Assignment 1 (Math418) Write a complete set of proofs for Theorem 4.5 in the Croom book.  – State the hypotheses – State the conclusions – Clearly and prec

Assignment 1 (Math418) Write a complete set of proofs for Theorem 4.5 in the Croom book. 
– State the hypotheses
– State the conclusions
– Clearly and precisely prove the conclusions from the hypotheses
– Results presented earlier in the text may be used and must be clearly documented
A few notes about format: use Microsoft Word; use Equation Editor for all mathematical symbols, e.g. x ∈ X or Cl(A) ⋂ Cl(X-A); and use the fonts Cambria and Cambria Math in size 11 so your typed work is the same font as your equations.
 

Theorem 4.5: Let Abe a subset of a topological space X. (1)bdy . (2)bdy A, int A, and int (XA) are pairwise disjoint sets whose union is X. (3)bdy A is a closed set. (4) = int A ∪ bdy A. (5)A is open if and only if bdy A ⊂ (XA). (6)A is closed if and only if bdy A ⊂ A. (7)A is open and closed if and only if bdy A = Ø. Proof: Properties (1) through (4) follow immediately from the definitions. To prove (5), note that if A is open, then A = int A by Theorem 4.3, part (2). Since int A and bdy A are disjoint by (2), then A and bdy A are disjoint, so bdy A must be a subset of XA. For the reverse implication, suppose bdy A ⊂ XA. Then no point of A is a boundary point of A, so every point of A is an interior point. Thus A = int A, so A is open. Statement (6) follows from the duality between open sets and closed sets: A is closed if and only if XA is open. By (5), this is equivalent to saying that or Statement (7) is proved by combining (5) and (6): A is both open and closed if and only if bdy A is contained in both A and XA. Since A and XA are disjoint, this occurs if and only if bdy A = Ø According to Theorem 4.5, the points of a subset A of a space X may be of two types, interior points and boundary points. The set A may have additional boundary points outside A, however; the union of all interior points and boundary points of A is . The points of X are of three non-overlapping types: (1) interior points of A, (2) interior points of XA, and (3) boundary points of A, which are identical with the boundary points of XA. (Of course, any of these three sets may be empty.) The following examples are an attempt to spare the reader some of the common misconceptions about boundaries and closures in metric spaces.

MLA 8th Edition (Modern Language Assoc.)
Croom, Fred H. Principles of Topology. Dover Publications, 2016.

APA 7th Edition (American Psychological Assoc.)
Croom, F. H. (2016). Principles of Topology. Dover Publications. 99

(b) Show that if is an infinite subset of , then every point of is a limit point of .A X X A

9. Let be a set. The for consists of Ø and all subsets of for which X countable complement topology X O X
is a countable set.X O

(a) Show that is actually a topology for .X

(b) For the space , show that a countable subset of X has derived set = Ø and that anA A
uncountable set has = B B X

(c) Show that the intersection of any countable family of members of is a member of .

10. Let = { , } be a two-element set and let . Show that is a topology andS a b
identify the limit points of each subset of . (The space is called .S S Sierpinski space )

11. How many different topologies are there for a set with three members?

4.2 INTERIOR, CLOSURE, AND BOUNDARY

The interior, closure, and boundary for subsets of a topological space are defined in complete
analogy with their counterparts for subsets of metric spaces.

Definition: . Let A be a subset of a topological space X A point x in A is an interior point of A if there is
. , . an open set O containing x and contained in A Equivalently A is called a neighborhood of x The

, , .interior of A denoted int A is the set of all interior points of A
The closure of A is the union of A with its set of limit points:

where A .is the derived set of A

A point x in X is a . boundary point of A ifx belongs to both and The set of boundary
.points of A is called the boundary of A and is denoted bdy A

Theorem 4.3: , For any subsets A B of a topological space X:

(1) The interior of A is the union of all open sets contained in A and is therefore the largest open set
contained in A.

(2) A is open if and only if A = .int A

(3) If A , .B then int A int B

(4) int ( = .) int A int B

Proof: , Statements (1) and (2) carry over from Chapter 3 and (3) is an immediate consequence of the

. , , definition of interior To prove (4) note first that since A B is a subset of both A and B then int (A B) is

. , a subset of int A int B by (3) For the reverse inclusion note that int A int B is an open set and is a

. , subset of Since int (A B) is the largest open set contained in A B then

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100

Example 4.2.1

It is not true in general that int equals int int . As a counterexample, consider the real line(A B) A B

with = [0, 1] and = [1, 2]. ThenA B

while

so int contains 1 while int int does not. The reader is asked in one of the exercises to(A B) A B

prove that the inclusion

is always valid.

Theorem 4.4: , For any subsets A B of a space X:

(1) The closure of A is the intersection of all closed sets containing A and is therefore the smallest
closed set containing A.

(2) A is closed if and only if .

(3) If A , .B then

(4) .

Proof: . , Again statements (1) and (2) carry over from ( and )Chapter 3 Theorems 3.10 3.11 For (3) note

, . that if A B then the definition of limit point guarantees that A B Then

To prove (4), . note first that is a closed set which contains A B Since is the

, smallest closed set containing A B then

For the reverse inclusion, use (3) and the fact that both A and B are subsets of A B

Example 4.2.2

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101

It is not true in general that equals . For example, let = (0, 1) and = (1,A B
2) on the real line. Then

but

The reader is left the easy exercise of showing that the inclusion

is always valid.

Theorem 4.5: .Let Abe a subset of a topological space X

(1) bdy .

(2) bdy A, , .int A and int (XA) are pairwise disjoint sets whose union is X

(3) bdy A is a closed set.

(4) = .int A bdy A

(5) A is open if and only if bdy A .(XA)

(6) A is closed if and only if bdy A .A

(7) A is open and closed if and only if bdy A = Ø.

Proof: . , Properties (1) through (4) follow immediately from the definitions To prove (5) note that if A is
, = , . , open then A int A by Theorem 4.3 part (2) Since int A and bdy A are disjoint by (2) then A and bdy A

, . , . are disjoint so bdy A must be a subset of XA For the reverse implication suppose bdy A XA Then no

, . = , .point of A is a boundary point of A so every point of A is an interior point Thus A int A so A is open
Statement (6) follows from the duality between open sets and closed sets: A is closed if and only if

XA is open. , By (5) this is equivalent to saying that

or

Statement (7) is proved by combining (5) and (6): A is both open and closed if and only if bdy A is
contained in both A and XA. , = ØSince A and XA are disjoint this occurs if and only if bdy A

According to , the points of a subset of a space may be of two types, interior pointsTheorem 4.5 A X

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102

and boundary points. The set may have additional boundary points outside , however; the union of allA A
interior points and boundary points of is . The points of are of three non-overlapping types: (1)A X
interior points of , (2) interior points of , and (3) boundary points of , which are identical with theA XA A
boundary points of . (Of course, any of these three sets may be empty.)XA

The following examples are an attempt to spare the reader some of the common misconceptions
about boundaries and closures in metric spaces.

Example 4.2.3

For an open ball ( , ) in a metric space (X, ), may be the closed ball [ , ], andB a r d not B a r

bdy ( , ) may be { ( , ) = }.B a r not x X: d x a r

(a) Consider first the case of a discrete metric space , and an open , 1) of radius 1:(X d) bail B(a

Note also that

(b) These phenomena are not restricted to discrete spaces. Let be the subspace of shadedY

in : .Figure 4.1

FIGURE 4.1

In , while [, 1] is the union of {} with the unit circle. Also, bdy (,Y B

1) = Ø and { ( , ) = 1} is the unit circle.x Y: d x

Definition: . A subset A of a space X is dense in X provided that If X has a countable denseCo
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103

, .subset then X is a separable space

It is a simple consequence of the definitions of closure and dense set that a subset of is dense in A X
if and only if every non-empty open set in contains at least one point of .X X A

Example 4.2.4

(a) The real line is separable. The set of rational numbers is countable and dense in .

(b) Euclidean -space is separable. The set of points of having only rationaln R
coordinates is dense in by . This set is countable since it is theExample 3.3.4(c)
product of the set of rational numbers (a countable set) taken as a factor times.n

(c) Hilbert space is separable. Let denote the set of all points = ( , . . ., , . . .) all ofH C x x1 xn
whose coordinates are rational and for which only finitely many coordinates arexi
non-zero. In other words,

where = { = ( , . . ., , 0, 0, . . .) is rational for = 1, . . ., and = 0 for }.Cn x x1 xn H: xi i n xi i > n

Since each set is countable, then is the union of a countable family of countable setsCn C
and is hence countable. To see that is dense in , consider a non-empty open set 0. Let (C H B

, ) be a ball with center = ( . . ., . . .) and positive radius contained in . Since a r a a1 an r O

converges, there is a positive integer such thatN

For = 1, . . . , there is a rational number between and i N xi

. Then = ( , . . . , , 0, 0, . . .) belongs to andx x1 xN C

so

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104

Thus and is separable.H

(d) The real line with the finite complement topology is separable since, by ,Example 4.1.2
every countably infinite subset is dense.

Definition: .A subset B of a space X is nowhere dense provided that int

The relations between dense sets and nowhere dense sets are explored in the exercises for this
section.

EXERCISE 4.2

1. Let and be subsets of a space . Show thatA B X

(a) int int int .A B (A B)

(b) .

(c) int (int = int .A) A

(d) .

2. Prove statements (1) through (4) of .Theorem 4.5

3. Let , ) be a metric space, a point of , and a positive number. Prove that(X d a X r

(a) .

(b) bdy ( , ) { ( , ) = r}.B a r x X: d x a

4. Identify int , bdy , int , , and the derived set in each of the following cases:A A (XA) A

(a) in ;

(b) = [0, 1], as a subset of with the finite complement topology;A

(c) = { } where = { , } with the discrete topology;A a X a b

(d) = { } where { , } with the trivial topology.A a X= a b

5. (a) If ( , ) and ( , ) are separable metric spaces, prove that the product metric space × is separable.X1 d1 X2 d2 X1 X2
(b) Use (a) to prove that is separable for each positive integer .n

6. Let be a subset of a space . Prove that is dense in if and only if int ( ) = Ø.A X A X XA

7. Let be a subset of a space . Prove that the following statements are equivalent.B X

(a) is nowhere dense.B

(b) is dense in .X

(c) .

(d) .

8. For , . Prove that ext = int Definition: a subset A of a space X the exterior of A is the set ext A (X
.A)

9. Prove:

(a) Every finite subset of is nowhere dense.

(b) The set of points of all of whose coordinates are integers is nowhere dense.

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105

(c) is nowhere dense when considered as a subset of .

10. The purpose of this problem is to show that the concept of topology for a set can be defined in terms of theX
closure operation.

Definition: . Let X be a set A closure operator on X is a function c which associates with each subset A of X a subset
c(A) of X satisfying the following properties:

(1) c(Ø = Ø,)

(2) A ,c(A)

(3) c(c(A)) = ,c(A)

(4) c(A = ,B) c(A) c(B)

for all subsets A, .B of X

A subset A of X is = , .c-closed provided that c(A) A and a subset B of X is c-open provided that XB is c-closed

Assume that is a closure operator for a given set . Prove that:c X

(a) The family of -open sets is a topology for .c X

(b) For each subset of , , where is the closure of in the topology .A X A

4.3 BASIS AND SUBBASIS

A topology for a set can be a very large and complicated family of subsets. Often it simplifiesX
matters to deal with a smaller collection which generates the topology by taking unions. Such a
subcollection is called a the precise definition follows.basis;

Definition: . Let be a topological space A base or basis for is a sub-collection of
. with the property that each member of is a union of members of Reference to the topology is

, . sometimes omitted and we speak of basis for X rather than a basis for the topology of X The members of
, .are called basic open sets and is the topology generated by

Example 4.3.1

(a) The collection of all open intervals is a basis for the usual topology of .

(b) For any metric space ( , ), the collection of all open balls ( , r), , 0, is a basisX d B a a X r >

for the topology generated by .d

(c) For any set , the collection of all singleton sets { }, , is a basis for the discreteX x x X

topology.

(d) For any space , the topology is a basis for itself. This fact is of little use
because the point of defining a basis is to produce a smaller collection of open sets with
which to work.

Definition: . Let be a space and let a be a member of X A local base or local basis at a is aC
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