## Basic Concepts of Point Set Topology

### Notes for OU course Math 4853

### Spring 2011

### A. Miller

### 1. Introduction.

The definitions of ‘metric space’ and ’topological space’ were developed in the early 1900’s, largely through the work of Maurice Frechet (for metric spaces) and Felix Hausdorff (for topological spaces). The main impetus for this work was to provide a framework in which to discuss continuous functions, with the goal of examining their attributes more thoroughly and extending the concept beyond the realm of calculus. At the time, these mathematicians were particularly interested in understanding and generalizing the Extreme Value Theorem and the Intermediate Value Theorem. Here are statements of these important theorems, which are well-known to us from calculus:

Theorem 1.1 (Extreme Value Theorem). Every continuous functionf : [a, b]→Rachieves

a maximum value and a minimum value.

Theorem 1.2 (Intermediate Value Theorem). Iff : [a, b]→R is a continuous function and

y is a real number betweenf(a)andf(b)then there is a real numberxin the interval[a, b] such thatf(x) =y.

In these statements we recall the following basic terminology. For real numbersaandb, [a, b] denotes theclosed finite interval

[a, b] ={x∈R|a≤x≤b}.

Ifc is a real number in the interval [a, b] such thatf(c)≥f(x) for everyx∈[a, b] then f(c)
is a maximum value for f on [a, b], and if d is a real number in the interval [a, b] such
that f(d)≤f(x) for every x∈[a, b] thenf(d) is a minimum value for f on [a, b]. The
functionf : [a, b]→_{R}is continuousprovided that for eachx0∈[a, b] and each >0 there

is a positive real number δ such that if x∈[a, b] and |x−x0|< δ then |f(x)−f(x0)| < .

As we proceed to develop the ideas of point set topology, we will review and examine these notations more thoroughly.

Notice that in the statement of these two theorems the fact that the domain of the function f is a closed finite interval [a, b] is crucial. Here are two examples that should convince you that this hypothesis is needed in both statements.

Example1.3. Consider the functionf : (0,2]→Rgiven byf(x) = 1/x. From our calculus

experience we know that f is a continuous function (for example f is a rational function whose denominator does not have any roots inside the interval (0,2]). This function has a minimum value of f(2) = 1/2 but it has no maximum value because limx→0+f(x) = +∞. So the conclusion of the Extreme Value Theorem fails, but of course the ‘half-open interval’ (0,2] ={x∈R|0< x≤2} is not a closed finite interval—so the example does not violate

the statement of the Extreme Value Theorem.

Example 1.4. Consider the function g : [−1,0)∪(0,2] → R given by g(x) = 1/x (whose

domain is the set [−1,0)∪(0,2] = {x∈R| −1 ≤x≤2 andx6= 0}). Again from calculus

experience we know that gis a continuous function. This function satisfiesg(−1) =−1 and

g(2) = 1/2 however there is no real numberxwithg(x) = 0 even thoughg(−1)<0< g(2). So the conclusion of the Intermediate Value Theorem fails, but again the set [−1,0)∪(0,2] is not a closed finite interval so the example does not contradict the statement of the Intermediate Value Theorem. [Sketch the graphs of the functions in these two examples so that you can better visualize what goes wrong.]

Mathematicians in the early twentieth century, such as Frechet and Hausdorff, came to realize that the validity of the Extreme Value Theorem and the Intermediate Value Theorem particularly relied on the fact that any closed finite interval [a, b] satisfies two key ‘topological properties’ known as ‘compactness’ and ‘connectedness’. Our goal in the point-set topology portion of this course is to introduce the language of topology to the extent that we can talk about continuous functions and the properties of compactness and connectedness, and to use these ideas to establish more general versions of the Extreme Value and Intermediate Value Theorems. We will end with a discussion of the difference in perspective between the Frechet and Hausdorff definitions, and how attempts to bridge these definitions contributed to the early development of the subject of point set topology.

### 2. Review of Set Theory.

A setAis a collection of objects or elements. In order to avoid certain paradoxes of set theory we will assume that an ‘object’ is always chosen from some universal set.

If x is an element and A is a set then ‘x∈ A’ means that x is an element of A, and ‘x /∈ A’ means x is not an element in A. Sets are usually described by either listing all of their elements or by using ‘set-builder notation’.

Example2.1. The setAconsisting of three elements labeled by 1, 2 and 3 might be described in one of the following three ways:

A ={1,2,3}

={x|xis a positive integer andx≤3}

={x|xis a positive integer and 0< x2<10}

The first description lists the elements of the set while the last two descriptions illustrate set-builder notation.

Examples. Here are some important special examples of sets that we will refer to frequently:

• The empty set ∅ is the set which contains no elements. This means that x /∈ ∅ for every elementxin the universal set.

• Z+={1,2,3, . . .} is theset of positive integers.

• Z={. . . ,−2,−1,0,1,2, . . .} is theset of integers.

• _{Q}={m/n|m, n∈_{Z}andn6= 0} is the set of rational numbers. Recall that every
rational number can be expressed asm/nwhere the greatest common divisor ofmand

nequals 1—when this happens we say thatm/nis in reduced form. (For example, the reduced form for the rational number 39/52 would be 3/4.)

• Ris theset of real numbers.

• R−Q={x∈R|x6=Q} is theset of irrational numbers.

• R+={x∈R|x >0} is the set of positive real numbers.

• Ifaandbare real numbers then we can define various ‘intervals fromatob’: [a, b] ={x∈R|a≤x≤b}called a closed interval

(a, b) ={x∈R|a < x < b}called a closed interval

[a, b) ={x∈_{R}|a≤x < b}

(a, b] ={x∈R|a < x≤b}

(a,∞] ={x∈R|a < x}

(−∞, b] ={x∈Rx≤b}

Notice that the ‘interval’ [a, b] is the empty set ifa > b, and that [a, a] ={a}.

Definition 2.2. We list some of the most important basic definitions in set theory. In the following letAandB be sets.

1. A is a subset of B (writtenA⊆B) provided that if x∈Athenx∈B.

2. A is equal to B (written A=B) if and only ifAis a subset ofB andB is a subset of A. This is equivalent to saying that x is an element ofA iff xis an element of B. (Notation: ‘iff’ is short for ‘if and only if’.)

3. A is a proper subset of B (written A(B orA⊂B) ifAis a subset ofB but Ais

not equal toB.

4. Theunion ofA and B is the set A∪B ={x|x∈Aorx∈B}.

5. Theintersection of A and B is the setA∩B={x|x∈Aandx∈B}.

6. The set difference of A and B is the setA−B ={x|x∈Aandx /∈ B}. Notice that typically A−B andB−Aare two different sets.

Definition2.3. In addition to defining the union and intersection of two sets we can define the union and intersection of any finite collection of sets as follows. Letnbe a positive integer (that is,n∈Z+), and letA1, A2, . . . , An be sets. Then define

A1∪A2∪ · · · ∪An = {x|x∈Ai for somei∈ {1,2, . . . n}} A1∩A2∩ · · · ∩An = {x|x∈Ai for every i∈ {1,2, . . . n}}

With a little more background we can also define unions and intersections of arbitrary families of sets. Let J be a non-empty set, and let Aj be a set for each j ∈J. In this situation we say that{Aj |j ∈J} is afamily (or collection) of sets indexed by J, and we refer to

J as theindex setfor this family. The union and intersection of this family are respectively defined by the following:

[

j∈J

Aj = [{Aj|j∈J} = {x|x∈Aj for somej∈J}

\

j∈J

Aj = \{Aj|j∈J} = {x|x∈Aj for everyj∈J}

In particular notice that a finite union A1∪A2∪ · · · ∪An is the same as Sj∈JAj where J ={1,2, . . . , n}, and similarly for intersections. Here are two examples of infinite unions and intersections.

Example 2.4. For eachn ∈ Z+ let An = [0,1/n] (which is a closed interval in R). Then

{An | n∈ Z+} is a family of sets indexed by the positive integers, with A1 = [0,1], A2 =

[0,1/2],A3= [0,1/3] and so on. Notice that each setAn in this family contains its successor An+1, when this happens we say the family is adecreasing family of nested sets. Here we have

[

n∈Z+

An= [

n∈Z+

[0,1/n] = [0,1]

and

\

n∈Z+

An=

\

n∈Z+

[0,1/n] ={0}.

Since the determination of infinite unions and/or intersections like these can sometimes be difficult, let’s give a formal proof of the second of these statements. As is often the case, being able to write down a formal proof is tantamount to being able to systematically determine the result in the first place.

Proof that T

n∈Z+ An={0}: First suppose thatxis an element of

T

n∈Z+ An. By definition

of intersection this means that x∈An = [0,1/n] for every positive integer n. In particular,

xis an element of A1 = [0,1] and so we must have x≥0. Suppose first thatx > 0. Then

by the Archimedean Principle1 _{there is a positive integer} _{N} _{such that 1}_{/N < x}_{. It follows}

thatx /∈[0,1/N] =AN which contradicts our choice ofxas being an element of T

n∈Z+ An.

1_{The}_{Archimedean Principle}_{asserts that for every positive real number}_{x}_{there is a positive integer}_{N}_{such}

Thus our supposition thatx >0 must be false (using ‘proof by contradiction’). Since we have already observed that x≥0 it must be that x= 0. So we have that x∈ {0}, and since x

was an arbitrary element ofT

n∈Z+An, this shows that

T

n∈Z+An⊆ {0}(by the definition of

subset).

Next, let’s suppose thatx∈ {0}. Thenx= 0 (since{0}is a set with only one element). For each positive integern, x= 0 is an element of An = [0,1/n]. Therefore x∈Tn∈Z+ An

by the definition of intersection, and this shows that{0} is a subset of T

n∈Z+ An. We have

shown that each of the sets{0}andT

n∈Z+ An is a subset of the other, and therefore the two

sets are equal by the definition of set equality.

[See if you can write down a formal proof of the first statement above, and of the two claims in the next example.]

Example2.5. For eachn∈Z+letBn = [0,1−1/n] (which is a closed interval inR). This is

a family of sets indexed by the set of positive integers, withB1= [0,0] ={0}, B2= [0,1/2], B3 = [0,2/3] and so on. Notice that each set Bn in this family is a subset of its successor Bn+1, when this happens we say the family is an increasing family of nested sets. Here we

have

[

n∈Z+

Bn=

[

n∈Z+

[0,1−1/n] = [0,1) and

\

n∈Z+

Bn=

\

n∈Z+

[0,1−1/n] ={0}.

The basic definitions of set theory satisfy a variety of well-known laws. For example, the

‘associative law for union’ says that for any three setsA,B andC, (A∪B)∪C=A∪(B∪C). And the‘commutative law for intersection’ says that for all sets Aand B, A∩B =B∩A. Of particular interest areDeMorgan’s Laws which can be expressed as follows: LetAbe a set and let{Bj |j∈J}be a family of sets then

A−[

j∈J

Bj = \

j∈J

(A−Bj)

A−\

j∈J

Bj = [

j∈J

(A−Bj)

We will not give a comprehensive list of all of the elementary laws of set theory here, but any book on set theory or discrete mathematics will contain such lists. In particular, the theory of sets can be completely described by certain lists of laws, one well-known such list is referred to as the ‘Zermelo-Fraenkel Axioms for Set Theory’.

Another definition involving sets that is very important and useful is the Cartesian prod-uct. Let n be a positive integer and let A1, A2, . . . , An be a family of n sets. Then the

Cartesian productof this family is the set

The formal symbol (a1, a2, . . . , an) is called anordered n-tuple, and we refer to ai as the ith coordinate of then-tuple. Notice that in this definition we don’t require that the sets

Ai all be distinct from each other. In fact, in the extreme case where all of theAi’s are the same setAwe obtain

An = A×A× · · · ×A = {(a1, a2, . . . , an)|ai ∈Afor eachi∈ {1,2, . . . , n}}.

For example,_{R}n _{denotes the set of all ordered}_{n}_{-tuples of real numbers. One can also define}

infinite Cartesian products but we will not need them in this course. IfX is a set then thepower set ofX is the set of all subsets ofX

P(X) = {A|A⊆X}.

(Some authors denote the power setP(X) by 2X.) Note that the empty set∅ and the setX

itself are always elements of P(X). If X is a finite set with n elements then P(X) is a set
with 2n _{elements.}

### 3. Open Sets in the Euclidean Plane.

A very familiar example of a Cartesian product is the set of ordered pairs of real numbers

R2 = {(x1, x2)|x1∈Randx2∈R}

which can be identified with the set of points in a plane using Cartesian coordinates. Writing

x= (x1, x2) andy= (y1, y2), the Pythagorean Theorem leads to the formula d(x, y) = p(x1−y1)2+ (x2−y2)2

measuring the distance between the two pointsxandy in the plane. This distance function
allows us to examine the standard Euclidean geometry of the plane, and for this reason we
refer to _{R}2 _{in conjunction with the distance formula} _{d} _{as the} _{Euclidean plane. Given}
x= (x1, x2) ∈R2 and a positive real number > 0 we define the open disk of radius

centered at xto be the set

B(x, ) = {y∈_{R}2_{|}_{d}_{(}_{x, y}_{)}_{< }}

(which is also sometimes called an open balland denoted byB(x)). Observe thatB(x, ) coincides with the set of all points inside (but not on) the circle of radiuscentered atx. We say that a subsetU ⊆R2of the Euclidean plane is anopen setprovided that for each element

x∈ U there is a real number > 0 so that B(x, ) ⊆U. This can be loosely paraphrased by sayingU ⊆R2 is an open set iff for each element x∈U all nearby points are contained in U. However note that the term ”nearby” must be interpreted in relative terms (which is equivalent to observing that different values ofmay be required for different pointsx∈U). Example3.1. The set{(x1, x2)∈R2|x1>3}, which consists of all points lying strictly to

the right of the vertical linex1= 3 in the x1x2-plane, is an open set in R2. However the set

{(x1, x2)∈R2|x1≥3}, which consists of points on or to the right ofx1= 3 is not an open

The next theorem provides an important description of the open sets in_{R}2_{.}

Theorem 3.2. The collection of open sets in the Euclidean planeR2 satisfies the following

properties:

(1) The empty set is an open set.
(2) The entire plane_{R}2 _{is an open set.}

(3) IfUj is an open set for eachj∈J thenSj∈J Uj is an open set.

(4) IfU1 andU2 are open sets thenU1∩U2is an open set.

Proof. Part (1) follows immediately from our definition of open set since the empty set does not contain any elements. To prove (2): supposex∈R2 then B(x, ) is contained inR2 for

any choice of >0. Now consider (3). Let Uj be an open set for eachj∈J and letxbe an element of the unionS

j∈JUj. Thenx∈Uj0 for somej0∈J (definition of union). SinceUj0 is an open set, there is a real number >0 so that B(x, )⊆Uj0. Since Uj0 ⊆

S

j∈J Uj, it

follows thatB(x, )⊆S

j∈J Uj, and this shows thatSj∈J Uj is an open set.

Finally, consider (4). LetU1andU2be open sets, and letx∈U1∩U2 (which means that x∈U1andx∈U2). Then there are real numbers1>0 such thatB(x, 1)⊆U1, and2>0

such thatB(x, 2)⊆U2. Letbe the smaller of the two numbers1 and2. Then >0, B(x, )⊆B(x, 1)⊆U1

and

B(x, )⊆B(x, 2)⊆U2.

Therefore,B(x, )⊆U1∩U2 and it follows thatU1∩U2 is an open set.

Note that the key observation in the final paragraph of the proof of the Theorem is that if≤1thenB(x, )⊆B(x, 1) which follows immediately from the definitions ofB(x, ) and B(x, 1). This Theorem suggests the following general definition which is the central focus

for ‘point set topology’ (this is essentially Hausdorff’s definition):

Definition 3.3. LetX be a set and let T be a family of subsets ofX which satisfies the following four axioms:

(T1) The empty set∅ is an element ofT. (T2) The setX is an element ofT.

(T3) IfUj∈ T for every j∈J then Sj∈J Uj is an element ofT.

(T4) IfU1andU2are elements ofT then the intersectionU1∩U2 is an element ofT.

then we say that T is a topology on the set X. Here the axiom (T3) is called closure of T under arbitrary unions and axiom (T4) is called closure of T under pairwise intersections.

Comments:

(i) Notice that the definition of topology seems unbalanced in the sense that unions and intersections are treated differently—we only require the intersection of two elements of

T to be in T, but we require arbitrary unions of elements of T to be in T. However it’s really this imbalance which contributes to the definition leading to a rich and useful theory.

(ii) WARNING: Every setXthat has at least two elements will have more than one different collection of subsets that form a topology. That is, one setXwill generally have MANY different possible topologies on it. So to specify a topology we have to specify both the set X and the collection of subsetsT.

Example3.4. By theorem 3.2, the collection

Teuclid = {U ⊂R2|U is an open set inR2}

forms a topology on the Euclidean plane. This is called theEuclidean topology on R2.

Notice that for each positive integernthe subsetUn⊆R2 defined by

Un = {(x1, x2)∈R2|x1>3−1/n}

is an open set in_{R}2 _{, however the intersection}

\

n∈Z+

Un = {(x1, x2)∈R2|x1≥3}

is not an open set. This shows that part (4) of Theorem 3.2 will not be true for arbitrary intersections, and justifies the imbalance mentioned in Comment (i) above.

To end this section, we will describe how the example of the Euclidean topology of the plane can be generalized to ‘Euclideann-space’.

Example 3.5. For any positive integer n, we can endow the set Rn of ordered n-tuples of

real numbers with a distance function

d(x, y) = p(x1−y1)2+ (x2−y2)2+· · ·+ (xn−yn)2

where x= (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn). The set Rn together with the distance

formuladis referred to asEuclidean n-space. For eachx= (x1, x2, . . . , xn)∈Rn and each

real number >0, let

B(x, ) ={y∈Rn|d(x, y)< }.

Then a subset U ∈ Rn is said to be an open set in Euclidean n-space provided that

for each x ∈ U there is a real number > 0 such that B(x, ) ⊆ U. With this definition theorem 3.2 extends easily to describe the collection of open sets in Euclideann-space (just replaceR2withRn in the statement and proof of that theorem). As a consequence, it follows

that the collectionTeuclid of open sets in Euclideann-space forms a topology on Rn. This is

called theEuclidean topology on Rn.

Example 3.6. Whenn= 1, Euclidean n-space is called theEuclidean lineR1=R. Here

the distance function is given by

d(x1, y1) =

p

(x1−y1)2=|x1−y1|,

forx1, y1∈R, and the open disks are open intervals:

Thus a subset U of the Euclidean line _{R}is an open set iff for each x∈U there is an >0
such that (x−, x+)⊆U. In other words, U ⊂Ris an open set iff for eachx∈U there is

an >0 such that ify is a real number with|x−y|< theny∈U. This collection of open
sets forms the Euclidean topologyTeuclid on the real line_{R}.

### 4. Review of Functions.

LetX andY be sets. Afunctionf fromX toY is a rule that associates each element ofX

with exactly one element ofY. For each elementx∈X, the unique element ofY associated with x is denoted by f(x) and called the image of x under f. We write f : X → Y to signify thatf is a function fromX to Y.

Given a functionf :X →T, the set X is referred to as thedomain of f, and the set

Y is called thecodomain of f. The set {y ∈ Y | y =f(x) for somex∈X} is called the range of the functionf. Notice that by definition the range of a function is a subset of its co-domain.

A functionf :X →Y is said to beone–to–one (orinjective) provided that ifx1 and x2 are elements ofX andf(x1) =f(x2) thenx1=x2. Thus a function is one–to–one iff each

element in the range of the function is associated with only one element of the domain. A functionf : X →Y isonto (or surjective) provided that for eachy ∈Y there is at least one x∈X such thatf(x) = y. In other words, f :X →Y is onto iff the range off equals the codomain of f. A function f : X → Y which is both one–to–one and onto is called a one–to–one correspondence(or abijection).

Iff :X →Y andg:Y →Z are functions (where the codomain of f equals the domain ofg) then we can form thecomposition off with gto be the functiong◦f :X →Z given byg◦f(x) =g(f(x)) for each x∈X. Note that the domain of g◦f equals the domain of

f, while the codomain ofg◦f equals the codomain of g. IfX is a set and A is a subset of

X then there is functioni:A→X called theinclusion function from A toX defined by

i(a) =afor eacha∈A. The inclusion function from a setX to itself is called theidentity function on X.

Definition 4.1 (Images and Inverse Images of Sets). Let f : X → Y be a function with domainX and codomainY. IfA⊂X then theimage of A underf is the set

f(A) = {y∈Y |y=f(a) for somea∈A} .

IfB is a subset ofY then theinverse image of B underf is the set

f−1(B) = {x∈X |f(x)∈B} .

These definitions can be summarized by saying that each function f : X → Y induces two functions on power sets. The first fromP(X) toP(Y) given byA7→f(A) for eachA∈ P(X), and the second fromP(Y) toP(X) given byB7→f−1(B) for each B∈ P(Y).

Example 4.2. Let f :R→Rbe the function defined by f(x) =x2+ 1 for allx∈R. Let

and codomain off. Then

f(A) =f((1,3]) ={y∈R|y =x2+ 1 for somex∈(1,3]}={x2+ 1|1< x≤3}= (2,10]

and

f−1(A) =f−1((1,3]) ={x∈R|x2+ 1∈(1,3]}={x|0< x2≤3}

={x|0< x≤√3 or −√3≤x <0}= [−√3,0)∪(0,√3].

Example 4.3. Let X be a set and let A be a subset of X. If i : A → X is the inclusion function andB ⊂X then

i−1(B) = {a∈A|i(a) =a∈B} = {aA|a∈Aand a∈B} = A∩B .

### 5. Definition of Topology and Basic Terminology.

Let’s start by repeating the definition of a topology on a setX:Definition 5.1. Let X be a set and let T be a family of subsets of T which satisfies the following four axioms:

(T1) The empty set∅ is an element ofT. (T2) The setX is an element ofT. (T3) IfUj∈ T for eachj∈J thenS

j∈J Uj is inT.

(T4) IfU1andU2are in T thenU1∩U2 is in T.

then we say that T is a topology on the set X. Here the axiom (T3) is referred to by saying T is closed under arbitrary unions and axiom (T4) is referred by saying T is closed under pairwise intersection.

A setX together with a topology T on that set is called a topological space, more formally we will refer to this topological space as (X,T). If (X,T) is a topological space then the elements ofT are calledopen sets in X.

Theorem 5.2 (T is closed under finite intersections.). Let T be a topology on a set X, and letU1, U2, . . . , Un be open sets in this topology wheren is a positive integer. Then the

intersectionU1∩U2∩ · · · ∩Un is an open set.

Proof. We prove this statement by induction onn. Ifn= 1 then the collection of open sets has only one setU1and the intersection is justU1 which is an open set. This shows that the

statement is true when n = 1. Now suppose the statement is true for a positive integer n

(this is theinduction hypothesis). To complete the proof by induction we need to show that the statement is true forn+ 1. So suppose thatU1, U2, . . . , Un+1 are open sets. Then we can

write U1∩U2∩ · · · ∩Un+1 =V ∩Un+1 where V =U1∩U2∩ · · · ∩Un. ThenV is an open

set by the induction hypothesis, and V ∩Un+1 is open by axiom (T4). This shows that the

statement is true forn+ 1 (that is,U1∩U2∩ · · · ∩Un+1 is open) and completes the proof by

Ifxis an element ofX andU is an open set which contains xthen we say thatU is a neighborhood of x.

Theorem 5.3. A subset V ⊆ X is an open set if and only if every element x ∈ V has a neighborhood that is contained inV.

Proof. (⇒) Suppose that V is an open set in X. Then for every element x ∈ V, V is a neighborhood ofxandV ⊆V.

(⇐) Suppose thatV is a subset ofX and each elementx∈V has a neighborhoodUx such

that Ux ⊆ V. Then V = S

x∈V Ux and since each Ux is open then the union is open by

axiom (T3). This shows thatV is an open set and completes the proof. [In the second part of the previous proof, carefully explain the equalityV =S

x∈V Ux?]

If T1 and T2 are two topologies on a set X and T2 ⊆ T1, then we say that T1 is finer

than T2, or thatT2 is coarser than T1. If T2(T1 then T1 is strictly finer than T2. If

neither of the topologiesT1andT2are finer than the other then we say that the two topologies

areincomparable. We end this section by describing a number of examples of topological spaces.

Example5.4. For any setX letTdiscrete=P(X) ={A|A⊆X}. Clearly axioms (T1) and (T2) hold since∅ andX are subsets ofX. The union and the intersection of any collection of subsets of X is again a subset of X and this shows that axioms (T3) and (T4) hold as well. ThereforeTdiscrete forms a topology onX and this is called thediscrete topology on

X. So in the discrete topology, every subset of X is an open set. Therefore Tdiscrete is the largest possible topology onX, which is to say that the discrete topology is finer than every topology onX.

Example 5.5. For any set X let Ttrivial = {∅, X}. It is easily verified that Ttrivial is a topology onX and this is called the trivial topology on X. This topology contains only two open sets (assuming thatX is a nonempty set). ThereforeTtrivialis the smallest possible topology onX, which is to say that every topology onX is finer than the trivial topology.

Example 5.6. Just a reminder that for each positive integer n,Teuclid is a topology onRn

called the Euclidean topology. In particular, taking n = 1 gives the Euclidean topology

Teuclid on the real lineR, referred to briefly as theEuclidean line. Here

Teuclid={U ⊂_{R}| for eachx∈U there is >0 such that (x−, x+)⊆U}.

With this definition it is not hard to show that every open interval inRis an open set in the

Euclidean topology. [Be sure you can write out an explanation for this.] But note carefully that there are open sets in the Euclidean topology that are not open intervals. For example, the union of two or more disjoint intervals (such as (−1,1)∪(√2, π)) will be an open set which is not an interval.

Example5.7. Consider the collection of subsetsT` of the real lineRgiven by

This set forms a topologyT`onRcalled thelower limit topology on R. It is not hard to

show that (1) each finite half-open interval of the form [a, b) where a < bis an open set in
the lower limit topology but not an open set in the Euclidean topology, while (2) every open
set in the Euclidean topology on_{R}is open in the lower limit topology. Thus the lower limit
topology on_{R}is strictly finer than the Euclidean topology on_{R}(that isTeuclid_{(}T`).

Example5.8. LetX be a set and letx0 be an element ofX. DefineT to be the collection

of subsets ofX consisting ofX itself and all subsets ofX which do not containx0. Thus T = {X} ∪ {U ⊆X|x0∈/U}.

This forms a topology onX which is called theexcluded point topology.

Example5.9. LetX be any set and define

Tcof inite={∅}∪{U ⊆X|X−U is a finite set} = {∅}∪{X−F| F is a finite subset ofX}.

Then it is not hard to show thatTcof initeforms a topology onX, called thecofinite topology on X. (We will explain this using closed sets in the next section.) Note that if the set X

itself is finite then every subset ofX will be finite, and so the cofinite topology on a finite set

X is the same as the discrete topology on X.

### 6. Closed Sets and the Closure Operation.

Throughout this section let (X,T) be a topological space. A subset C ⊆X is said to be a closed set in X provided that its complementX−C is an open set (that is,X−C ∈ T). Notice that taking complements inX defines a bijection between the collection of open sets in a topological space and the collection of closed sets. Basic properties of closed sets are described by the next theorem, whose proof depends principally on DeMorgan’s Laws. Theorem 6.1. The collectionC of all closed sets inX satisfies the properties

(1) The empty set∅ is an element ofC. (2) The setX is an element ofC.

(3) IfUj∈ C for eachj∈J thenTj∈J Uj is inC.

(4) IfU1 andU2 are inC thenU1∪U2 is in C.

Moreover, for any collection C of subsets of X which satisfies properties (1)–(4) there is a unique topology onX for whichCis the collection of closed sets.

Example6.2. For any setX consider the collection of subsets given by

C = {X} ∪ {F ⊆X|F is a finite set} .

Then it is easy to check thatC satisfies the four conditions in theorem 6.1 (for example, the intersection of any collection of finite sets is finite, and the union of two finite sets is finite). ThereforeCforms the collection of closed sets for some topology onX, and this is the cofinite topology

Tcof inite = {∅} ∪ {U ⊆X|X−U is a finite set}

If A is an arbitrary subset of X then the closure of A is the intersectioncl(A) of all closed sets in X which contain A. The next theorem is easily proved using this definition. The first two parts of this theorem can be summarized by saying thatthe closurecl(A) of a subsetA is the smallest closed set which containsA.

Theorem 6.3. For any subsetA ⊆X the closure cl(A)is a closed set. If C is a closed set containingA thencl(A)⊆C. The subset Ais a closed set if and only ifcl(A) =A.

Example6.4. Consider the trivial topology Ttrivial={∅, X}on a setX. The closed sets in this topology areX− ∅=X andX−X=∅. Then the closure of a subsetAofX is described by

cl(A) =

_{∅,} _{if}_{A}_{=}_{∅}

X, ifA6=∅.

To explain: IfAis the empty set then both closed sets∅andXcontainA, socl(A) =∅∩X =∅. If A is a nonempty subset of X then the only closed set containing A is X and so the intersection of all the closed sets containingAisX.

Example6.5. Consider the discrete topologyTdiscrete=P(X) on a setX. Then every subset ofX is closed, and so here the closure operation is described bycl(A) =A (by theorem 6.3).

Definition 6.6. Let Abe a subset of X. An elementx∈X is called a limit point of A provided that every neighborhood U ofxcontains an element of Aother than xitself. The set of all limit points ofAis called thederived set of Aand denoted byA0.

The next theorem describes a relationship between the set of limit points of a set and the closure of a set. The proof of this theorem is quite typical of proofs in point set topology. Theorem 6.7. For any subsetA⊆X,cl(A) =A∪A0.

Proof. The proof is broken into two parts where first we will show thatcl(A)⊆A∪A0 and then thatA∪A0 ⊆cl(A).

cl(A)⊆A∪A0: Supposex∈cl(A). Then we must show thatxis an element ofA∪A0. For contradiction, let us assume that x /∈ A∪A0. Thenx /∈A and x /∈A0 (by the definition of union). Sincex /∈A0,xis not a limit point ofAand this means that there is a neighborhood

U ofxwhich contains no element ofAother thanx. In fact, becausex /∈A,U cannot contain any elements ofA. ThusAis a subset ofX−U. SinceX−U is a closed set (asU is open) andcl(A) is the smallest closed set containingA(by theorem 6.3),cl(A) must be a subset of

X−U. On the other hand,xis an element ofcl(A) (by supposition) but x /∈X−U (since

U is a neighborhood ofx). This yields a contradiction and shows thatx∈A∪A0, as desired.

A∪A0 ⊆cl(A): Supposex∈A∪A0. For contradiction, let’s assume thatx /∈cl(A). This means thatx /∈A(sinceAis a subset of cl(A)) and thatx∈A0 (becausex∈(A∪A0)−A). Observe that X−cl(A) is an open set (being the complement of the closed setcl(A)), and thatX−cl(A) is a neighborhood ofx(x∈X−cl(A) since x /∈cl(A)). Asxis a limit point ofA, this neighborhood X−cl(A) must contain an element ofA. But sinceA is a subset of

cl(A) it can’t contain any elements ofX−cl(A). This contradiction leads us to conclude that

x∈cl(A), showing thatA∪A0 ⊆cl(A).

Theorem 6.7 is often helpful for determining the closurecl(A) of a given subsetA in a
topological space. However, it is not so useful for determining the derived setA0, in part this
is because there is no general relationship between the setsAandA0 _{(in particular, these two}
sets need not be disjoint).

Example6.8. In the Euclidean line (R,Teuclid) consider the subsets

A1= (0,3), A2= (0,3)∪ {5} and A3={1/n|n∈Z+}.

Then it is not difficult to determine the derived sets to be

A0_{1}= [0,3], A0_{2}= [0,3] and A0_{3}={0},

and from this we find the closures

cl(A1) = [0,3], cl(A2) = [0,3]∪ {5} and cl(A3) ={0} ∪ {1/n|n∈Z+} .

Notice that of these three setsAit’s only the last one where A∩A0=∅.

### 7. Continuous Functions and Subspaces.

Let (X,T1) and (Y,T2) be topological spaces. A functionf :X →Y is said to becontinuous

provided thatf−1_{(}_{U}_{)}_{∈ T}

1wheneverU ∈ T2. To summarize this situation we will sometimes

write thatf : (X,T1)→(Y,T2) is a continuous function.

Example 7.1. Consider the function f : R → R given by f(x) =x2, and the lower limit

topologyT` onR. The half-open interval [2,3) is open in the lower limit topology but

f−1([2,∞)) ={x∈_{R}|x2∈[2,∞)}={x∈_{R}|2≤x2}= (−∞,−√2]∪[√2,∞)

is not inT` (because −√2∈f−1_{([2}_{,}_{∞}_{)) but [}_{−}√_{2}_{,}_{−}√_{2 +}_{}_{) is not a subset of}_{f}−1_{([2}_{,}_{∞}_{))}

for any >0.) This shows that f : (_{R},T`)→ (_{R},T`) is not continuous. Notice that since

f−1_{([2}_{,}_{3)) is not open in the Euclidean topology, this also shows that}_{f} _{: (}

R,Teuclid)→(R,T`)

is not continuous. Butf : (R,Teuclid) →(R,Teuclid) is continuous (see theorem 7.4 below),

andf : (_{R},T`)→(R,Teuclid) is also continuous.

Theorem 7.2. Let (X,T1) and (Y,T2) be topological spaces. A function f : X → Y is

continuous if and only iff−1(C)is a closed set in X wheneverCis a closed set in Y.

Proof. Suppose thatf :X →Y is continuous and that C is a closed set inY. ThenY −C

is an open set inY, and sof−1_{(}_{Y} _{−}_{C}_{) is an open set in}_{X} _{by the definition of continuity.}

Then

X−f−1(Y −C) ={x∈X |x /∈f−1(Y −C)}={x∈X |f(x)∈/Y −C}

SinceX−f−1_{(}_{Y} _{−}_{C}_{) is a closed set (because}_{f}−1_{(}_{Y} _{−}_{C}_{) is open), this shows that}_{f}−1_{(}_{C}_{)}

is a closed set inX. The other direction of the if-and-only-if statement is proved in a similar fashion.

Example7.3. Again consider the functionf :R→Rgiven byf(x) =x2but this time with

respect to the cofinite topology Tcof inite on_{R}. In this topology a closed set is either _{R}or a
finite subsetF ⊂R. Sincef−1(R) =Randf−1(F) is finite [Can you explain this? How many

elements doesf−1(F) have?], the functionf : (R,Tcof inite)→(R,Tcof inite) is continuous by

the previous theorem.

Theorem7.4. Letf :R→Rbe a function. Thenf : (R,Teuclid)→(R,Teuclid)is continuous

if and only iff :R→Ris continuous with theδ–definition of continuity.

Proof. Let f : R→ R be a function. Before starting on the proof let’s recall that the δ–

definition of continuity for f means: for eachx0 ∈R and each > 0 there is aδ > 0 such

that|f(x)−f(x0)|< whenever|x−x0|< δ.

First assume that f : _{R} → Ris continuous with the δ–definition. Suppose that U ∈

Teuclid and that x0 ∈ f−1(U). By the definition of the Euclidean topology on R, there is

an > 0 such that B(f(x0, )) = (f(x0)−, f(x0) +) ⊆ U. By the δ–definition, there

is a δ > 0 so that if |x−x0| < δ then |f(x)−f(x0)| < . Thus if |x−x0| < δ then f(x)∈(f(x0)−, f(x0) +). It follows that

f−1(U)⊇f−1((f(x0)−, f(x0) +))⊇ {x∈R| |x−x0|< δ}= (x0−δ, x0+δ).

So we have shown that if x0 ∈f−1(U) then there existsδ >0 with B(x0, δ)⊆f−1(U) and

this implies thatf−1_{(}_{U}_{) is open in the Euclidean topology on}

R. Since U was an arbitrary

open set we have established thatf is continuous.

Now assume thatf is continuous (with the topology definition). Let x0 ∈R and let

be a positive real number. The open interval U = (f(x0)−, f(x0) +) is an open set in

the Euclidean topology. Thereforef−1(U) is an open set as well (applying the continuity of

f). Sincex0∈f−1(U) (becausef(x0)∈U) there is positive real number δwith B(x0, δ)⊆
f−1_{(}_{U}_{). If}_{x}_{is a real number with}_{|x}_{−}_{x}

0|< δ thenx∈B(x0, δ) and thereforex∈f−1(U).

This means thatf(x) is an element ofU = (f(x0)−, f(x0)+), in other words|f(x)−f(x0)|< . Thus we have shown that f is continuous with the δ–definition of continuity, and that completes the proof.

Consequence: Every functionf :R→Rknown to be continuous in Calculus I is continuous

with respect to the Euclidean topology on_{R}. For example, polynomial functions are
contin-uous as is f(x) = cos(x), and etc. This consequence even extends to functions f : X →R

where X is a subset of R (here we take the topology on X to be the ‘subspace topology’

coming from the Euclidean topology on _{R}as described below). Thus for example, rational
functions and the natural logarithm function are continuous when we take the domain of the
function to be the ‘domain of definition’ (that is, the set of all real numbers for which the
equation makes sense).

Let (X,T) be a topological space, and letAbe a subset ofX. We define a collectionTA

of subsets ofAby:

TA = {V ⊂A|V =U∩Afor some U ∈ T } = {U ∩A|U∈ T }.

Theorem7.5. Let(X,T)be a topological space, and letAbe a subset ofX. ThenTAforms a topology onA. This is called thesubspace topology onA.

Proof. I leave it to you to verify the four axioms (T1)–(T4) for a topology.

Theorem 7.6. Let (X,T)be a topological space and let (A,TA) be a subspace. Then the inclusion mapi:A→X is continuous.

Proof. IfU is an open set inX theni−1_{(}_{U}_{) =}_{U}_{∩}_{A}_{and}_{U}_{∩}_{A}_{∈ TA}_{.}

Example7.7. Consider the Euclidean line (R,Teuclid), and the subsets

A1= [0,4], A2= [0,1]∪[3,4] and A3=Z.

We describe some aspects of the subspace topology onA1,A2andA3.

Subspace topology on A1: Notice that the interval [0,1) ⊂A1 is open in the subspace

topology on A1 since [0,1) = (−1,1)∩A1 and (−1,1) (being an open interval) is an open

set in the Euclidean topology. On the other hand [0,1) isnot an open set in the Euclidean topology onR.

Subspace topology onA2: Since [0,1] = (−1,2)∩A2and [3,4] = (2,5)∩A2, the sets [1,2]

and [3,4] are open sets in the subspace topology onA2. Moreover, observe thatA2−[1,2] =

[3,4] and it follows that [0,1] is a closed set in the subspace topology on A2.

Subspace topology on A3: Let n an element of A3 =Z. Then the open interval (n−

1/2, n+ 1/2) is an open set in the Euclidean line. Thus (n−1/2, n+ 1/2)∩Z={n} is an

open set in the subspace topology on_{Z}. This shows that every singleton subset {n} of _{Z}is
open in the subspace topology. IfV is any subset ofZthen V =∪n∈V{n} so V is an open

set by axiom (T3). This shows that the subspace topology onZis the discrete topology.

WhenAis a subset of Euclidean n-spaceRn then we refer to the subspace topology as

theEuclidean topology onA.

Theorem 7.8. Let X and Y be topological spaces and let f : X → Y be a continuous function. LetA be a subset of X. Then the function g: A→f(A)defined by g(a) =f(a) for eacha∈Ais continuous whenA⊆X andf(A)⊆Y have the subspace topologies.

Proof. Let f : X → Y be a continuous, let A ⊆ X and let g : A → f(A) be as defined. SupposeU is an open subset off(A) with the subspace topology. Then there is an open set

U0 inY so thatU =U0∩f(A). Then

g−1(U) ={a∈A|g(a)∈U}={a∈A|g(a)∈U0}

={x∈X |f(x)∈U0} ∩A=f−1(U0)∩A .

Since f−1_{(}_{U}0_{) is open in} _{X} _{by the continuity of} _{f}_{, it follows that} _{g}−1_{(}_{U}_{) is open in the}

### 8. Connectedness and the Intermediate Value Theorem

Definition8.1. A topological space (X,T) is said to beconnectedprovided that the only subsets ofX which are both open and closed are the empty set and the whole spaceX. (By axioms (T1) and (T2) we know that the sets∅ andX are always both open and closed.)Many authors give the definition of connectedness ofX as: if X =U∪V where U and

V are open sets which are disjoint then eitherU orV is the empty set. [Convince yourself of the equivalence of these definitions.]

Example8.2. The trivial topologyTtrivialon a setX is connected since in this topology the only open sets are∅andX. IfX is a set with at least two distinct elements then the discrete topology Tdiscrete on X is not connected: because if x0 ∈ X then the singleton set {x0} is

both open and closed (in the discrete topology, every subset of X is both open and closed) but{x0} 6=∅ and{x0} 6=X.

Example 8.3. Consider the setA2 from example 7.7. Since [0,1] is an open and closed set

which is neither∅norA2, we conclude thatA2with the subspace topology is not a connected

topological space. This example should give you an idea of where the term ‘connected’ comes from. (Draw a picture ofA2 inR, does it look like it should be ‘connected’ ?) The discussion

of this exampleA2⊂Rleads directly to the next result.

Theorem 8.4. If Ais a subset ofRwhich is not an interval then A is not connected in the

Euclidean topology.

Proof. IfA is not an interval then there is a real number x0∈ R−A such that Acontains

at least one real number smaller than x0 and at least one number larger than x0. Then

(−∞, x0)∩A is both open and closed in the subspace topology on A but it is neither the

empty set nor all ofA.

Theorem 8.5. Every interval in the real line is connected in the Euclidean topology.

Proof. We will give the proof only in the case where the interval is a closed finite interval
[a, b] where a < b. It is not difficult to adapt the proof to handle the other possible intervals
such as_{R}= (−∞,+∞), (−∞, b], [a, b) and so on.

Suppose that U is a nonempty subset of [a, b] which is both open and closed in the Euclidean topology. In particular, this means that [a, b]−U is open (sinceU is closed). By replacingU with [a, b]−U if necessary, we may assume thata∈U. The goal of the proof is to show that U = [a, b]; if we can show this then it follows that [a, b] is connected. Define a subsetS of [a, b] by

S={s∈[a, b]|[a, s]⊆U}

Observe that S ⊆U, that a∈S (since [a, a] = {a}), and that if s∈ S and a < t < s then

t∈S. This shows thatS must be an interval of the form [a, m) or [a, m] for somem∈[a, b]. We now consider three different cases.

subset of [a, b]. It follows that there is an >0 such that (m−, m+)∩[a, b]⊆[a, b]−U. Ifsis an element of (m−, m+)∩[a, b] which is smaller thanmthens∈S⊆U, which is impossible sincem /∈U. ThereforeCase 1can’t happen.

Case 2: S= [a, m) andm∈U. Then there is an >0 such that (m−, m+)∩[a, b]⊆U. Letsbe an element of (m−, m+)∩[a, b] which is smaller thanm. Then [a, s]⊆U (since

s∈S) and [s, m]⊆U. This implies that [a, m] = [a, s]∪[s, m]⊆U, and that m∈S, which is impossible. ThereforeCase 2can’t happen.

Case 3: S= [a, m] andm < b. In this case m∈U sincem∈S. So there is an >0 with (m−, m+)⊆U. Lets be an element of this open interval which is larger thanm. Then

s /∈S (since s > m) buts∈S (since [a, s] = [a, m]∪[m, s]⊆U). This contradiction shows
that_{Case 3}can’t happen.

Since none of the three cases can happen, we conclude thatS= [a, m] wherem=b. By the definition of the setS this means that U = [a, b], and the proof is complete.

Theorem 8.6. LetX andY be topological spaces and letf :X →Y be a function which is continuous and onto. IfX is connected thenY is connected.

Proof. LetU be a subset of Y which is both open and closed. Then f−1(U) is open in X

sincef is continuous. Alsof−1_{(}_{U}_{) is closed in} _{X} _{by theorem 7.2. So}_{f}−1_{(}_{U}_{) is both open}

and closed. AsX is connected we conclude that either (1)f−1_{(}_{U}_{) =} _{∅}_{or (2)} _{f}−1_{(}_{U}_{) =}_{X}_{.}

Using the hypothesis that f is onto, in case (1)U must be empty and in case (2), U must equal Y. Thus the only subsetsU ofY that are both open and closed are ∅ andY, so Y is connected.

We can now prove the_{Intermediate Value Theorem}:

Theorem 8.7. If f : [a, b]→Ris continuous andc is a real number between f(a) andf(b)

then there is anx∈[a, b]such thatf(x) =c.

Proof. Letf : [a, b] → Rbe continuous. Then the function g : [a, b]→ f([a, b]) defined by

g(x) =f(x) forx∈[a, b] is continuous by theorem 7.8, and g is onto. We know that [a, b] is connected with the Euclidean topology by theorem 8.5, and sog([a, b]) =f([a, b]) is connected by theorem 8.6. Thereforef[a, b] must be an interval by theorem 8.4. Sincef(a) andf(b) are elements of the intervalf[a, b], any real numbercbetweenf(a) andf(b) must be inf([a, b]). This completes the proof.

This proof of the Intermediate Value Theorem readily extends to a more general state-ment, which has been applied in many different situations since Hausdorff’s time. Most of these applications have to do with being able to guarantee that there will exist solutions to certain types of equations.

Theorem 8.8. Let X be a connected topological space and let f : X → R be continuous

with respect to the Euclidean topology on _{R}. Ifa and b are elements of X and c is a real
number betweenf(a)andf(b)then there is anx∈X such thatf(x) =c.

### 9. Compactness and the Extreme Value Theorem

Definition 9.1. Let (X,T) be a topological space. Anopen cover of X is a collectionU
of open subsets ofX such thatS_{U} _{=}_{X} _{where}S_{U} _{=}S

U∈UU. In other words,U is an open cover ofX iffU ⊆ T and for each x∈X there is at least oneU ∈ U such thatx∈U. IfU

is an open cover ofX then asubcover of U is a subsetV ⊆ U which also coversX (that is

S_{V}_{=}_{X}_{). An open cover is}_{finite}_{if it contains only finitely many elements.}

Definition 9.2. We say that the topological space (X,T) is compactprovided that every open cover ofX has a finite subcover.

Example9.3. The cofinite topologyTcof inite on a nonempty set X is compact. Recall that

Tcof inite={∅} ∪ {X−F |F is a finite subset ofX} .

SupposeU is an open cover ofX. Then one of the elements ofU has the formX−F where

F ={x1, . . . , xk}is a finite subset ofX. For eachi= 1, . . . , k, letUibe an element ofU with xi ∈Ui. Then{X−F, U1, . . . , Uk} is a finite subcover ofU.

Example 9.4. The Euclidean line R is not compact. For example, the collection of open

intervals U ={(n−1, n+ 1)| n ∈_{Z}} is an open cover of the Euclidean line which has no
finite subcover. (This is because each integernis contained in the interval (n−1, n+ 1) but
not in any other interval inU. So any proper subset of U will not coverR.)

For a different explanation, it can also be shown that the collection of open intervals{(−n, n)| n∈Z+}forms an open cover ofRwith no finite subcover.

Example 9.5. The half-open interval [0,3) is not compact with the Euclidean topology. To see this observe that U = {[0,3− 1

n) | n ∈ Z

+_{}} _{is an open cover of [0}_{,}_{3) which has}

no finite subcover. (Note that [0,3− 1

n) is open in the subspace topology on [0,3) since

[0,3− 1

n) = (−1,3− 1

n)∩[0,3). The union of any finite subset ofU will not include 3− 1 N

for some positive integerN.)

Theorem 9.6. Every closed finite interval [a, b] in the real line is compact in the Euclidean topology.

Proof. LetU be an open cover of [a, b] with respect to the Euclidean topology on [a, b]. Let

S be the subset of [a, b] defined by

S = {s∈[a, b]|there is a finite subsetV ⊆ U with [a, s]⊆[V}.

Thena∈S and ifs∈S anda < t < sthent∈S. It follows thatS is an inerval of the form [a, m] or [a, m) for some m∈ [a, b] (in the second form [a, m) we assume that m > a). We consider two cases:

Case 1: S = [a, m] where m < b. In this case m is an element of S so there is a finite subset V ⊆ U with [a, m] ⊂S

an > 0 such that (m−, m+)∩[a, b] ⊆U. Ift is an element of (m−, m+)∩[a, b]
larger thanmthen [a, t] = [a, m]∪[m, t]⊆S_{V}_{. Therefore there is an element}_{t} _{in} _{S} _{which}

is larger thanm but this can’s happen sincem was assumed to be the largest element ofS.
This contradiction shows that_{Case 1}can’t happen.

Case 2: S= [a, m) wherem > a. LetU be an element of the open coverU that contains

m. Then there is an >0 such that (m−, m+)∩[a, b]⊆U. Chooset∈(m−, m+)∩[a, b] smaller thanm. Thent∈S so that there is a finite subsetV in U with [a, t]⊆S

V. Then
[a, m] = [a, t]∪[t, m] is contained inS_{(}_{V ∪ {U}_{}}_{). Since} _{V ∪ {U}} _{is a finite subset of} _{U}_{, this}

implies thatm∈S, which is a contradiction. Therefore_{Case 2}can’t happen.

As neitherCase 1norCase 2can happen, we conclude thatS = [a, m] wherem=b. This means thatU has a finite subcover, which shows that [a, b] is compact.

Theorem 9.7. If C is a subset of R which does not have a largest (or a smallest) element

thenC is not compact with the Euclidean topology.

Proof. LetC be a subset ofRwhich does not have a largest element. We consider two cases

according to whether or notC is ‘bounded above’. A subset ofRisbounded aboveif it is

a subset of (−∞, H) for some real numberH.

Case 1: Suppose that C is not bounded above. Then {(−∞, n)∩C |n∈_{Z}+_{}} _{is an}

open cover ofCwhich can’t have any finite subcover (sinceC is not bounded above). Case 2: Suppose that C is bounded above. ThenC has a least upper boundm. Note thatm∈CsinceCdoes not have a largest element. In this case{(−∞, m−1

n)∩C|n∈Z
+_{}}

is an open cover ofC which has no finite subcover.

In either case, we conclude that C is not compact. A similar argument using lower bounds works ifC does not have a smallest element.

The next theorem categorizes exactly which subsets ofRare compact with the Euclidean

topology (taken= 1). A subsetCof_{R}n _{is}_{bounded}_{if there exists a positive real number}_{N}

so thatC⊂B(0, N). Note that C⊂Ris bounded iff there are real numbers LandH such

that L≤x≤H for all x∈C. HereL is called alower bound for C andH is an upper bound forC.

Theorem 9.8 (Heine-Borel Theorem). A subset C of Euclidean n-space Rn is compact in

the Euclidean topology if and only ifC is closed and bounded.

One part of the proof of the Heine-Borel Theorem follows from the next result.

Theorem 9.9. If X is a compact topological space andC is a closed subset ofX thenC is compact with the subspace topology.

Proof. Assume that X is a compact topological space and that C is a closed subset of X. Let U be an open cover of C. Then for each U ∈ U there is a setU0 which is open in X

such that U = U0∩C. Since U covers C, C is a subset of S_{{U}0 _{|} _{U} _{∈ U }}_{. Since}_{C} _{is a}
closed subset of X, X−C is open and therefore{U0 _{|}_{U} _{∈ U } ∪ {X} _{−}_{C}} _{is an open cover}
of X. Since X is compact this open cover has a finite subcover which can be taken to be

{U0 _{|}_{U} _{∈ V} ∪ {X}_{−}_{C}} _{where} _{V} _{is a finite subset of}_{U}_{. Since} _{X}_{−}_{C} _{doesn’t contain any}
elements of C, V must coverC. This shows that V is a finite subcover of U and that C is
compact.

Whenn= 1 it is not difficult to complete the proof of the Heine-Borel Theorem, however we will not pursue it here as we won’t need the result in our subsequent discussions. It is also not difficult to prove Heine-Borel Theorem whenn >1 although this does rely on some knowledge of ‘product topological spaces’, we will describe this concept later.

Theorem 9.10. Let X and Y be topological spaces and let f : X → Y be a continuous function. If X is compact then range(f) = f(X) is compact with the subspace topology coming fromY.

Proof. Letf :X →Y be a continuous function whereX is compact. LetU be an open cover of f(X) using the subspace topology that f(X) inherits as a subset ofY. For each U ∈ U

there is an open setU0 in Y such that U =U0∩f(X). Because f is continuous, each of the
sets f−1_{(}_{U}0_{) where}_{U} _{∈ U} _{is open in} _{X}_{. Moreover the collection} _{{U}0 _{|}_{U} _{∈ U }}_{is an open}
cover of X since f(X) ⊂ S

U∈UU0. Since X is compact this open cover of X has a finite
subcoverV. Then the collection{U ∈ U |f−1_{(}_{U}0_{)}_{∈ V}_{)}_{}}_{is a finite subcover of}_{U}_{. This shows}
thatf(X) is compact.

We can now assemble a proof of the_{Extreme Value Theorem}.

Theorem 9.11. Letf : [a, b] → R be a continuous function with respect to the Euclidean

topologies on[a, b]andR. Thenf has a maximum value and a minimum value.

Proof. Letf : [a, b]→ Rbe a continuous function with respect to the Euclidean topologies

on [a, b] and R. By theorem 9.6 the closed finite interval [a, b] is compact. By continuity of

f (using theorem 9.10) this implies the range of f is compact with the Euclidean topology. By theorem 9.7 the range off must have a largest and a smallest element, and by definition these are respectively maximum and minimum values forf on [a, b].

This proof of the Extreme Value Theorem readily extends to a more general statement Theorem 9.12. Let X be a compact topological space and letf : X →Rbe a continuous

function with respect to the Euclidean topology on _{R}. Then there are elements L, H ∈ X

such thatf(L)≤f(x)andf(H)≥f(x)for allx∈X.

### 10. Metric Spaces

Definition10.1. LetX be a set. Ametric onX is a functiond:X×X →[0,+∞) which satisfies the following properties:

(M1) d(x, y) = 0 if and only if x=y. (M2) d(x, y) =d(y, x) for all x, y∈X.

(M3) d(x, z)≤d(x, y) +d(y, z) for allx, y, z∈X.

A metricdonX is sometimes referred to as adistance functiononX, andd(x, y) is called the distance from xto y. The third axiom (M3) is known as the triangle inequality. A fixed setX can have many different metrics on it. Ifdis metric onX then we say that (X, d) is ametric space. When it is clear which metricd is being considered then we sometimes say thatX is a metric space.

Example 10.2. TheEuclidean metric on the real line R is the functiond: R×R→

[0,∞) defined by d(x, y) = |x−y| for x, y ∈ R. [Verify that this forms a metric.] More generally, theEuclidean metric on Rn is the functiondgiven by

d(x, y) = (x1−y1)2+· · ·+ (xn−yn)2 1/2

wherex= (x1, . . . , xn) andy= (y1, . . . , yn).

Definition 10.3. Let (X, d) be a metric space. Let x ∈ X and let be a positive real number. Theopen ball of radius centered at xis the set

B(x, ) = {y∈X|d(y, x)< }.

If there is any doubt as to which metricdis used to define an open ball, we writeBd(x, ) in place ofB(x, ).

Theorem 10.4. Let(X, d)be a metric space. The collection of subsets ofX given by

Td={U ⊆X | for eachx∈U there is >0 such thatB(x, )⊆U}

forms a topology on X. This topology is called the metric topology on X associated with d.

Proof. We need to verify the four topology axioms (T1)-(T4). This is proof is exactly the same as the proof of theorem 3.2.

For eachx∈Xand each >0, the open ballB(x, ) is an open set in the metric topology

Td. To prove this suppose that y∈B(x, ), and putδ=−d(x, y) (which is a positive real

number sinced(x, y)< ). Ifz∈B(y, δ) then

d(z, x)≤d(z, y) +d(y, x)< δ+d(x, y) =−d(x, y) +d(x, y) =.

This shows thatB(y, δ)⊆B(x, ) and completes the proof thatB(x, ) is an open set in the metric topology. On the other hand, it is not true that every open set in the metric topology is an open ball. So the open balls are very special examples of open sets. However, the definition of the metric topology in theorem 10.4 can be seen to be equivalent to saying that every open set inTd is a union of open balls. [Can you verify this? Look back at theorem 5.3.] IfBis a subset of a topologyT with the property that every element of T is a union of elements of

Bthen we say thatB is a basis for the topology T. So we can say that the collection of open balls

B={B(x, )|x∈X and >0}

forms a basis for the metric topology onX.

Here are some more examples of metric spaces.

Example10.5. LetX be any set and letd:X×X→[0,∞) be defined by takingd(x, y) = 0 ifx=y andd(x, y) = 1 ifx6=y. It is easy to verify that (X, d) is a metric space. Moreover notice that forx∈X, the open ballB(x, ) is the singleton set{x}if≤1. This shows that every singleton subset ofX is open in the metric topology, and soTd=Tdiscrete. This metric onX is called the discrete metric.

Example10.6. OnR2, a metricρcan be defined by the equation:

ρ(x, y) = max{|x1−y1|,|x2−y2|}

where x= (x1, x2) andy = (y1, y2). [Verify the metric space axioms.] With this metric, an

open ballBρ(x, ) is actually an open square with horizontal and vertical sides whose center

is at x(the side length of the square is 2). Since every circle has an inscribed square and vice-versa, it can be shown that the metric topologyTρ onR2 is the same as the Euclidean

topology. The moral of this example is that different metrics on a set X may very well generate the same topology.

Here’s a more careful explanation thatTρ coincides with the Euclidean topology on_{R}2_{.}

Letdbe the standard Euclidean metric on_{R}2 _{given by}
d(x, y) = (x1−y1)2+ (x2−y2)2

1/2

where x= (x1, x2) andy = (y1, y2). Then it is not hard to verify that, for each > 0 and x= (x1, x2)∈R2, we have

Bρ(x, )⊃Bd(x, ) and Bd(x, )⊃Bρ(x, /√2).

Now suppose thatU is a subset of_{R}2_{which is open in the}_{ρ}_{–metric topology}_{T}

ρ. By definition

this means that for each x ∈ U there is > 0 such that Bρ(x, ) ⊆ U. Then Bd(x, ) is contained in U (since Bd(x, ) is a subset of Bρ(x, )), and this shows that U is open in

Td =Teuclid. On the other hand, suppose that U is open in the Euclidean topology on_{R}2_{.}

This means that for eachx∈U there is >0 so thatBd(x, )⊆U. ThenBρ(x, /√2)⊆U

and soU is open in theρ–metric topology Tρ. ThusTρ =Td =Teuclid, as claimed.

Example 10.7. If (X, d) is a metric space and A is a subset of X. Then the function

d0:A×A→[0,∞) defined byd0(x, y) =d(x, y) forx, y∈Ais a metric onA. Note that, for

x∈A,Bd0(x, ) =Bd(x, )∩A. From this it follows that the metric topologyTd0 onAis the

same as the subspace topology coming from the metric topologyTd onX.

There are many advantages to being able to work with a metric space topology rather than an arbitrary topology on a space. One of these is that in metric spaces, many aspects of the topology can be described in terms of ‘convergent sequences’. For example, ifX is a metric space andAis a subset ofX then an element x∈X will be a limit point of Aif and only if there is a sequence of elements inAwhich converges tox. Of course we haven’t defined what a convergent sequence is but it can be defined just like in calculus. That is, a sequence (xn) in the metric space (X, d) converges tox∈X iff for each >0 there is a positive integer

N such that d(xn, x)< for all n≥N. In metric topologies, sequences can also be used to describe the notion of compactness—so in metric spaces certain basic topological definitions can be formulated in different ways.

### 11. The Metrizability Problem

The definition of ‘metric space’ and its associated topology was first introduced by Frechet about ten years before Hausdorff’s more general abstract definition of ‘topological space’. The

early history of topology was dominated by a desire to understand the relationship between these two definitions. To explain more precisely, we say that a topological space (X,T) is metrizableif there is a metric don X so that T =Td. So the basic questions became: is every topological space metrizable? if not, are there some additional axioms on a topological space that guarantee metrizability? In order to study these ‘metrizability problems’, many fundamental properties of topological spaces were discovered and interrelationships developed. For example these studies led to various ‘separation axioms’ and ‘countability axioms’ which play a role throughout the subject. In the 1920’s, a Russian mathematician Pavel Urysohn used some of these ideas to solve a special version of the metrizability problem which we will describe below (without proof). Ultimately, in the early 1950’s, a complete (although complicated) solution was found to the most general version of the metrizability problem, bringing an end to the early history of the subject.

Here are two of the ‘separation axioms’ which play a role in the metrizability problem. Let (X,T) be a topological space. ThenX is aHausdorff spaceprovided that wheneverx1

andx2 are distinct elements of X then there are neighborhoodsU1 of x1 andU2 of x2 such

thatU1∩U2=∅. The topological space X is said to be normalprovided that wheneverC1

andC2are disjoint closed sets inX then there are disjoint open setsU1andU2withC1⊆U1

andC2⊆U2. One can show that in a Hausdorff topological space every singleton set is closed,

and this weaker property is commonly referred to as the T1–axiom. Many authors take as

part of the definition of a normal space that the space satisfy theT1–axiom. In the presence

of this axiom (but not otherwise) it is easy to see that every normal space is Hausdorff. Theorem 11.1. Let(X, d) be a metric space. The metric topology on X is Hausdorff. In fact, it is normal andT1.

Proof. We will only prove the Hausdorffness. Suppose that (X, d) is a metric space and consider the associated metric topology onX. Letx1andx2be distinct elements ofX. Set

to be half the distance fromx1tox2. That is=d(x1, x2)/2 which is a positive real number

(by (M1) since x1 6= x2). Then x1 ∈ B(x1, ) and B(x1, ) is a neighborhood of x1 (since

open balls are open in the metric topology). Likewise, B(x2, ) is a neighborhood of x2. If z∈B(x1, )∩B(x2, ) then

d(x1, x2)≤d(x1, z) +d(z, x2)< += 2=d(x1, x2),

and this shows thatd(x1, x2) is strictly less thand(x1, x2) which is clearly impossible. From

this contradiction we conclude thatB(x1, )∩B(x2, ) is empty, thus showing that the metric

topology is Hausdorff.

One consequence of this theorem is that a topological space which is not Hausdorff (or normal) can’t possibly be metrizable. For example, the trivial topology on a set with more than one elements is not metrizable. Also, the cofinite topology on an infinite set is T1 but

not Hausdorff, so it is not metrizable either.

Urysohn proved the following remarkable theorem:

Theorem 11.2 (Urysohn’s Lemma). LetX be a normal topological space, and letAandB be disjoint closed sets inX. Then there is a continuous functionf :X →[0,1]withf(A) = 0 andf(B) = 1where the interval[0,1]has the Euclidean topology.

And he used this result to establish:

Theorem 11.3 (Urysohn’s Metrization Theorem). LetX be a normalT1 topological space

which has a basisB with countably many elements in it. ThenX is metrizable.

The condition thatX has a basis with countably many elements is an example of one of the ‘countability axioms’ for a topological space. This particular condition is often called second countability of X. For the record, a set B is countable (or B has ‘countably many elements’) if and only if there is a surjective function from Z+ to B. Another of the

countability axioms (one that is weaker than second countability) is known as ‘separability’: a topological space isseparableprovided that there is a countable subset ofX whose closure equalsX. Urysohn’s Metrization Theorem only solves the metrizability problem for separable topological spaces— the complete solution took another thirty odd years for mathematicians to complete.

[More precisely, what we are claiming here is that a separable metric space is second countable. In fact, a metric space is separable if and only if it is second countable. See if you can prove this.]

A key role in the proof of Urysohn’s Metrization Theorem is played by the construction of ‘product of topological spaces’. Since this definition is very important in any graduate level course in point set topology, we include it here for completeness. Let (X,T1) and (Y,T2) be

topological spaces. Then the collection of subsets ofX×Y given by

{Z⊆X×Y | for each (x, y)∈Z there areU ∈ T1 andV ∈ T2 with (x, y)∈U×V ⊆Z)}

forms a topologyTproductcalled theproduct topologyonX×Y. Equivalently, the product topology is the topology for which the collection of sets

{U×V |U ∈ T1 andV ∈ T2}

forms a basis. The setsU ×V where U ∈ T1 andV ∈ T2 are often calledopen rectangles

in X×Y. Therefore, each open rectangle inX×Y is an open set in the product topology, but not every open set need be an open rectangle. In general, a subset ofX×Y is open in the product topology iff it can be expressed as a union (possibly infinite) of open rectangles.

### 12. The Homeomorphism Problem

As the basic principles of topology evolved another fundamental problem which is philosph-ically very different than the metrization problem emerged as an important central problem in the field. The problem is frequently referred to as ‘the homeomorphism problem’. It has its roots in nineteenth century (pre-topology) mathematics and continues to have great rele-vance in today’s mathematics. In order to describe the problem we first need to define what a ‘homeomorphism’ is.

Recall that a functionf :X →Y is abijectionif and only if it is both one-to-one and
onto. Associated with each bijectionf :X →Y there is aninverse function f−1_{:}_{Y} _{→}_{X}

defined for eachy∈Y by takingf−1_{(}_{y}_{) to equal}_{x}_{where}_{f}_{(}_{x}_{) =}_{y}_{. (Note that such an element}
x∈X exists sincef is onto and it is unique sincef is one-to-one.) The inverse functionf−1

is characterized by the facts that the compositionf−1_{◦}_{f} _{is the identity function 1}

X on X

and that the compositionf ◦f−1 _{is the identity function 1}

Y onY. In other words, for each x∈X,f−1(f(x)) =x, and for eachy∈Y,f(f−1(y)) =y.

Ahomeomorphism between topological spaces X andY is a bijectionf :X →Y

such that both f and f−1 are continuous. Another way to define what a homeomorphism between X and Y is is to say that it is a bijection f : X → Y for which a subset U is open inX if and only iff(U) is open inY. Thus a homeomorphism fromX toY induces a bijection between the open sets inX and the open sets inY, and this captures the idea that

X andY are essentially the same topological space with elements labelled differently. If there is a homeomorphism between X and Y then we say that X and Y are homeomorphic spaces. Being homeomorphic is easily verifed to be an equivalence relation on the collection of all topological spaces. This means that: (1) each topological space X is homeomorphic to itself (the identity function 1X is a homeomorphism); (2) if X is homeomorphic to Y

then Y is homeomorphic to X (iff : X → Y is a homeomorphism then f−1 _{:} _{Y} _{→} _{X} _{is}

a homeomorphism; and (3) if X is homeomorphic to Y and Y is homeomorphic to Z then

X is homeomorphic to Z (if f : X → Y and g : Y → Z are homeomorphisms then so is

g◦f :X →Z).

The Homeomorphism Problem can now be phrased as: Given two topological spaces determine whether or not they are homeomorphic.

A propertyPof a topological spaceX which is expressed entirely in terms of the topology of X is called a topological property. If X and Y are homeomorphic spaces and P is a topological property thenX satisfies P if and only ifY does. So one way to show that two topological spaces are not homeomorphic is to find a topological property satisfied by one of the spaces but not the other. For example, the intervals [−1,1] and (−1,1) (with the Euclidean topology) cannot be homeomorphic because [−1,1] is compact but (−1,1) is not compact and being compact is a topological property. We have encountered a number of topological properties in our discssions. For example each of the following is a topological property: compactness, connectedness, Hausdorffness, metrizability.

### Problem List:

Problem 1. For each positive integern(that is, n∈Z+) let

Un= (−n, n) and let

Vn = [n, n+ 1].

(So, for eachn,Unis an open interval inR, andVnis a closed interval inR.) Determine each

of the following: 1. U5∪V6

2. U5∩V6

3. U1∪U3

4. U1∩U3

5. V1∪V2∪V3∪V4

6. V1∩V2∩V3∩V4

7. S

{Vn |n∈Z+}

8. T

n∈Z+Vn

9. S_{{Un}_{|}_{n}_{∈}

Z+}

10. T

{Un|n∈Z+}

Problem 2. IfA,B andC are the sets A={1,2}, B={1,3,5}and C={4} then list all of the elements of the Cartesian productA×B×C.

Problem3. LetAandBbe sets. IfA∪Bis a subset ofA∩B what does that imply about the relationship between the sets A and B? Clearly state your result as a Theorem, and then prove it using the definitions of intersection and union. (Suggestion: Use Venn diagrams as an aid to make your conjecture but leave them out of your final proof.)

Problem 4. Find the domain, codomain and range of each of the following functions: a) f :R→Rgiven byf(x) = 2 + sin(x).

b) g: [0,4]→Rgiven byg(x) = 3−2x2.

c) h:R→[−4,+∞) given byh(x) =

−3 ifx <−1

−x2 _{if}_{−}_{1}_{≤}_{x}_{≤}_{1}
x+ 3 ifx >1