a top ological space X thenX B is an op en set

containing A ConverselyifA is closed and U is an op en set containing A then A and X U are disjoint

closed sets

Thus Urysohns lemma can b e expressed in another form In a normal space given a closed set and an

op en set containing it there is a continuous realvalued function which is on the closed set and outside

the op en set It is this version that Ill prove the discussion ab ove shows that the other version is an easy

corollary

The idea is that for such a function to exist there should b e level sets a set where the function is

a set where the function is equal to and so on from to Going the other waymayb e I can equal to

dene such a function by constructing the level sets

A 1/4 1/2

3/4

There are various ways of constructing the level sets they dier in the way they index the level sets I

will take the approach of and which index the level sets using the dyadic rationals in the

rationals which can b e written in the form The rst step is to show that given an appropriate collection

of such level sets one may dene a continuous function by using the level set indices in the obvious way

Lemma Let X and Y b e top ological spaces let f X Y andletS b e a subbasis for the top ology on Y

f is continuous if and only if f U isopeninX for every U S

Pro of Since subbasic sets are op en if f is continuous then f U isopeninX for every U S

Conversely supp ose that f U isopeninX for every U S An arbitrary op en set in Y is a union

of nite intersections of elements of S Sincef preserves arbitrary unions and arbitrary intersections the

is op en in X Therefore f is continuous inverse image of an arbitrary op en set in Y

Lemma Let X b e a top ological space and let

n k S

Supp ose that for each s S there is an op en set U in X such that if rs S and rsthen U U

s r s

The function f X R dened by

inf fr S j x U g if x U

f x

if x U

is continuous

Note that some elements of S are represented by more than one fraction of the form but rep etitions

are eliminated by implication since S is a set

Pro of The op en rays a b for a b R form a subbasis for the standard top ology on RIt

therefore suces to show that f a and f b are op en for all a b R

Consider the set f a The range of f lies in Thus if a then f a X

and if a then f a In either case f a is op en

Supp ose then that a I will show that f a U

Let x f a so f x a Since f x is the greatest lower b ound of the indices r such

U that x U and since f x a there is an index ra suchthatx U Hence x

r r r

U Supp ose x U where ra Then f x rasox Conversely supp ose x

r r

f a

a U so f a is op en Thus f

Now consider the set f b The range of f lies in Thus if b then f b

and if b then f b X In either case f b is op en

X Supp ose then that b I will showthat f b U

Let x f b so f x b Since in addition b I maynd rs S suchthatbrs

f x by construction U U Sincef x is the greatest lower b ound of the indices t such that x U

r s t

and since f x sitfollows that x U Hence x U so x X U and x X U

s r r r

Conversely supp ose x X U Supp ose x X U where rbThenf x U so the

r r r

fact that the U s are nested implies that f x U for all s S with s r Therefore f x rbso

x f b

Thus f b X U so f b is op en

Therefore f is continuous

Theorem Urysohns Lemma Let X b e a normal space let C b e a closed subset of X and let U be an

op en set containing C There is a continuous function f X such that f C andf X U

Pro of Ill dene a sequence of sets U indexed by

n k S

and satisfying U U for rs

r s

Let U U By normalityI may nd an op en set U such that C U U U

By normalityI may nd an op en set U such that

U U U U

By normalityI may nd op en sets U and U such that

U U U U

and

U U U U

Keep going By construction the U s satisfy the hyp otheses of the lemma The function f constructed

by the lemma satises the conclusion of the theorem

Corollary Let X b e a normal space and let A and B b e disjoint closed subset of X There is a continuous

function f X suchthatf A and f B

Glen Bredon Topology and GeometryNew York SpringerVerlag New York Inc ISBN

Serge Lang Real Analysis edition Reading Massachusetts AddisonWesley Publishing Company

Inc ISBN

by Bruce Ikenaga

The Tietze Extension Theorem

The Tietze Extension Theorem says that a continuous realvalued function on a closed subset of

a normal space may b e extended to the entire space Ill show that this extension condition is essentially

equivalent to normality

The construction of the extension involves building a sequence of functions whose sum agrees with the

given function on the subspace

Lemma Let X b e a normal space let A b e a closed subset of X and let f A R be a continuous function

satisfying jf xjc for all x A and some c R Then there is a continuous function g X R suchthat

c for all x X jg xj

jf x g xj c for all x A

Pro of The sets

c and f c f

are disjoint closed sets in A since A is closed these sets are also closed in X By Urysohns lemma I may

nd a continuous function g X c c such that

c c and g f c c g f

By construction jg xj c for all x X

Now consider cases If c f x cthen g x cso jf x g xj c

c f x c then g x cso jf x g xj c If

Finally supp ose cfx c Since c g x c I again have jf x g xj c

Theorem Let X b e Hausdor The following are equivalent

X is normal

If A is a closed subset of X anycontinuous function f A R extends to a continuous function

F X R

Tosay that F extends f means that F j f

Remark R maybereplacedby a closed interval a b

Pro of Supp ose that if A is a closed subset of X anycontinuous function f A R extends to a

continuous function F X R Let C and D b e disjoint closed subsets of X Dene f C D R by

if x C

f x

if x D

C D is closed so by assumption I may extend f to a function F X RPick disjoint op en sets U

and V in R such that U and V ThenF U andF V are disjointopensetsinX C F U

and D F V Therefore X is normal

Supp ose rst that jf xj c for all x A and some c R Ill extend f to F X R suchthat

jF xj c

By the Lemma I may nd a continuous function g X R suchthat

c for all x X jg xj

jf x g xj c for all x A

Next apply the Lemma to f g A R to nd a function g X R such that

jg xj c for all x X

jf x g x g xj c for all x A

Supp ose n andg X R has b een dened and satises

jg xj c for all x X

jf x g x g x g xj c for all x A

Apply the Lemma to nd g X R satisfying

jg xj c for all x X

jf x g x g x g xj c for all x A

Iwant to dene F x g x Imust showthatthisseriesconverges To see this note that

n n

g x jg xj c and that c is a convergent geometric series Therefore

n n

converges absolutely by direct comparison

Next I want to show that F xiscontinuous Let x X and let

s x g x

k n

b e the k partial sum

If kjthen

k k k k

X X X X X

jg xj js x s xj g x g x c g x

n k j n n n

nj nj n n nj

k j

j n j j k

B C

c c c

B C

Fix j and let k

c jF x s xj

Since the right side go es to as j independent of x it follows that the partial sums converge

uniformly to F x Since the partial sums are continuous it follows that F xiscontinuous

For x AI have

f x g x c

As k the right side go es to and the sum go es to F x Thus F agrees with f on A

Before I consider the case where f is unb ounded I need to observe something ab out the preceding

construction I claim that if jf xj c I can construct an extension F x satisfying jF xj c

To see this rst construct F xasabove Note that

X X X

jF xj g x jg xj c c

n n

n n n

Let B F fcgfcg This is a closed set disjoint from A since if x A jF xj jf xj cLet

b e a continuous function suchthatA and B Dene F xx F x F still extends

f but now jF xj c

Now Ill consider the unb ounded case where f A R

b e the arctangent function x arctan xThen f A satises Let R

Construct an extension F X j f xj Then F X R is a continuous function

which extends f

by Bruce Ikenaga