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Thus

By (b), there is a member B ofx with

Then O1 . O is the union of such members B of2 x and is therefore a member of Thus is a

.topology for X

Example 4.3.4

Let be the family of all intervals in of the form [ , ), . It is easily observed that a b a < b satisfies the conditions of and is a basis for a topology, called the Theorem 4.8 half-open interval

for . It is left as an exercise for the reader to show that is firsttopology countable and separable but not second countable. (The real line with the half-open interval topology is sometimes called the .)Sorgenfrey line

Definition: . Let and be bases for topologies and for a set X Then S and are .equivalent bases provided that the topologies and are identical

The proof of the following theorem is left as an exercise.

Theorem 4.9: Bases and for topologies on a set X are equivalent if and only if both of the following conditions hold:

(a) For each , .and x B there is a member such that x B B

(b) For each , .and x B there is a member such that x B B

In some instances it is advantageous to have a smaller collection of sets which generates a basis by the process of forming finite intersections. Such a family, called a , is defined as follows:subbasis

Definition: . Let be a space A subcollection of is a subbasis or subbase for if the .family of all finite intersections of members of is a basis for

Example 4.3.5

The collection of all open intervals of the form ( , ) and (–, ), , , is a subbasis for thea b a usual topology for .

EXERCISE 4.3Co py ri gh t © 2 01 6. D ov

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1. Let be a space and a subcollection of . Suppose that for each in , the set ofa X

members of which contain is a local base at . Show that is a basis for .a a

2. Prove part (a) of .Theorem 4.7

3. Give an example different from of a space that is first countable but not second countable.Example 4.3.4 X

4. Let be a first countable space and a limit point of a subset of X Show that there is a sequence of points ofX x A { } which converges to .A\ x x

5. Describe the bases and for determined by the open balls of the taxicab metric and the maxd metric respectively. Show that and are both equivalent to the basis of open balls in thed usual metric .d

6. Let be a first countable space and a member of . Prove that there is a local nested basis at X x X a

(i.e., a local basis such that for each positive integer ).Sn+1 Sn n

7. Let be the real line with the half-open interval topology of .Example 4.3.4

(a) Find the closure, interior, and boundary of the set = [0, 2] and the set = (0, 2).A B

(b) Prove that is separable and first countable but not second countable.

8. Let and be metrics for a set , and let and denote, respectively, the families of all open ballsd1 d2 X

generated by and . Show that and are equivalent metrics if and only if and ared1 d2 d1 d2 equivalent bases.

9. Prove .Theorem 4.9

10. Let be a set and any family of subsets of whose union equals . Show that is a subbasis for aX X X topology for .X

4.4 CONTINUITY AND TOPOLOGICAL EQUIVALENCE

Since distance between points is not defined for general topological spaces, how should one define continuity for a function from one topological space to another? Keep in mind that the definition, when applied to metric spaces, should agree with the definition of continuity given in . Recall that aChapter 3 function : from metric space ( , ) to metric space ( , ) is continuous at a point in if andf X Y X d Y d a X

only if for each open ball ( ( ), ) in centered at ( ), there is an open ball ( ) in centered at Bd f a Y f a Bd a, X a

such that ( ( , ) is contained in ( ( ) ). Continuity at for a function : on topologicalf Bd a Bd f a a f X Y

spaces and y is defined by simply replacing the open balls centered at ( ) and by open setsX f a a containing ( ) and , respectively.f a a

Definition: , , Let and be topological spaces f: X Y a function and a a point of . X Then f is continuous at a provided that for each open set V in Y containing f(a) there is an open set U

. in X containing a such that f(U) V The function f is continuous if it is continuous at each point of its

.domain

We shall deal most often with continuity of a function on its entire domain rather than at each point separately, so we examine the definition a bit more closely. The function is continuous simultaneouslyƒ for all in means that for every open set in and every point with ( ) in , there is an open set a X V Y a ƒ a V U

in with in and ( . Since ( ) is equivalent to ( , this means that ( containsa X a Ua f U )a V f Ua V Ua f 1 V) f1 V)

an open set about each of its members; in other words, ( is open in for each open set in .f1 V) X V YC op yr ig ht ©

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Alternate Definition: A function f: is continuous means that for each open ( .set V in Y f1 V) is an open set in X

The following theorem restates the definition of continuity in several equivalent forms.

Theorem 4.10: . Let f: X Y be a function on the indicated topological spaces and let a X The

following statements are equivalent:

(1) f is continuous at a.

(2) For each open set V in Y containing f(a), .there is an open set U in X such that a U and U f (V)1

(3) For each neighborhood V of f(a), .f1(V) is a neighborhood of a

(4) For each subset Vof Y with f(a) , .int V a belongs to int f1(V)

A proof of can be formulated directly from the definitions of the terms involved; this isTheorem 4.10 left as an exercise for the reader.

Theorem 4.11: . Let f: X Y be a function on the indicated topological spaces The following .statements are equivalent

(1) f is continuous.

(2) For each closed subset C of Y, .f1(C) is closed in X

(3) For each subset A of X, .

(4) There is a basis ( for the topology of Y such that f1 B) is open in X for each basic open set B in .

(5) There is a subbasis for the topology of Y such that f (S) is open in X for each subbasic open1

.set S in

Proof: We use the open set formulation to describe continuity: f is continuous if and only if for each , . open set V in Y f (V) is open in X1 The equivalence of (1) and (2) follows from the duality between open

, .sets and closed sets precisely as in the proof of Theorem 3.13

[(2) , . , (3)]: Suppose that (2) holds and let A be a subset of X Then is a closed subset of Y so

. its inverse image is closed in X Since

and the latter set is closed, then

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and (3) holds.

[(3) . (2)]: Assume (3) and let C be a closed subset of Y Then

and f1(C) must be a closed set. .Thus (3) implies (2) We have completed the proof that (1), , . (2) and (3) are equivalent Since a basis and subbasis

, . , for Y consist of open sets it should be clear that (1) implies both (4) and (5) Similarly since a basis is a , . .subbasis (4) implies (5) The proof will be completed by showing that (5) implies (4) and (4) implies (1)

[(5) (4)]: Suppose (5) holds and consider the basis generated from by taking finite

. ,intersections For any basic open set

for some finite collection of members S1, . . . , . S ofn Then

by . Since each set f (S ) is open in X and the intersection of any finite collection of openTheorem 1.7 1 i sets is open, . .then f (B) is open in X1 Thus (5) implies (4)

[(4) , . ,(1)]: Assuming (4) let O be an open set in Y By the definition of basis

for some subcollection {B : a . I} of the basis Then

Since each set f () is open in X and the union of any family of open sets is open1 , then f (O) is open in X1

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Since each set f () is open in X and the union of any family of open sets is open1 , then f (O) is open in X1

.and f is continuous

Conditions (4) and (5) of will be useful when we want to deal with a basis or subbasisTheorem 4.11 rather than with the entire topology of the range space. Condition (3) is the description of continuity which seems to fit best with the intuitive idea of closeness. We visualize continuity of at by thinkingf x that whenever is “close” to a set , ( ) is “close” to ( ). In topological language, is “close” to x A then f x f A x A

means , and ( ) is “close” to ( ) means .f x f A

Theorem 4.12: Z , If f: X Y and g: Y are continuous functions on the indicated spaces then the .composite function g f: X° Z is continuous

The proof of is left as an exercise.Theorem 4.12

Example 4.4.1

It should be emphasized that continuity for a function : is expressed in terms of f X Y inverse of open sets in is continuous if and only if for each open set in , ( ) is open in .images Y: ƒ O Y f1 O X

This is not to be confused with mapping open sets in to open sets in , which is quite a differentƒ X Y

property. Consider, for example, the function from the real line to a discrete two-point space = { , } defined byY a b

This function does map open sets in to open sets in , because every subset of y is open. But isX Y f not continuous; { } open in buta is Y

is not open in .

Definition: . Let f: X Y be a function on the indicated spaces Then f is an open function or open , . mapping if for each open set O in X f(O) is open in Y The function fis a closed function or closed , .mapping if for each closed set C in X f(C) is closed in Y

Example 4.4.1 shows that an open mapping may fail to be continuous. The reader is asked in the exercises to find examples to illustrate the following:

(a) A closed mapping may not be continuous.

(b) An open mapping may not be closed, and conversely.

(d) A mapping that is both open and closed may not be continuous.

Definition: Topological spaces X and Y are topologically equivalent or homeomorphic if there is a .one-to-one function f: X Y from X onto Y for which both f and the inverse function f are continuous1

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.The function f is called a homeomorphism

Topological equivalence is an equivalence relation for topological spaces. A homeomorphism : between spaces and a one-to-one function from onto forƒ X Y X Y is X Y

which both and are continuous. Since ( ) = for a bijection, continuity of can be expressed by thef f1 f1 1 ƒ f1

fact that for each open set in , ( ) is open in . In other words, a homeomorphism is just a bijectionO X f O Y which is also an open, continuous function. Since continuity of can also be expressed by the fact thatf1

for each closed set in , ( ) is closed in , then is a homeomorphism if and only if it is a closed,C X f C Y f continuous bijection.

Definition: A property P of topological spaces is a topological property or topological invariant , provided that if space X has property P then so does every space Y which is topologically equivalent to

The next three theorems give examples of topological properties.

Theorem 4.13: .Separability is a topological property

Proof: . Let X be a separable space with countable dense subset A and Y a space homeomorphic to X Let . . f: X Y be a homeomorphism The obvious candidate for a countable dense subset of Y is f(A) To see

, . . that f(A) is dense in Y let O be a non-empty open set in Y Then f (O) is a non-empty open set in X1 Since , . , A is dense in X f (O) contains some member a of A1 Then O contains the member f(a) of f(A) so every

. .non-empty open set in Y contains at least one member of f(A) Thus and Y is separable

Proofs of the next two theorems are left as exercises.

Theorem 4.14: .First countability and second countability are topological properties

Definition: .A topological space X is metrizable provided that the topology of X is generated by a metric

Theorem 4.15: .The property of being a metrizable space is a topological property

Hint: Let ( , ) be a metric space, topological space homeomorphic to , and aX d Y a X f:X Y homeomorphism. Define × byd on Y Y

Show that is a metric and that generates the topology of .d d Y

Since and (0, 1) are homeomorphic, we note that the property of being a bounded metric space is not a topological property.

Until this point, it has not been possible to give a precise definition of the branch of mathematics known as topology. is the branch of mathematics which deals with topological properties. OneTopology objective of topology is to give criteria in terms of topological invariants which allow one to determine whether or not two given spaces are homeomorphic. Ideally, one would want a list of topological properties which are easy to check and for which two spaces are homeomorphic if and only if they share the same properties from the list. Theorems of this type are called because theyclassification theorems divide topological spaces into classes, with two members of the same class being homeomorphic. Mathematicians have experienced only limited success in classifying topological spaces. There is no known list of topological properties which completely classifies topological spaces.

A second objective of topology is to establish relations among various topological properties and to

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show which combinations of topological properties are equivalent. Theorems of this type are called since they completely describe or characterize spaces of a certain type. Wecharacterization theorems

shall see one of the most famous characterization theorems, the Urysohn characterization, which gives necessary and sufficient conditions for a space to be homeomorphic to a separable metric space (

), in .Theorem 8.18 Chapter 8

EXERCISE 4.4

1. Let : be a function and let . Prove that is continuous at if and only if for each subset of ƒ X Y a X ƒ a A X

with .

2. Let : y be a function and let . Prove that is continuous at if and only if there is a local basis f X a X f a

at ( ) such that for each , ( ) is a neighborhood of .f a f1 B a

3. Find an example of each of the following:

(a) A closed mapping that is not continuous

(b) An open mapping that is not closed and a closed mapping that is not open

(d) A function that is both open and closed but not continuous

4. Prove that the composition of continuous functions is continuous ( ).Theorem 4.12

5. Prove that topological equivalence is an equivalence relation.

6. Let : be a one-to-one correspondence from space onto space . Prove that the following statementsf X Y X Y are equivalent:

(a) is a homeomorphism.ƒ

(b) and are both open mappings.f ƒ1

7. Prove .Theorem 4.14

8. Prove .Theorem 4.15

9. Let : be a one-to-one function from space onto space .f X Y X Y

(a) Show that the following statements are equivalent:

(1) is continuous.ƒ1

(2) is an open mapping.ƒ

(3) is a closed mapping.ƒ

(b) Show that the following statements are equivalent:

(1) is a homeomorphism.ƒ

(2) is an open, continuous mapping.ƒ

(3) is a closed, continuous mapping.ƒ

10. Let be a separable space, a space, and : . continuous function from onto . Prove that isX Y f X Y a X Y Y separable.

11. . Definition: Let X be a space and a real-valued function on X Then f is upper , , semicontinuous if f (–1 a) is open for each a in ; fis lower semicontinuous if f (a1 ) is open for each a in

(a) Prove that a function is continuous if and only if it is both upper and lower semicontinuous.

(b) Give an example of an upper semicontinuous function that is not continuous.

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12. (a) Prove that a function from a space into is upper semicontinuous if and only if { X x X

: } is closed in for each in R.f(x) a X a

(b) State and prove the condition analogous to (a) for lower semicontinuous functions.

13. . Definition: Let A be a subset of a given space X The characteristic function of A is the function .having value 1 at each point of A and value 0 at each point of X\A

Let be a space with subspace . Prove:X A

(a) The characteristic function of is lower semicontinuous if and only if is open.A A

(b) The characteristic function of is upper semicontinuous if and only if is closed.A A

14. (a) Let be a space and { } a family of lower semicontinuous functions for which { ( ): X f A f x

} has an upper bound for each in . Prove that the function defined byA x X

is lower semicontinuous.

(b) State and prove the corresponding result for upper semicontinuous functions.

4.5 SUBSPACES

A subspace of a metric space ( , ) is simply a subset of with the metric of used to measureX d A X X distances between points of . In other words, the metric of is essentially “cut down” to . The mainA X A definition of this section defines subspaces of a general topological space in an analogous manner byX “cutting down” the open sets of .X

Definition: , . Let (X T) be a topological space and A a subset of X The relative topology or subspace

topology for A determined by consists of all sets of the form O A for which O is an open set of

The members of , are called relatively open sets or simply open sets in A and is

.called a subspace of

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FIGURE 4.2 Open sets in are of the form A , .where is open in O X

When no confusion is likely, it is common practice to refer to as a subspace of , omitting mentionA X of the topologies and .

It is a simple matter to show that what we have called the subspace topology for a subset of aA space is actually a topology for A:

(a) Ø and are relatively open sets sinceA

and both Ø and are open in .X X

(b) For any family { } of relatively open sets, where each is open in ,O A O X

is relatively open because the union of any family of open sets in is open.X

is relatively open because the intersection of any finite family of open sets in is open.X A subset of is if it is a closed set in the subspace topology for is relativelyD A relatively closed A: D

closed if and only if

for some open set in . The next theorem shows that a relatively closed set could be definedO X equivalently as the intersection with of a closed set in .A XCo

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Theorem 4.16: . Let be a subspace of a topological space A subset D of A is

= .closed in the subspace topology for A if and only if D C A for some closed subset C of X

Proof: . Suppose first that D is a relatively closed set Then

for some open set O in X, and

so D is the intersection of A with the closed set C = .X\O in X

For the reverse implication, = . = suppose that D C A for some closed set C in X Then O X\C is open

in X and

so A\D is open in the subspace topology of A, .and D is a relatively closed set

Proofs of the next two theorems are left as exercises.

Theorem 4.17: , , Let be a subspace of a space a a member of A and N a subset of

. = A Then N is a neighborhood of a with respect to the subspace topology for A if and only if N U A

.where U is a neighborhood of a with respect to the topology on X

Theorem 4.18: , , . Let (X d) be a metric space and (A d) a metric subspace Let be the topology for X , , generated by d the subspace topology for A determined by and the metric topology for A . .determined by d Then

Definition: , A property P of topological spaces is hereditary provided that if X has property P then .every subspace of X has property P

Example 4.5.1

First countability and second countability are hereditary properties. If has a countable local basis X

at point and is a subset of containing , then is a locala X A X a

base at in the subspace topology for . If is second countable, the same method of proof showsa A X that every subspace is second countable.

Example 4.5.2

Separability is not hereditary. Consider, for example, the subset of consisting of the real axisX and the one additional point = (0, 1). Define a topology on to consist of the empty set Øa X

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dense. However, the subspace topology for as a subspace of is the discrete topology, so X is not separable.

Definition: , A topological space X is a Hausdorff space if for each pair a b of distinct points of X there

.exist disjoint open sets U and V such that a U and b V

Example 4.5.3

Every metric space ( , ) is Hausdorff. To see this, note that if , are distinct points of , then = X d a b X r ( , ) is a positive number. Thus = ( , 2) and = ( , 2) are disjoint open sets containing d a b U B a r/ V B b r/ a

and , respectively. Thus , Hubert space, and discrete spaces are examples of Hausdorffb spaces.

The space of is not Hausdorff since every non-empty open set in Example 4.5.2 X contains the point (0, 1). The real line with the finite complement topology and trivial spaces with more than one point are also not Hausdorff.

Example 4.5.4 The Zariski Topology

Let be a positive integer and consider the family of all polynomials in real variables , ,n n x1 x2 ..., . For such a polynomial , let ( ) denote its solution set in :xn P Z P

It is left as an exercise for the reader to show that the set of all complements of the sets ( ), Z P , is a basis for a topology for . This topology is called the for .Zariski topology

For = 1, the Zariski topology equals the finite complement topology on . The reason is thatn the finite subsets of coincide precisely with the solution sets of polynomials in one real variable. To see this, note that if = { , . . . , } is a finite subset of , thenA a1 an

is a polynomial for which ( ) = . Furthermore, the solution set of a polynomial in one realZ P A variable is always a finite subset of .

For 1, the Zariski topology does not coincide with the finite complement topology. Then > reason is that the finite subsets of do not coincide with the solution sets of polynomials in n real variables, 1. For example, the line = 1 in is the solution set of the polynomialn > y

but this solution set is not finite. It is left as an exercise for the reader to show that with the Zariski topology is not

Hausdorff.

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Theorem 4.19:

(1) The property of being a Hausdorff space is a topological and hereditary property.

(2) A sequence .in a Hausdorff space cannot converge to more than one point

Proof:

(1) . Suppose that X is Hausdorff and Y is homeomorphic to X by homeo-morphism f: X Y For , , distinct points a and b in Y f (a) and f (b) are distinct points of X1 1 so there are disjoint open

sets U and V in X such that

Then f(U) and f(V) are disjoint open sets in Y containing a and b, .respectively

The proof that the Hausdorff property is hereditary is even easier. If A is a subspace of X and , , , a b are distinct points of A then there are disjoint open sets U and V in X containing a and b

. , respectively Then U A and V A are disjoint relatively open sets in A containing a and b

, .respectively so A is Hausdorff

(2) Suppose that . converges to two distinct limits a and b in a Hausdorff space X Then . , there are disjoint open sets U and V containing a and b But by the definition of convergence

, . there are positive integers N and N such that if n1 2 N then x1 n U and if n N2 then xn V If n

, . is greater than or equal to the larger of N and N1 2 then x belongs to the empty set Un V This

contradiction shows that cannot converge to two distinct limits in a Hausdorff .space

Note that the preceding proof for the uniqueness of the limit of a convergent sequence in a Hausdorff space is essentially the same as the proof of the corresponding property for metric spaces ( ).Theorem 3.8

The Hausdorff property is sometimes called a since it states that any two distinctseparation property points can be “separated” by disjoint open sets. Additional separation properties will be studied in

and .Chapters 6 Chapters 8

Definition: If X is a space which is homeomorphic to a subspace A of a space Y, then X is said to be . .embedded in Y The homeomorphism f: X A is called an embedding of X in Y

The isometric embeddings of in and of in Hubert space discussed in Examples and are topological embeddings.3.6.1 3.6.3

Following the outline of our earlier work on metric spaces, it would be natural to introduce the product of topological spaces, the second method of forming new spaces from old ones, at this point. This is postponed until , however, in order to study the most important topological properties,Chapter 7 connectedness and compactness, first. Any reader who cannot wait to learn about product spaces may read that chapter now and then return to .Chapter 5

EXERCISE 4.5

1. (a) Give an example to show that if is a subspace of a space , then a relatively open set in may fail to beA X A an open subset of . Prove that if is an open subset of , then every relatively open set in is openX A X A

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in .X

(b) Repeat (a) for closed sets.

2. Give an example of a subspace of a topological space and subsets and of for whichA X B C X

(a) The closure of in the subspace topology for does not equal .B A A

(b) The interior of in the subspace topology for does not equal (int ) .C A A C A

3. Let be a subspace of a space and let be a subset of . Prove that:A X B A

(a) A point in is a limit point of in the subspace topology for if and only if is a limit point of inx A B A x B the topology for .X

(b) The closure of in the subspace topology for equals .B A

4. Let : be a continuous function on the indicated spaces and a subspace of . Prove that theƒ X Y A X restriction : of is continuous.ƒ |A A Y f to A

5. Prove .Theorem 4.17

6. Prove .Theorem 4.18

7. Let be a space, a subset of and a subset of . Then has a subspace topology and Y X Z Y Y Z

can be assigned a subspace topology in two ways: has a subspace topology as a subspace of Z

, and a subspace topology as a subspace of . Prove that .

8. Give an example of a separable Hausdorff space which has a non-separable subspace.

9. Prove that a finite subset of a Hausdorff space has no limit points. Conclude that must be closed.A X A

10. Let be a Hausdorff space, a subset of , and a limit point of . Prove that every open set containing X A X x A x contains infinitely many members of .A

11. Prove: If there is an embedding of in and an embedding of in Z, then there is an embedding of in Z.X Y Y X

12. Give an example of spaces and for which can be embedded in and can be embedded in , but and A B A B B A A are not homeomorphic. ( Simple examples can be found in .)B Hint:

13. (a) Let be a space and a sequence of separable subspaces of for which isX X dense in . Prove that is separable.X X

(b) Use (a) to prove that Hilbert space is separable.H

14. (a) Show that the family of sets of is a basis for a topology for .Example 4.5.4

(b) Show that the Zariski topology for is not Hausdorff.

15. Definition: A space X is locally Euclidean of dimension n provided that each point in X belongs to an open .set homeomorphic to n-dimensional Euclidean space

This exercise shows that a locally Euclidean space may not be Hausdorff.

Let be the subset of defined byX

Let consist of all subsets of of the following types:X

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EBSCO Publishing : eBook Collection (EBSCOhost) - printed on 9/26/2021 6:19 PM via AMERICAN PUBLIC UNIVERSITY SYSTEM AN: 1565225 ; Croom, Fred H..; Principles of Topology Account: s7348467