math 671 : topology homework solutions

Byeongho Ban [email protected]

MATH 671 : Topology Homework Solutions

:Taught by R. Inanc BaykurLastly updated on May 10, 2019

Mathematics & Statistics

University of Massachusetts, Amherst

Byeong Ho Ban

1

MATH 671 Editor : Byeongho Ban

Due date : September 18th, 2018

2.4 Let X be a topological space and let A be a collection of subsets of X. Prove the following containment.(d)

Int

( ⋃A∈A

A

)⊇⋃A∈A

IntA

Proof. .

Suppose that

x ∈⋃A∈A

IntA =⇒ ∃A ∈ A such that x ∈ IntA

=⇒ ∃U 3 x such that U is open and U ⊂ A.

=⇒ ∃U 3 x such that U is open and U ⊂⋃A∈A

A

=⇒ x ∈ Int

( ⋃A∈A

A

).

Therefore,

Int

( ⋃A∈A

A

)⊇⋃A∈A

IntA.

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2.1 Let X be an infinite set.(a) Show that

T1 = {U ⊆ X : U = ∅ or X \ U is finite.}is a topology on X, called the finite complement topology .(c) Let p be an arbitrary point in X, and show that

T3 = {U ⊆ X : U = ∅ or p ∈ U}is a topology on X, called the particular point topology.(e) Determine whether

T5 = {U ⊆ X : U = X or X \ U is infinite}is a topology on X.

Proof. .

(a)Clearly ∅ ∈ T1. Observe that X \X = ∅ is finite, so X ∈ T1.Suppose that {Uα}α∈A ⊆ T1, where A is an arbitrary set.If Uα = ∅ for every α ∈ A, ⋃

α∈AUα = ∅ ∈ T1.

Suppose ∃β ∈ A such that Uβ 6= ∅, then observe that

X \

(⋃α∈A

Uα

)=⋂α∈A

(X \ Uα) ⊂ X \ Uβ : finite.

Thus, ⋃α∈A

Uα ∈ T1

and T1 is closed under arbitrary union.

Lastly, suppose {Un}Nn=1 ⊂ T1.If X \ Un is finite for all n = 1, 2, . . . , N , observe that

X \

(N⋂n=1

Un

)=

N⋃n=1

(X \ Un) : finite

since finite union of finite sets is finite. Thus,

N⋂n=1

Un ∈ T1.

If ∃j ∈ {1, 2, . . . , N} such that Uj = ∅, then

N⋂n=1

Un = ∅ ∈ T1.

Thus, T1 is closed under finite intersection.Therefore, T1 is a topology.

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Proof. (c)

Clearly ∅ ∈ T3 and X ∈ T3 since p ∈ X.Let {Uα}α∈A ⊆ T3 where A is an arbitrary set be given.If Uα = ∅ ∀α ∈ A, ⋃

α∈AUα = ∅ ∈ T3.

If ∃β ∈ A such that Uβ 6= ∅, since Uβ ∈ T3, we have p ∈ Uβ . Thus,

p ∈ Uβ ⊆⋃α∈A

Uα =⇒⋃α∈A

Uα ∈ T3.

Thus, T3 is closed under arbitrary union.Let {Un}Nn=1 ⊆ T3 be given.If Un 6= ∅ ∀ n = 1, 2, . . . , N , then p ∈ Un ∀n, thus

p ∈N⋂n=1

Un =⇒N⋂n=1

Un ∈ T3.

If ∃j ∈ {1, 2, . . . , N} such that Uj = ∅,

N⋂n=1

Un = ∅ ∈ T3.

Thus, T3 is closed under finite intersection.Therefore, T3 is a topology.

(e)It is not a topology.Let y ∈ X be given, and note that for each x ∈ X, {x} ∈ T5 since X \ {x} is infinite. However,⋃

x∈X\{y}

{x} = X \ {y} 6∈ T5

since X \ {y} 6= X and X \ (X \ {y}) = {y} is not infinite set. Thus, T5 is not closed under arbitrary union so T5 isnot a topology.

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2.5 For each of the following properties, give an example consisting of two subsets X,Y ⊆ R2, both considered as topologicalspaces with their Euclidean topologies, together with a map f : X → Y that has the indicated property.(a) f is open but neither closed nor continuous.(b) f is closed but neither open nor continuous.(c) f is continuous but neither open nor closed.(d) f is continuous and open but not closed.(e) f is continuous and closed but not open.(f) f is open and closed but not continuous.

Proof. .

(a)Consider fa : A = B2(0, 0) \ ∂B1(0, 0)→ B = B4(0, 0) defined by

fa(x, y) =

{(x, y) if (x, y) ∈ B1(0, 0),

(2x, 2y) if (x, y) ∈ A \B1(0, 0).

Clearly, fa is not continuous. Note that fa(B 32(0, 0)) = B1(0, 0) ∪

(B3(0, 0) \B2(0, 0)

)is not closed. Thus, fa is not closed.

Observe that, for any U ⊂ A open, f(U) = (U ∩B1(0, 0)) ∪ (2U ∩ (2A \B2(0, 0)) is open. So, fa is open.(b)Consider fb(x, y) : R2 → R2 defined by

fb(x, y) =

{(0, 0) x ≥ 0,

(1, 0) x < 0.

Clearly, fb is not continuous. Note that only possible images of fb are {(0, 0), (1, 0)}, {(0, 0)}, {(1, 0)} and ∅ which are closed.Thus, fb is closed. However, since fb(R2) = {(0, 0), (1, 0)} is not open in R2 even if R2 is open, fb is not open.(c)Consider the function fc : R2 → R× {0} defined by

fc(x, y) =

(1

1 + x2 + y2, 0

).

Clearly, fc is continuous. However, note that fc(R2) = (0, 1] × {0} is neither open nor closed even if R2 is open and closedset. Thus, fc is neither open nor closed.(d)Consider X = B1(0, 0) and Y = B2(0, 0) and let fd : X ↪→ Y be an inclusion map. Then clearly, fd is continuous open map.However, it is not closed since fd(X) = B1(0, 0) which is not closed even if X is closed and open.(e)Consider fe : R2 → R2 defined by fe(x, y) = (a, b) where (a, b) is a constant point in R2. Then, clearly fe is continuous.Also, since any possible image of fe is {(a, b)} or ∅, it is closed map. However, since fe(R2) = {(a, b)} is not open even if R2

is open, fe is not open.(f)Consider ff : R2 → {(0, 0), (1, 0)} = A defined by ff = fb. Clearly, ff is not continuous. Note that any images of ff areopen and closed in A at the sametime. Thus, ff is open and closed.

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2.6 Prove Proposition 2.30 (characterization of continuity, openness, and closedness in terms of closure and interior.)

Proposition 2.30Suppose X and Y are topological spaces, and f : X → Y is any map.(a) f is continuous if and only if f(A) ⊆ f(A) for all A ⊆ X.

(b) f is closed if and only if f(A) ⊇ f(A) for all A ⊆ X.

(c) f is continuous if and only if f−1(B) ⊆ ˚f−1(B) for all B ⊆ Y .

(d) f is open if and only if f−1(B) ⊇ ˚f−1(B) for all B ⊆ Y .

Proof. .

(a)

Suppose that f is continuous. Let A ⊆ X be given. Then, f−1(f(A)

)is closed. Also, observe that A ⊆ f−1

(f(A)

), so

A ⊆ f−1(f(A)

)since f−1

(f(A)

)is closed. Thus, observe that

f(A) ⊆ f(f−1

(f(A)

))= f(A).

Conversely, suppose that f(A) ⊆ f(A) ∀A ⊆ X. And let a closed subset C ⊆ Y be given. Now, observe that

f(f−1(C)

)⊆ f (f−1(C)) = C = C

and that

f−1(C) ⊆ f−1(C) ⊆ f−1(f(f−1(C)

))= f−1(C).

Therefore, f−1(C) is closed for any closed subset C ⊆ Y and so f is continuous.

(b)Suppose that f is closed and let A ⊆ X be given. Observe that A ⊆ A, so f(A) ⊆ f(A). Note that f is closed map so f(A)is closed. Thus,

f(A) ⊆ f(A).

Conversely, suppose that f(A) ⊇ f(A) ∀A ⊆ X and let a closed subset C ⊆ X be given. Now, note that

f(C) ⊆ f(C) = f(C) ⊆ f(C).

Thus, f(C) = f(C) and f(C) is closed. Therefore, f is closed map.

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Proof. .

(c)

Suppose that f is continuous and let a subset B ⊆ X be given. Since B ⊆ B, note that

f−1(B) ⊆ f−1(B).

Since ˚f−1(B) is a union of all open subsets of f−1(B) and f−1(B) is open,

f−1(B) ⊆ ˚f−1(B).

Conversely, suppose that f−1(B) ⊆ ˚f−1(B) ∀B ⊆ X. Let an open subset U ⊆ X be given. And observe that

f−1(U) = f−1(U) ⊆ ˚f−1(U) ⊂ f−1(U).

Therefore, f−1(U) = ˚f−1(U), so f−1(U) is open, It implies that f is continuous.

(d)

Suppose that f is open and let B ⊆ Y be given. Then note that f( ˚f−1(B)) is open and

f( ˚f−1(B)) ⊆ f(f−1(B)) = B, so f( ˚f−1(B)) ⊆ B.Since f is open, observe that

˚f−1(B) ⊆ f−1(f( ˚f−1(B))) ⊆ f−1(B).

Conversely, suppose that ˚f−1(B) ⊆ f−1(B) ∀B ⊆ Y and let an open subset U ⊆ X be given. Now observe that

U = U ⊆ ˚f−1(f(U)) ⊆ f−1( ˚f(U)).

And it implies that

˚f(U) ⊆ f(U) ⊆ f(f−1( ˚f(U))) = ˚f(U).

Thus, f(U) = ˚f(U) and it means that f(U) is open. Therefore, f is open.�

2.8 Let X be a Hausdorff space, let A ⊆ X and let A′ denote the set of limit points of A. Show that A′ is closed in X.

Proof. .

In Hausdorff space, if A is finite set, then A has no limit point. Then, since ∅ is closed, A′ = ∅ is closed.Thus, let’s assume that A is an infinite set.Let x ∈ X \A′ be given. Then ∃ a neighborhood, U , of x such that (U \ {x}) ∩A = ∅. Assume that ∃y ∈ U ∩A′, then U isa neighborhood of a limit point y, so U has infinitely many points in A. However, U ∩ A ⊆ {x}, so it is a contradiction. Itimplies that U ∩A′ = ∅ and so U ⊆ X \A′.

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2.10Suppose f, g : X → Y are continuous maps and Y is Hausdorff. Show that the set {x ∈ X : f(x) = g(x)} is closed in X.Give a counterexample if Y is not Hausdorff.

Proof. .

Let A = {x ∈ X : f(x) = g(x)} and let x ∈ X \ A. Then f(x) 6= g(x) so there are two disjoint open sets U 3 f(x) andW 3 g(x). Since f and g are continuous, note that f−1(U) and g−1(W ) are open. Now, let Q = f−1(U) ∩ g−1(W ). Sincex ∈ Q, Q is nonempty open subset. Let y ∈ Q be given. Then observe that f(y) ∈ U and g(y) ∈ W . Since U and W aredisjoint, f(y) 6= g(y), so y ∈ X \A. Thus, Q is an open subset contained in X \A which mean that X \A is open.

Consider X = Y = R and X and Y with trivial topology. Let f be an identity map and g be a trivial map with x 7→ 0 forall x ∈ X. Then note that A = {0}. Note that A is not closed in X since only closed set is X and ∅. Thus, if Y is notHausdorff, it is not true.

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Additional : Write a homeomorphism between the unit closed disk A = {(x, y ∈ R2 : x2 + y2 ≤ 1)} and the unit closedsquare B = {(x, y) ∈ R2 : max {|x|, |y|} ≤ 1}, equipped with metric topologies induced by the standard Euclidean metric onthe plane R2. Verify that your map is indeed a homeomorphism.

Proof. .

Consider ϕ : A→ B and ψ : B → A defined by

ϕ(x, y) =(x, y)

max (|x|, |y|)√x2 + y2 and ψ(x, y) =

(x, y)√x2 + y2

max (|x|, |y|).

Observe that ϕ and ψ are clearly continuous and that

ϕ ◦ ψ(x, y) = ϕ(ψ(x, y)) = ϕ

(max (|x|, |y|)√

x2 + y2x,

max (|x|, |y|)√x2 + y2

y

)

=

(max (|x|, |y|)√

x2 + y2x,

max (|x|, |y|)√x2 + y2

y

) √(max (|x|,|y|)√x2+y2

x

)2

+

(max (|x|,|y|)√

x2+y2y

)2

max

(∣∣∣∣max (|x|,|y|)√x2+y2

x

∣∣∣∣ , ∣∣∣∣max (|x|,|y|)√x2+y2

y

∣∣∣∣)=

(max (|x|, |y|)√

x2 + y2x,

max (|x|, |y|)√x2 + y2

y

) √x2 + y2

max (|x|, |y|)

= (x, y).

Similarly, observe that

ψ ◦ ϕ(x, y) = ψ(ϕ(x, y)) = ψ

( √x2 + y2

max (|x|, |y|)x,

√x2 + y2

max (|x|, |y|)y

)

=

( √x2 + y2

max (|x|, |y|)x,

√x2 + y2

max (|x|, |y|)y

) max

(∣∣∣∣ √x2+y2

max (|x|,|y|)x

∣∣∣∣ , ∣∣∣∣ √x2+y2

max (|x|,|y|)y

∣∣∣∣)√( √x2+y2

max (|x|,|y|)x

)2

+

( √x2+y2

max (|x|,|y|)y

)2

=

( √x2 + y2

max (|x|, |y|)x,

√x2 + y2

max (|x|, |y|)y

)max (|x|, |y|)√

x2 + y2

= (x, y).

Therefore, ϕ and ψ are bijection and thus homeomorphisms between A and B.�


Due date: September 25th, 2018

2-14.Prove Lemma 2.48 (the sequence lemma).

Lemma 2.48(Sequence Lemma)Suppose X is a first countable space, A is any subset of X, and x is any point of X.(a) x ∈ A if and only if x is a limit of a sequence of points in A.(b) x ∈ IntA if and only if every sequence in X converging to x is eventually in A.(c) A is closed in X if and only if A contains every limit of every convergent sequence ofpoints in A.(d) A is open in X if and only if every sequence in X converging to a point of A is eventuallyin A.

Proof. .

(a) Suppose that x ∈ A. Then let {Ui} be a nested neighborhood basis at x. Then note that Ui ∩A 6= ∅ ∀i ∈ Z+. Thus, let’spick ak ∈ Uk ∩ A for each k. Let U be a given neighborhood of x. Since {Ui} is a neighborhood basis at x, ∃N ∈ Z+ suchthat UN ⊂ U . Since xn ∈ UN ⊂ U for any n ≥ N . Since U is arbitrary, xk converges to x.

Conversely, suppose that there is a sequence {xk} ⊂ A such that xk → x. If x 6∈ A, then x ∈ Ext(A), so there exists aneighborhood, U , of x such that U ⊂ Ext(A) since Ext(A) is open. But then , for any N ∈ Z+, xn 6∈ U even if n ≥ N . Andit means xk 6→ x and it is a contradiction.

(b) Suppose that x ∈ Int(A) and let {xk} ⊂ X be a sequence such that xk → x. Since Int(A) is a neighborhood of x,∃N ∈ Z+ such that xn ∈ Int(A) ∀n ≥ N . Thus, the sequence is eventually in in Int(A), so is in A.

Conversely, suppose that every sequence in X converging to x is eventually in A and let {Uk} be a neighborhood basisat x. If x ∈ ∂A, then Uk ∩ Ac 6= ∅ for each k ∈ Z+. Let’s build a sequence such that xk ∈ Uk ∩ Ac for each k ∈ Z+.Then, xk → x but {xk}∩A = ∅. Thus, {xk} is not eventually in A even if xk → x. It is a contradiction. Therefore, x ∈ Int(A).

(c) Suppose that A is closed in X. And let x be a limit of a convergent sequence of points in A. Then by (a), x ∈ A = Asince A is closed. Thus, A contains every limit of every convergent sequence of points in A.

Conversely, suppose that A contains every limit of every convergent sequence of points in A. Let x ∈ A be given. Note that,by (a), x ∈ A. Thus, A ⊂ A, so A = A and A is closed.

(d) Suppose that A is open and let {xk} ⊂ X be a sequence converging to x ∈ A. Then, since A is a neighborhood, ∃N ∈ Z+

such that xn ∈ A for all n ≥ N which mean that {xk} is eventually in A.

Conversely, suppose that every sequence in X converging to a point of A is eventually in A. And let x ∈ A be given and leta nested neighborhood basis at x, {Uk} be given. If x ∈ ∂A, Uk ∩ Ac 6= ∅, so we can choose xk ∈ Uk ∩ Ac for each k ∈ Z+.But then, even if xk → x in X, {xk} ∩A = ∅. Since it is a contradiction, x ∈ Int(A). Thus, A ⊂ Int(A) and Int(A) = A, soA is open. �


2-18This problem refers to the topologies defined in Problem 2-1.(a) Show that R with the particular point topology is first countable and separable but not second countable or Lindelof.(b) Show that R with the excluded point topology is first countable and Lindelof but not second countable or separable.(c) Show that R with the finite complement topology is separable and Lindelof but not first countable or second countable.

Proof. .

(a) Let p ∈ R be given. Then note that

Tp = {U ⊆ R : U = ∅ or p ∈ U}is the particular point topology on R.Let x ∈ R be given. Then

B = {{p, x}}is a countable neighborhood basis at x in (R, Tp) since ∀U ∈ Tp with x ∈ U , {x, p} ⊂ U . Thus, it is first countable space.Secondly, note that the smallest closed set containing p is X. Thus, {p} is the countable dense subset of X. Thus, (R, Tp) isseparable.Note that U = {{x, p} : x ∈ R} is an open cover of R. Observe that for any countable subcollection S = {{xk, p} : k ∈ Z+} ⊂ Uthere is x ∈ R \ ({xk}k∈Z+) and so x 6∈

⋃k∈Z+{xk, p}. Thus, any countable subcollection of U cannot be open cover of R and

it means that (R, Tp) is not a Lindelof space.By the argument in Exercise 2-20, (R, Tp) is not second countable.

(b)Let q ∈ R be given. Then note that

TE,q = {U ⊆ R : p 6∈ U or U = R}is the excluded point topology.Let x ∈ R be given. Then let

Bx = {{x}} if x 6= q,

Bx = {X} if x = q.

Then, Bx is clearly a countable neighborhood basis at x since ∀U ⊂ TE,q with x ∈ U , {x} ⊂ U if x 6= q.( If x = q, the smallestopen set containing x is X.) Thus, (R, TE,q) is first countable.Suppose that U is any open cover of (R, TE,q). Then ∃U ∈ U such that q ∈ U . But, since only open set containing q is X,X ∈ U . Note that {X} ⊂ U is the countable open subcover. Thus, (R, TE,q) is Lindelof space.Suppose that {xk}k∈Z+ is any countable subset of R. Note that the smallest closed set containing {xk}k∈Z+ is {x ∈ R : x =q or x ∈ {xk}k∈Z+} 6= R(since R is uncountable). Thus, (R, TE,q) is not a separable space.Due to the argument in Exercise 2− 20, (R, TE,q) is not second countable. �


(c)Note that

T = {U ⊆ R : U = ∅ or R \ U is finite }is the finite complement topology.Let an open cover of R, {Uα}α∈A, be given. Note that ∃α0 ∈ A such that Uα0

6= ∅. Then since R \ Uα0= {x1, x2, . . . , xk},

finite, and since {Uα}α∈A is an open cover, let’s pick a αi ∈ A such that xi ∈ Uαi for each i ∈ {1, 2, . . . , k}. Then {Uαi}ki=0

is the countable open subcover of {Uα}α∈A. Thus, (R, T ) is Lindelof space.Note that only possible closed sets of this space are finite sets, emptyset or R. Therefore, the smallest closed set containinginfinitely many countable element is R. Thus, any infinitely many countable subset of R is dense in R. Thus, this space isseparable.Let x ∈ R be given. Assume that Bx = {Bn}n∈Z+ is a countable neighborhood basis at x. Then observe that⋃

n∈Z+

(R \Bn) = R \⋂n∈Z+

Bn ⊆ R \ {x}.

Suppose that y ∈ R \ {x}, then x ∈ R \ {y}, so there exists BN such that x ∈ BN ⊆ R \ {y}. It means that y 6∈ BN , soy ∈ R \BN ⊆

⋃n∈Z+(R \Bn). Thus,

⋃n∈Z+

(R \Bn) = R \⋂n∈Z+

Bn = R \ {x}.

Note that (R \Bn) is finite for each n ∈ Z+, thus the countable union of them is countable. However, R \ {x} is uncountablewhich is not equivalent to countable. Therefore, it is a contradiction. Thus, (R, T ) is not first countable.Since second countable implies first countable and (R, T ) is not first countable, it is not second countable.


2-20Show that second countability, separability, and the Lindelof property are all equivalent for metric spaces.

Proof. .

(Second Countability → Separability)Suppose that X is second countable topological space. Then there is a countable basis {Bn} of X. Let’s choose xk ∈ Bk for

each k ∈ Z+ and let A = {xn}n∈Z+ . Then assume that x ∈ Ac. Since Ac

is open, note that ∃ a nbhd U of x such that U ∩A.But then there exists k ∈ Z+ such that Bk ⊂ U so xk 6∈ U . Since it is a contradiction, A = X. And since A is countable, Xis separable.

(Second Countability → Lindelof Property)Suppose that X is a countable space and {Bk}k∈Z+ is a countable basis. Let an open cover {Uα} of X be given. Then let’spick {Uk}k∈J ⊂ {Uα} where J ⊂ Z+ such that ∀k ∈ J , Bk ⊂ Uk. Then {Uk} ⊂ {Uα} is a countable open cover of X. If not,

∃x ∈(⋃

k∈Z+ Uk)c ⊂ X so ∃k ∈ Z+ such that x ∈ Bk ⊂ Uk and it is a contradiction. Thus, every second countable space is

Lindelof space.

(Separability → Second Countable)Suppose (X, d) is a separable metric space and {xk}k∈Z+ ⊂ X is a countable dense subset. Let’s set

B = {Bd(xk, r) : k ∈ Z+ and r ∈ Q+}.In order to show that B is the basis for X, it suffice to check the basis criterion. Let U ⊂ X be an open set and let x ∈ U .Since U is open, there exists an open ball Bd(x, ε) with small enough ε > 0. Now, note that ∃k ∈ Z+ such that xk ∈ Bd(x, ε),otherwise, Bd(x, ε) ⊂ A

cwhich means that A is not dense in X. Let l = d(x, xk). Since ∃N ∈ Z+ such that ε− l > 1

N ∈ Q+,

Bd(xk,1N ) ⊂ Bd(x, ε) and Bd(xk,

1N ) ∈ B. Thus, B is a basis of the metric space. Since it is clear that B is countable, X is

second countable.

(Lindelof Property → Second Countable)Suppose that (X, d) is a Lindelof metric space. Let

Bn =

{Bd

(x,

1

n

): x ∈ X

}.

Since X is Lindelof and Bn is an open cover for each n ∈ Z+, there exists {xk,n}k∈Z+ ⊂ X such that

B′n =

{Bd

(xk,

1

n

): k ∈ Z+

}.

is a countable open cover of X.Then let

B =⋃n∈Z+

B′n.

Now, let’s verify the basis criterion. Let U ⊂ X be an open set and let x ∈ U . Then ∃ε > 0 such that Bd(x, ε) ⊂ U . Notethat ∃N ∈ Z+ such that ε > 1

N and, since B′4N is an open cover of X, ∃Bd(xk,

14N

)∈ B′4N ⊂ B such that x ∈ Bd

(xk,

14N

).

Then, Bd(xk,

14N

)⊂ Bd(x, ε) ⊂ U . Thus, B is a countable basis for (X, d). (B is a countable union of countable collections

which means that B is countable.) �


3.12 (c),(d),(f)Suppose S is a subspace of the topological space X.(c) If (pi) is a sequence of points in S and p ∈ S, then pi → p in S if and only if pi → p in X.(d) Every subspace of a Hausdorff space is Hausdorff.(f) Every subspace of a second countable space is second countable.

Proof. .

(c)Suppose that (pi) is a sequence of points in S and p ∈ S.

Suppose that pi → p in S. Let U be a neighborhood of p open in X. Then U ∩ S is a neighborhood of p open in S. Thus,∃N ∈ Z+ such that pn ∈ S ∩ U ∀n ≥ N . And it means that pn ∈ U ∀n ≥ N . Thus, pi → p in X.

Conversely, suppose that pi → p in X. Let US be a neighborhood of p open in S. Note that, then ∃ a neighborhood U openin X such that US = U ∩ S. Then ∃N ∈ Z+ such that pn ∈ U ∀n ≥ N . Since (pi) is a sequence in S, pn ∈ U ∩ S = US∀n ≥ N , so pi → p in S.

(b)Let S ⊆ X be a subspace of Hausdorff space X. Let x, y ∈ S be given with x 6= y. Then, there are disjoint open setsU, V ⊆ X such that x ∈ U and y ∈ V . But then S ∩ U and S ∩ V are disjoint open sets in S such that x ∈ S ∩ U andy ∈ S ∩ V . Thus, S is Hausdorff space.

(f)Let S ⊆ X be a subspace of second countable space X. Then, there is a countable basis B = {Bn}n∈Z+ of X. Let

BS = {S ∩Bn : Bn ∈ B} .Then BS is a basis of S because of following argument.Let x ∈ S, and let US be a neighborhood of x open in S. Then there exists an open set, U , in X such that x ∈ US = U ∩ S.Since x ∈ U , there exists n ∈ Z+ such that x ∈ Bn ⊆ U . Thus, x ∈ U ∩ Bn ⊆ US . Therefore, B is basis. Since B is clearlycountable, S has countable basis, so is second countable space.

�


3-1Suppose M is an n−dimensional manifold with boundary. Show that ∂M is an (n− 1)−manifold (without boundary) whenendowed with the subspace topology. You may use without proof the fact that Int(M) and ∂M are disjoint.

Proof. .

Note that M is n−manifold, so is Hausdorff and second countable. Since subspace topology preserves the properties, ∂M issecond countable and Hausdorff space. Thus, we only need to prove that ∂M is locally Euclidean of dimension n− 1.Let x ∈ ∂M be given. Then there is a boundary chart (U,ϕ) such that

∗ x ∈ U∗ ϕ : U → V ⊂ Hn

∗ ϕ(x) ∈ V ∩ ∂Hn.

Let ϕr : ϕ−1(V ∩ ∂Hn)→ V ∩ ∂Hn be the restriction of ϕ to ϕ−1(V ∩ ∂Hn). Note that

ϕ−1(V ∩ ∂Hn) = ϕ−1(V ) ∩ ϕ−1(∂Hn) = U ∩ ∂Mand that U ∩ ∂M is open with the subspace topology. Also, note that ϕr is a homeomorphism again. Thus, ∂M is locallyhomeomorphic to ∂H. Since ∂H is n− 1 dimensional Euclidean space, ∂M is locally n− 1 Euclidean space thus, it is (n− 1)manifold.

�


3-2Suppose X is a topological space and A ⊆ B ⊆ X. Show that A is dense in X if and only if A is dense in B and B is densein X.

Proof. .

Suppose that A is dense in X. Then A = X. Note that the closure of A in B is B ∩A since any closed sets in B containingA is B ∩ C for some closed set C in X and A ⊆ C so B ∩A ⊆ B ∩ C.Note that A ∩B = X ∩B = B. Thus, A is dense in B. Also, note that X = A ⊆ B. Thus, B = X so B is dense in X.

Conversely, suppose that A is dense in B and B is dense in X. Then A ∩ B = B and B = X. Thus, B ⊆ A and soX = B ⊆ A. Therefore, A = X and A is dense in X. �


3-4Show that every closed ball in Rn is an n−dimensional manifold with boundary, as is the complement of every open ball.Assuming the theorem on the invariance of the boundary, show that the manifold boundary of each is equal to its topologicalboundary as a subset of Rn, namely a sphere. [Hint : for the unit ball in Rn, consider the map π ◦ σ−1 : Rn → Rn, where σis the stereographic projection and π is a projection from Rn+1 to Rn that omit some coordinate other than the last.]

Proof. .

Note that every closed ball is homeomorphic to a closed ball centered at origin and having radius 1 by translation x 7→ x+ pfor p ∈ Rn and dilation x→ rx for r ∈ R. Thus, it suffices to prove thatB(0, 1) ⊆ Rn is n dimensional manifold with boundary.

Firstly, observe that B(0, 1) is Second countable Hausdorff space with the subspace topology of Rn. And further note that ev-ery neighborhood of every points x ∈ B(0, 1) contained in B(0, 1) is homeomorphic to a neighborhood of Rn by inclusion map.

Therefore, it suffices to prove that every point in ∂B(0, 1) has a neighborhood homeomorphic either to an open subset of Rnor to an open subset of Hn.

Note that stereographic projection σ : ∂B(0, 1) → Rn−1 is a homeomorphism from class. And note that ψ : Rn−1 → ∂Hndefined by ψ(x1, x,2 , . . . , xn−1) = (x1, x2, . . . , xn−1, 0) is homeomorphism. Therefore, ψ ◦ σ : ∂B(0, 1) → ∂Hn is a homeo-morphism. Thus, every neighborhood U of a given point x ∈ ∂B(0, 1) is homeomorphic to an open subset ψ ◦ σ(U) ⊆ ∂Hn.

Note that we can find two open sets V and W such that U = ∂B(0, 1) ∩ V and ψ ◦ σ(U) = ψ ◦ σ(∂B(0, 1) ∩ V ) = W ∩ ∂Hnwhere W and V are homeomorhpic by some homeomorphism. Then, V ∩B(0, 1) is a neighborhood of x ∈ ∂B(0, 1) and it ishomeomorphic to W ∩Hn by the homeomorphism.

Lastly, note that every points x ∈ ∂B(0, 1)(topological boundary) is in boundary chart. Since every points in boundary ofmanifold is in topological manifold, they are same.

�


Due date: October 4th, 2018

3.26 Show that the product topology on Rn = R × · · · × R is the same as the metric topology induced by the Euclideandistance function.

Proof. .

Let

B = {U1 × · · · × Un : Ui is an open interval in R ∀i}.Note that B is a basis for Rn since any open set in R is countable union of open intervals.Let U = U1 × · · · × Un ∈ B and let x = (x1, . . . , xn) ∈ U . And let

εi = min

(|xi − inf

y∈Uiy|, |x− sup

y∈Uiy|)

Further let ε = min1≤i≤n εi.Then, note that the open ball B(x, ε) is a subset of U . Thus, any U ∈ B is an open set in the metric topology induced byEuclidean distance function.

Conversely, let an open (with respect to Euclidean topology) set U ⊂ Rn be given. Suppose that x ∈ U , then ∃r > 0 suchthat B(x, r) ⊂ U . Then observe that(

x1 −r

2√n, x1 +

r

2√n

)× · · · ×

(xn −

r

2√n, xn +

r

2√n

)⊂ B(x, r)

since√∑n

i=1r2

4n < r.

Therefore, any open set in Euclidean is open in the product topology.

Thus, the two topologies are same.�


3.32 Let X1, . . . , Xn be given topologies.(b) For each i ∈ {1, . . . , n} and any point xj ∈ Xj , j 6= i, the map f : Xi → X1 × · · · ×Xn given by

f(x) = (x1, . . . , xi−1, x, xi+1, . . . , xn)

is a topological embedding of Xi into the product space.(e) If Si is a subspace of Xi ∀i ∈ {1, . . . , n}, then the product topology and the subspace topology on S1 × · · · × Sn ⊆X1 × · · · ×Xn are equal.(f) If each Xi is Hausdorff, so is X1 × · · · ×Xn.(h) If each Xi is second countable, so is X1 × · · · ×Xn.

Proof. .

(b)Note that if f(x) = f(y), then (x1, . . . , xi−1, x, xi+1, . . . , xn) = (x1, . . . , xi−1, y, xi+1, . . . , xn), so x = y. Thus, f is injection.Let Uj ⊆ Xj is open set for any j 6= i. If xj ∈ Uj for all j 6= i, f−1(U) = Ui and it is open. If ∃k 6= i such that xk 6∈ Uk, thenf−1(U) = ∅ and it is also open. Thus, f is continuous.Note that f(Xi) = {x1} × · · · {xi−1} × Xi × {xi+1} × · · · × {xn} and that f(Ui) is open in subspace topology of f(Xi) inX1 × · · · ×Xn. Therefore, f is a homeomorphism between {x1} × · · · {xi−1} ×Xi × {xi+1} × · · · × {xn} and Xi so f is anembedding.

(c)Let U = U1 × · · · × Un with Ui ⊆ Si is open relatively to Si for any i. Then for any i, there is a open set Vi ⊆ Xi such thatSi ∩ Vi = Ui. Then Note that

(U1 × · · · × Un) = (S1 × · · · × Sn) ∩ (V1 × · · · × Vn).

Thus, U1 × · · · × Un is open in the subspace topology.Conversely, suppose that U ⊂ S1 × · · · × Sn is open in the subspace topology. Then there is an open set V ⊆ X1 × · · · ×Xn

such that V ∩ (S1 × · · · ×Sn) = U . Since V is open in the product topology, for each i, there is sequence of open sets {W ji }j

such that

V =⋃j

W j1 × · · · ×

⋃j

W jn.

It implies that

U =

⋃j

W j1 ∩ S1

× · · · ×⋃

j

W jn ∩ Sn

.

Since each⋃jW

ji ∩ Si is open in Si, U is open in the product topology on S1 × · · · × Sn.

Therefore, the two topologies are same.

(f)Let x, y ∈ X1 × · · · ×Xn with x = (x1, . . . , xn) and y = (y1, . . . , yn) with x 6= y be given. Since x 6= y, ∃i ∈ {1, . . . , n} suchthat xi 6= yi. Since Xi is Hausdorff, there are two disjoint open sets Ui and Vi with xi ∈ Ui and yi ∈ Vi. Then

(X1 × · · · ×Xi−1 × Ui ×Xi+1 × · · · ×Xn) ∩ (X1 × · · · ×Xi−1 × Ui ×Xi+1 × · · · ×Xn) = ∅and the two sets are open with containing x and y exclusively. Therefore, X1 × · · · ×Xn is Hausdorff.

(h)Since Xi is Hausdorff for each i, there exists countable basis Bi for each Xi. Then let

B = {B1 × · · · ×Bn : Bi ∈ Bi ∀i ∈ {1, . . . , n}}.Note that B is a basis for X1 × · · · ×Xn due to following reasoning.Let U = U1 × · · · × Un with Ui is open in Xi for each i and x = (x1, . . . , xn) ∈ U be given. Since ∀i ∃Bi ∈ Bi such thatxi ∈ Bi ⊆ Ui, x ∈ B1 × · · · × Bn ⊆ U1 × · · · × Un. Thus, B is a basis for X1 × · · · × Xn ans it is countable. Therefore,X1 × · · · ×Xn is second countable. �


3.45 Let X be any space and Y be a discrete space. Show that the Cartesian product X × Y is equal to the disjoint union∐y∈Y X, and the product topology is the same as disjoint union topology.

Proof. .

Observe that, by definition, ∐y∈Y

X = {(x, y) : x ∈ X and y ∈ Y } = X × Y.

So they are equal.(It suffices to use basis for product topology.) Suppose that U is open in X and V is open in Y . Then for each y ∈ Y ,(X × {y}) ∩ (U × V ) = U × {y} and U is open in X. Thus, U × V =

∐y∈V U is open in the disjoint union topology. Thus,

every elements in basis for product topology are open in disjoint union topology.

Conversely, Let a subset U × V ⊆∐y∈Y X open with respect to disjoint union topology be given. By definition, U is open

in X. Since any set in discrete topology is open, V is also open. Thus, by the definition of product topology, U × V is anelement of basis, so is open.Therefore, the two topologies are same.

�

3-6Let X be a topological space. The diagonal of X × X is the subset ∆ = {(x, x) : x ∈ X} ⊆ X × X. Show that X isHausdorff if and only if ∆ is closed in X ×X.

Proof. .

Suppose that X is Hausdorff. Let a = (a1, a2) ∈ (X ×X) \∆ be given. Since a1 6= a2, there exist two disjoint open subsetsUa, Va ⊆ X such that a1 ∈ Ua and a2 ∈ Va. Then observe that (Ua × Va) ∩∆ = ∅ since U ∩ V = ∅. Now, let Wa = Ua × Vaand Y = (X ×X) \∆. Note that each Wa is open. And let

W =⋃a∈Y

Wa.

Then note that W ∩∆ = ∅ and W is open. Thus, ∆ is closed.

Conversely, suppose that ∆ is closed in X × X. Let x, y ∈ X with x 6= y be given. Then note that (x, y) 6∈ ∆. Since(X ×X) \∆ is open, there exists an open set U ∈ (X ×X) \∆ such that (x, y) ∈ U . By the definition of product topology,there exist two open sets B1, B2 ⊆ X such that B1 × B2 ⊆ U with (x, y) ∈ B1 × B2. Thus, x ∈ B1 and y ∈ B2. SinceB1 ×B2 ⊆ (X ×X) \∆, we also have B1 ∩B2 = ∅. Therefore, B1 and B2 are the tow disjoint open sets containing x and yexclusively, so X is Hausdorff. �

3-8Let X denote the Cartesian product of countably infinitely many copies of R (which is just the set of all infinite sequencesof real numbers), endowed with the box topology. Define a map f : R → X by f(x) = (x, x, x . . . ). Show that f is notcontinuous, even though each of its component functions is.

Proof. .

Observe that each component function fi(x) = x ∀i ∈ N is continuous since they are identity function. However, note thateven if

U =∏n∈N

(x− 1

n, x+

1

n

)is an open set in X, we have

f−1(U) =⋂n∈N

(x− 1

n, x+

1

n

)= {x}

is not open set.Thus, f is not continuous even if each component function is continuous which means that the characteristic property doesnot work here.

�


3-10Prove Theorem 3.41 (the characteristic property of disjoint union spaces).

Proposition 3.41(Characteristic Property of Disjoint Union Spaces)Suppose that (Xα)α∈A is an indexed family of topological spaces, and Y is any topological space. A map f :

∐α∈AXα → Y

is continuous if and only if its restriction to each Xα is continuous. The disjoint union topology is the unique topology on∐α∈AXα with this property.

Proof. .

Suppose that f :∐α∈AXα → Y is continuous. And let an open subset U ⊆ Y be given. If f |Xα is the function f restricted

to Xα, observe that

(f |Xα)−1(U) = f−1(U) ∩Xα ∀α ∈ A.

Since f−1(U) is open in∐α∈AXα, we have f−1(U) ∩ Xα is open for any α ∈ A by definition of disjoint union topology.

Thus, (fXα)−1(U) is open so f |Xα is continuous ∀α ∈ A.

Conversely, suppose that f |Xα is continuous ∀α ∈ A. Then observe that, for any open set U ⊆ Y ,

(f |Xα)−1(U) = f−1(U) ∩Xα is open ∀α ∈ A.

Thus, by definition of disjoint union topology, f−1(U) is open so f is continuous.

Suppose that Tg be any given topology on∐α∈AXα with the property. Consider the identity function

Id1 : (∐α∈AXα, Tg) →

(∐α∈AXα, TD

). Let U ∈ TD be given, Then U ∩ Xα is open in Xα ∀α ∈ A. Observe that

(Id1|Xα)−1(U) = Xα ∩ U is open in Xα so in ∀α ∈ A . Thus, Id1|Xα is continuous ∀α ∈ A. Therefore, by the property, Id1is continuous.

Conversely, consider the identity function Id1 : (∐α∈AXα, TD) →

(∐α∈AXα, Tg

). let V ∈ Tg be given. Observe that

(Id2|Xα)−1(V ) = V ∩ Xα is open in Xα since Xα’s are disjoint. Thus, Id2|Xα is continuous ∀α ∈ A. Therefore, Id2 iscontinuous.In conclusion, the two topology are same.

�

Additional ProblemShow that the n−torus Tn embeds into Rn+1 for any integer n. (Hint: find a way to induct on n.)

Proof. .

Observe that T 1 = S1 embeds into R2 by inclusion map ι1 : T 1 ↪→ R2. Also, observe that S1 × R embeds into R2 because acylinder S1 × R and R2 \ (0, 0) are homeomorphic.(We can think of R2 \ {(0, , 0)} as a cylinder having a hole at origin andanother hole at ∞ )

Assume that Tn−1 embeds into Rn and Tn−1 × R embeds into Rn. Then, note that Tn−1 × R2 = (Tn−1 × R) × R embedsinto Rn × R = Rn+1. Therefore, since S1 embeds into R2, Tn = Tn−1 × S1 embeds into Tn−1 × R2 so is embeds into Rn+1.And further note that Tn×R = S1× · · · (S1×R) embeds into S1× · · ·×S1×R2 = S1× · · ·× (S1×R)×R and so it embedsinto S1 × · · · × S1 × R3. By repeating this, we get Tn × R embeds into Rn+1.

By mathematical induction principle, we can conclude that Tn embeds into Rn+1 for all n ∈ N.�


Due date: October 16th, 2018 Byeongho Ban

3.63 Prove Proposition 3.62.

Proposition 3.62(a) Any composition of quotient maps is a quotient map.(b) An injective quotient map is a homeomorphism.(c) If q : X → Y is a quotient map, a subset K ⊆ Y is closed if and only if q−1(K) is closed in X.(d) If q : X → Y is a quotient map and U ⊆ X is a saturated open or closed subset, then the restriction q|U : U → q(U) isa quotient map.(e) If {qα : Xα → Yα}α∈A is an indexed family of quotient maps, then the map q :

∐αXα →

∐α Yα whose restriction to

each Xα is equal to qα is a quotient map.

Proof. .

(a)Let q1 : X1 → X2 and q2 : X2 → X3 be quotient maps. Since q1 and q2 are surjective, q2 ◦ q1 is also surjective. Also, observethat

U is open in X3 ⇐⇒ q−12 (U) is open in X2

⇐⇒ q−11 (q−12 (U)) = (q2 ◦ q2)−1(U) is open in X1.

Therefore, q2 ◦ q1 : X1 → X3 is a quotient map.(b)Suppose that q : X → Y is an injective quotient map. By definition of quotient map, q is surjective, so is bijective continuousmap. Now, suppose that U ⊆ X is open. We need to show that q(U) is open in order to prove that q−1 is continuous. Notethat q(U) is open if and only if q−1(q(U)) is open in X. Since q is bijective, note that q−1(q(U)) = U . Therefore, q(U) isopen therefore q is a homeomorphism.

(c)

Suppose that K ⊆ Y is closed, then Kc is open. Then, by definition of quotient space, q−1(Kc) =(q−1(K)

)cshould be open

in X. Therefore, q−1(K) is closed in X.Conversely, suppose that q−1(K) is closed in X. then q−1(Kc) is open in X. Thus, by the definition of quotient space, Kc

is open in Y . Thus, K is closed in Y .

(d)Note that restriction map of continuous map is continuous and q|U : U → q(U) is surjective. Firstly, suppose that U ⊆ Xis a saturated open subset then U = q−1(W ) for some subset W and W is open since q is quotient map . We only need toprove that q sends any saturated open subset of U to open subset of q(U). Suppose that V ⊆ U be a saturated open subset.Then V = q|−1U (O) = q−1(O) ∩ q−1(W ) = q−1(W ∩ O). Note that V is open in X since U is open. Thus, W ∩ O is open inY . Note that

q|U(V ) = q(V ) ∩ q(U) = q(q−1(W ∩O)) ∩ q(U) = W ∩O ∩ q(U).

Since W ∩O ∩ q(U) is open in q(U), q|U sends saturated open set to open set so q|U is quotient map.

(e)Let y ∈

∐α Yα be given. Then y ∈ Yα for some β. Since qα : Xβ → Yβ is surjective, there is x ∈ Xβ ⊆

∐αXα such that

q(x) = qβ(x) = y. Thus, q is surjective.Let an open set U ⊆

∐α Yα be given. Then U ∩ Yα is open for any α. Observe that

q−1(U) ∩Xα = q|−1Xα(U) ∀α.

Since restriction of q to Xα is quotient map, the above inverse image is open. Therefore, q−1(U) is open in∐αXα. �


3-14 Show that real projective space Pn is an n-manifold.[Hint: consider the subsets Ui ⊆ Rn+1 where xi = 1.]

Proof. .

Second countableLet Ui = {l(x1, x2, . . . , xn) : xi = 1} and ϕi : Ui → Rn such that

ϕi(l(x1, x2, . . . , x3)) =

(x1xi,x2xi, . . . ,

xi−1xi

,xi+1

xi, . . . ,

xnxi

)= (x1, . . . , xi−1, xi+1, . . . , xn).

Observe that ϕ is a composition of quotient map and projection, so is clearly continuous. And let V be a countable basis ofRn. Then

W = {ϕ−1(V ) : V ∈ V}is a countable basis of Pn so it is second countable.

HausdorffLet l1, l2 ∈ Pn with l1 6= l2 be given. Let l1 ∈ Ui and l2 ∈ Uj .If i = j, note that ϕi(l1) 6= ϕi(l2), so there exist two disjoint open sets V1 and V2 such that ϕi(li) ∈ V1 and ϕi(l2) ∈ V2. SinceV1 and V2 are disjoint, ϕ−1i (V1) and ϕ−1i (V2) are two disjoint open sets containing l1 and l2 respectively.If i 6= j, without loss of generality, suppose that l1 ∈ U1 \ U2 and l2 ∈ U2 \ U1. Note that x2 coordinate of l1 is zero and x1coordinate of l2 is zero. Then let ϕ1(l1) = (0, u12, . . . , u1n) and ϕ2(l2) = (0, u22, . . . , u2n). Let

V1 =

{(x1, . . . , xn) + ϕ1(l1) :

n∑i=1

x2i <1

2

}

V2 =

{(x1, . . . , xn) + ϕ1(l2) :

n∑i=1

x2i <1

2

}Note that ϕ1(l1) ∈ V1 and ϕ2(l2) ∈ V2. Let W1 = ϕ−11 (V1) and W2 = ϕ−12 (V2). Then W1 ∩W2 = ∅ by following reason. Ifl ∈W1 ∩W2, then

l = l(1, x1, x2 + u12, . . . , xn + u1n) = l(y1, 1, y2 + u22, . . . , yn + u2n).

Note that x1 = 1y1

so x1y1 = 1. Then |x1| ≤ 1 or |y1| ≤ 1. Without loss of generality, let |x1| ≥ 1 and observe that

1 = x21 ≤n∑i=1

x2i <1

2

and it is a contradiction. Therefore, W1 ∩W2 = ∅.

Locally EuclideanObserve that ϕi(Ui) = Rn for i = 1, 2, . . . , n + 1. Let p ∈ Pn be given. Then p ∈ Uj for some j. Note that ϕj is a bijectionand it is homeomorphism. Thus, Ui ∼= Rn. Therefore, Pn is locally Euclidean.

In conclusion, Pn is n-manifold.�


3-16Let X be the subset (R × {0}) ∪ (R × {0}) ⊆ R2. Define an equivalence relation on X by declaring (x, 0) ∼ (x, 1) if x 6= 0.Show that the quotient space X/ ∼ is locally Euclidean and second countable, but not Hausdorff. (This space is called theline with two origin)

Proof. .

Let q : (R×{0})∪ (R×{0}) be the quotient map and M = R×{0}∪R×{1} and let p ∈ X/ ∼ be given. If q−1({p}) = (0, 1)or (0, 0), without loss of generality let q−1({p}) = (0, 0). Let ϕ : R × {0} → (X/ ∼) \ {(0, 1)}. Note that ϕ is a bijection.Since ϕ is a restriction of q, and it is a bijection. Thus, by 3.62, ϕ is a homeomorphism. Thus

X/ ∼ \{(0, 1)} ∼= R× {0} ∼= R.

If q−1({p}) = {(x, 0), (x, 1)} with x 6= 0, we can use ϕ defined above as the homeomorphism since p ∈ (X/ ∼) \ {(0, 1)}.Therefore, X/ ∼ is locally Euclidean.And note that M has countable basis, namely, V = {U × {i} : U ∈ B, i = 1, 2} where B is a countable basis of R. Then

W = {q(V ) : V ∈ V}is a countable collections of open sets since q|R×{i} is injective for i = 1, 2. Let W ⊂ X/ ∼ be an open neighborhood of a

given point p. Since q is a quotient map, q−1(W ) is open so it is a union of the elements in a subcollection V0 of V such that

q−1(W ) =⋃N∈V

N.

Let’s choose one element A ∈ V0. Then q(A) ∈ W then q(A) ∈ W is open contained in W . Therefore, W is a countablebasis of X/ ∼ thus X/ ∼ is second countable.

Observe that (0, 0) 6= (0, 1). Suppose that (0, 0) ∈ q(A) and (0, 1) ∈ q(B) for some A,B ∈ W. Then A = N1 × {0} andB = N2 × {1} for some N1, N2 ∈ B. Note that since N1 and N2 are open containing 0, (N1 ∩N2) \ {0} 6= ∅. It implies thatq(A)∩ q(B) 6= ∅. Thus, any two open sets containing (0, 0) and (0, 1) respectively shares common element, so they are neverdisjoint. It means that X/ ∼ is never Hausdorff. �


3-18Let A ⊆ R be the set of integers, and let X be the quotient space R/A obtained by collapsing A to a point as in Example3.52. (We are not using the notation R/Z for this space because that has different meaning, described in Example 3.92. )(a) Show that X is homeomorphic to a wedge sum of countably infinitely many circles. [Hint : express both spaces asquotients of a disjoint union of intervals.](b) Show that the equivalence class A does not have a countable neighborhood basis in X, so X is not first or secondcountable.


3-19Let G be a topological group and let H ⊆ G be a subgroup. Show that its closure H is also a subgroup.

Proof. .

Let m : G × G → G and i : G → G be the product function and inverse function of the group respectively. Since H is asubgroup, note that m(H ×H) ⊆ H and i(H) ⊆ H. Observe that

m(H ×H) = m(H ×H) ⊆ m(H ×H) ⊂ H ∵ m(H ×H) ⊆ H,

i(H) ⊆ i(H) ⊆ H ∵ i(H) ⊆ H.

Thus, H is closed under product and inverse. Since the associativity is endowed from G and 1 ∈ H ⊆ H, H is a subgroup.�

3-22Let G be a group acting by homeomorphisms on a topological space X, and let O ⊆ X ×X be the subset defined by

O = {(x1, x2) : x1 = g · x2 for some g ∈ G}.It is called the orbit relation because (x1, x2) ∈ O if and only if x1 and x2 are in the same orbit.(a) Show that the quotient map X → X/G is an open map.(b) Conclude that X/G is Hausdorff if and only if O is closed in X ×X.

Proof. .

(a)Let U ⊆ X be a given open set. And let q : X → X/G be the quotient map. Note that

q(U) = {[Ox] : x ∈ Uand Ox is the orbit of x.}.

So it suffices to show that q−1(q(U)) is open. Note that

q−1(q(U)) =⋃x∈UOx.

Since each Ox is open, q(U) is open. Thus, q is an open map.(b)Suppose that q : X → X/G is the quotient map. Suppose that X/G is a Hausdorff space. Let (y1, y2) ∈ (X × X) \ O begiven. Since X/G is Hausdorff space, there are disjoint open sets Vy and Uy such that Vy contains class of G-orbit of y1,[Oy1 ] , and Uy contains the class of G-orbit of y2, [Oy2 ]. Note that q−1(Vy) and q−1(Uy) are open since q is a quotient map.Ant further note that (q−1(Vy)× q−1(Uy)) ∩O = ∅ and Wy = q−1(Vy)× q−1(Uy) is an open set in X ×X. Therefore, Wy isa open neighborhood of (y1, y2). Since (y1, y2) is arbitrary,

X \ O = W =⋃

(y1,y2)∈X\O

Wy.

since each Wy is open, X ×X \ O is open, so O is closed in X ×X.

Conversely, suppose that O is closed in X ×X and let the orbit of x, Ox , and the orbit of y, Oy, in X/G be given. Notethat Ox ×Oy ⊆ (X ×X) \O so there exists an open set Ox ×Oy ⊆ Vxy such that Vxy ∩O = ∅. If pri is a projection map toith coordinate for i = 1, 2, [Ox] ∈ pr1(Vxy) ⊆ X/G and [Oy] ∈ pr2(Vxy) ⊆ X/G. Note that pr1(Vxy) and pr2(Vxy) are opencontaining [Ox] and [Oy]. If [Oα] ∈ pr1(Vxy)∩pr2(Vxy), then ∃(a, b) ∈ Oα such that a = g · b for some g ∈ G and (a, b) ∈ Vxy.And it is a contradiction. Therefore, pr1(Vxy) and pr2(Vxy) are disjoint open sets. Thus, X/G is Hausdorff.

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Due date : October 25th, 2018 Byeongho Ban

4.4 Prove that a topological space X is disconnected if and only if there exists a nonconstant continuous function from X tothe discrete space {0, 1}.

Proof. .

Suppose that X is disconnected. Then there are two nonempty open sets U, V ⊆ X such that X = U ∪ V . Let’s define

f(x) =

{0 , x ∈ U1 , x ∈ V.

Then note that {{1}, {0}} is a basis of {0, 1} and that f−1({0}) = U and f−1({1}) = V are open. Therefore, f is continuousfunction from X to {0, 1}.

Conversely, suppose that f : X → {0, 1} is a non-constant continuous function. Then note that f−1({1}) and f−1({0}) aredisjoint open sets. And they are non empty since f is non constant. Since X = f−1({0, 1}) = f−1({0}) ∪ f−1({1}), X isdisconnected.

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4.22Prove Proposition 4.21.Proposition 4.21(a) The path components of X form a partition of X.(b) Each path component is contatined in a single component, and each component is a disjoint union of path components .(c) Any nonempty path-connected subset of X is contained in a single path component.

Proof. .

(a) Suppose that Bα and Bβ are two different path components of X. If Bα ∩Bβ 6= ∅, then Bα ∪Bβ is also path-connectedwhich contradicts to the fact that Bα and Bβ are path components. Thus, Bα ∩Bβ = ∅.Now we show that X is a union of the path components. Let x ∈ X be given. Let A = {P ⊆ X : P is path-connected and x ∈P}. Note that {x} is path connected so A is non empty collection. Let

B =⋃U∈A

U.

Then B is path connected. Also it is a path component, otherwise there is another path connected set, V , properlycontaining B and it is contradiction since V ∈ A so V ⊆ B. Thus, B is a path component and contains x.

(b) Assume that P is a given path component. Then P is path connected and so connected. Since each connected set iscontained in a single component, say E, thus, P ⊆ E.Moreover, suppose that E is a given connected component. Since each path connected components are mutually disjoint,and each point in E is contained in at least one path component of X by (a), it suffice to prove that each path componentis contained in a single component. Since we have prove it already, E is a disjoint union of path components.

(c) Suppose that P ⊆ X is a nonempty path connected subset. Then ∃x ∈ P . Let

Px = {B ⊆ X : x ∈ B and B is path connected.}Since {x} ∈ P, P 6= ∅. Let

Px =⋃B∈P

B.

Then as in the proof of (a), Px is a path component. Since P ∈ P, P ⊆ Px. Thus, we are done. �


4.24 Prove Proposition 4.23.Proposition 4.23Every manifold (with or without boundary) is locally connected and locally path-connected.

Proof. .

Let Mn be a manifold and let p ∈ Mn be given. of p ∈ int(Mn), let U ⊆ Mn be an open neighborhood of p. Then thereexists a chart (V, ϕ) such that p ∈ V . Note that U ∩ V is homeomorphic to an open subset W ⊆ Rn and ϕ|U∩V is thehomeomorphism. Then there exists ε > 0 such that B(ϕ(p), ε) ⊆ W . Then ϕ−1(B(ϕ(p), ε)) ⊆ U is an open neighborhoodwhich is path connected, so is connected.

On the other hand, if p ∈ ∂Mn and A ⊆ Mn is a neighborhood of p, then let (T, φ) be a chart such that p ∈ T . Thenφ(T ∩A) ∩ ∂Hn 6= ∅.(Hn ⊆ R is a set of all points of the form (x1, . . . , xn) with xn ≥ 0. ) Then there exists ε > 0 such thatB(p, ε) ∩Hn ⊆ φ(T ∩A) and it is path connected so is connected. Therefore, φ−1(B(p, ε) ∩Hn) ⊆ A is path connected so isconnected.

Therefore, Mn is locally path connected and locally connected regardless of having boundary.�

4-3Invariance of the boundary, 1-Dimensional case: Suppose M is a 1-dimensional manifold with boundary. Show

that a point of M cannot be both a boundary point and an interior point.

Proof. .

Assume that p ∈M is the point that is both a boundary and an interior point. Then there are a boundary chart (U,ϕ) andan interior chart (V, ψ). Then note that p ∈ U ∩ V . Now, we can pick a component W ⊆ U ∩ V that contains p . Since W isconnected and p ∈W , ϕ(W ) = [0, c) and ψ(W ) = (a, b) for some a, b, c ∈ (0,∞). Thus, we have

(0, c) ∼=ϕ W \ {p} ∼=ψ (a, ψ(p)) ∪ (ψ(p), b).

And it is a contradiction since left hand side has one component and right hand side has two component which means thatthey are not homeomorphic. Therefore, any point in 1 dimensional manifold M cannot be both a boundary point and aninterior point. �

4-4Show that the following topological spaces are not manifold.(a) The union of the x−axis and the y−axis in R2.

Proof. .

Let the union of x-axis and y-axis be M . And assume that M is a manifold. Note that (5, 0) ∈ M has a neighborhood(4, 6) × {0} ⊆ M and it is a homeomorphic to (4, 6) ⊆ R with the homeomorphism ϕ : M → R with ϕ(x, 0) = x. Thus, Mshould be 1- manifold.

Then there exists a chart (V, φ) such that (0, 0) ∈ V ⊆M and φ is a homeomorphism between V and an open subset O ⊆ R.Note that, if you remove (0, 0) from V , we have 4 components. However, if we remove φ(0, 0) from O, there should be twocomponent in O since O ⊆ R. Thus, φ is not a homeomorphism and it is a contradiction. Therefore, M is not a manifold.

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4-10Let S be the square I × I with the order topology generated by the dictionary order (see Problem 4-6).(a) Show that S has the least upper bound property.(b) Show that S is connected.(c) Show that S is locally connected, but not locally path-connected.

Proof. .

Recall that the order topology generated by the dictionary order ≤ is the topology that generated by sets

G(a,b) = {(x, y) ∈ S : (x, y) ≥ (a, b)} L(a,b) = {(x, y) ∈ S : (x, y) ≤ (a, b)}for each (a, b) ∈ I × I.(a)Let T ⊆ S be a given non empty set which is bounded above. Then there exists (p, q) ∈ S such that

(s, r) ≤ (p, q) ∀(s, r) ∈ T.Let Uq = {p ∈ I : (p, q) ≥ (s, r) ∀(s, r) ∈ T}. Since (p, q) ∈ Uq and Uq ⊆ I is bounded below, by the least upper boundproperty of I, there is a greatest lower bound of Uq, let’s say p′. And let U = {q ∈ I : (p′, q) ≥ (s, r) ∀(s, r) ∈ T} . Similarly,since U is nonempty bounded below subset of I, we have greatest upper bound of U , let’s say q′. Note that (s, r) ≤ (p′, q′)∀(s, r) ∈ T . Otherwise, there is (s, r) ∈ T such that (p′, q′) ≤ (s, r). Observe that if p′ < s or p′ = s and q′ < r, it is acontradiction to the condition for (p′, q′). Thus, (p′, q′) is an upper bound for T .Now, suppose that (x, y) ∈ S is an upper bound for T . If (x, y) < (p′, q′), then x < p′ or x = q′ and y < q′. But it is alsoa contradiction to the condition of (p′, q′). Thus, (p′, q′) is the least upper bound of T . Therefore, S has the least upperbound property.

(b)Assume that S is not connected. Then there are two nonempty disjoint open sets U and V such that S = U ∪ V . Then∃x ∈ U ∃y ∈ V . Without loss of generality, let x < y. Let

G = {a ∈ U : a < y}.Note x ∈ G and is non empty and G is bound above by y. Since S has least upper bound property, there is a least upperbound of G, let’s say g. Since U = V c is closed, g ∈ U .Let

L = {b ∈ V : g < b}.Since y ∈ L, L is nonempty set that is bounded below by g. Thus, by least upper bound property, there is a greatest lowerbound of L, let’s say it l. Since V = U c is closed, l ∈ V . Since V ∩U = ∅, l 6= g, so g < l. Note that for any (p, q), (p′, q′) ∈ Swith (p, q) < (p′, q′), if p < p′, then ∃m such that p < m < p′, so (p, q) < (m, p) < (p′, q′) or if p = p′ and q < q′, then ∃m′such that q < m′ < q′ so (p, q) < (p,m′) < (p′, q′). Thus, for any two different points, there is a element between them. So,∃k ∈ S such that g < k < l. However, by the setting, k 6∈ U and k 6∈ V . Thus, it is a contradiction. Therefore, S is connected.

(c)Let p ∈ S be given and let a neighborhood U ⊆ S of p be given. Note that Lp ∩ Up = {p} ⊆ U and it is open since each ofthem are two of the generator of this topology and the union is a finite intersection of open sets. Note that singleton set isclearly connected since we cannot find any two disjoint nonempty set of it. Also, it is path connected. Since any constantcurve is continuous connecting the same point. Thus, S is locally path-connected so is locally connected.

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4-11Let X be a topological space, and let CX be the cone on X (see Example 3.53.)(a) Show that CX is path-connected.

Proof. .

Recall that CX = (X × I/X ×{0}). Let (x1, t1), (x2, t2) ∈ CX be given. And let γ1 : [0, 1]→ CX and γ2 : [0, 1]→ CX suchthat

γ1(t) = (x1, t1(1− t)) γ2(t) = (x2, tt2).

Then γ1 and γ2 are paths. And observe that

γ3(t) =

{γ1(2t) t ∈ [0, 12 ]

γ2(2t) t ∈ ( 12 , 1].

Since γ1(1) = (x1, 0) = (x2, 0) = γ2(1) in CX, γ3 is continuous in CX. Since γ3 connect (x1, t1) and (x2, t2), γ3 is the path.Therefore, CX is path connected.

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4-13Let T be the topologist’s sine curve (Example 4.17).(a) Show that T is connected but not path-connected or locally connected.(b) Determine the components and the path components of T .

Proof. .

Recall that T = T0 ∪ T+ = {(x, y) : x = 0 and y ∈ [−1, 1]} ∪ {(x, y) : x ∈ (0, 2π ] and y = sin( 1

x )}.

(a)T is connected.Assume that T is disconnected. Then there are two disjoint open sets U and V such that U ∪ V = T . It is clear that T0and T+ are connected. Thus, without loss of generality, we can have T0 ⊆ U and T+ ⊆ V . Since U ⊆ T is an open set,there exists W ⊆ R2 such that U = W ∩ T . Then there is an open neighborhood B ⊆ U of (0, 0). Note that any openneighborhood of (0, 0) intersects with T+ and it mean that U ∩ T+. And it is a contradiction since T+ ⊆ V and U ∩ V = ∅.Therefore, T is connected.

T is not path-connected.Assume that T is path-connected. Then there is a path γ : [0, 1] → T such that γ(0) = ( 2

π , 0) and γ(1) = (0, 0). Note that,

in T , there is only one path to connect the two points. Thus, γ(t) =(

2π −

2π t, sin

π2(1−t)

). However, then γ is not continuous

since sin π2(1−t) is not continuous at t = 1. It is contradiction since path is continuous by definition. Therefore, T is not

path connected.

T is not locally connected.Assume T is locally connected. Then there is an connected open set V such that (0, 0) ∈ V ⊆ B((0, 0), 1/2) ∩ T . Since V isopen, there is an open set U ⊆ R2 such that U ∩ T = V . And further note that U ⊆ B

((0, 0), 12

)so that U ∩ T has infinitely

many components. Therefore, it is contradiction since V was connected. Thus, T is not locally connected.

(b)Since T is connected, there is only one component of T , namely, T .Observe that T0 and T+ are path connected. Since T is the union of the two path connected sets and T is not path connected,there are two path components, namely, T0 and T+.

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November 6th, 2018

4.49 Prove the preceding three theorems. (Special note: Prove only theorem 4.47 and 4.48.)

Theorem 4.47Endowed with the Euclidean metric, a subset of Rn is a complete metric space if and only if it is closed in Rn. In particular,Rn is complete.

Theorem 4.48Every compact metric space is complete.

Proof. .

Theorem 4.47Suppose that a subset A ⊆ Rn be a complete metric space. And let p be a limit point of A, then ∃(pi) ⊆ A such that pi → p.Since A is complete and (pi) is Cauchy sequence, (pi) converges to a point in A , so p ∈ A. Therefore, A contains all of itslimit points, so A is closed.

Conversely, suppose that a subset A ⊆ Rn is a closed set and let (pi) ⊆ A be a Cauchy sequence. Let ε > 0 be given. Then

∃N1 ∈ N such that ‖pN − pm‖ < ε2 ∀m ≥ N1. Then let R = max1≤i≤N{‖pi‖, ‖pn‖ + ε

2}. Then (pi) ⊆ BR(0). Observe that

K = BR(0) ∩ A is closed and bounded and (pi) ⊆ K, thus K is a compact space containing (pi). Since K is compact, itis sequentially compact. So (pki) ⊆ (pi) ⊆ K such that pki → p in K. Then we can find N1 ∈ N such that ‖pki − p‖ < ε

2∀i ≥ N2. Now, letting N = maxN1, N2, observe that

‖pi − p‖ ≤ ‖pi − pki‖+ ‖pki − p‖ < ε ∀i > N.

Therefore, pi → p so (pi) is a convergent sequence. Therefore, A is complete.

In particular, since Rn is closed subset of Rn, Rn is complete.

Theorem 4.48Suppose that a compact metric space K is given with the metric d. Let a Cauchy sequence (pi) ⊆ K and ε > 0 be given.Then ∃N1 ∈ N such that d(pi, pN1

) < ε2 . Since, in metric space, K is a sequentially compact. Therefore, (pki) ⊆ (pi) ⊆ K

such that pki → p ∈ K. Thus, ∃N2 ∈ N such that d(pki , p) <ε2 . Then, letting N = maxN1, N2, observe that

d(pi, p) ≤ d(pi, pN ) + d(pN , p) < ε ∀i ≥ N.Therefore, the Cauchy sequence (pi) is convergent in K so K is complete.

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4.67Show that any finite product of locally compact spaces is locally compact.

Proof. .

Let X and Y be any given locally compact spaces. Let (x, y) ∈ X × Y be given. Since X and Y are locally compact, thereare compact sets K ⊆ X and C ⊆ Y such that U ⊆ K and V ⊆ C where U and V are neighborhoods of x and y respectively.Since any finite product of compact spaces is compact, K ×C ⊆ X × Y is compact and it contains a neighborhood U × V of(x, y). Therefore, X × Y is locally compact space.

Then by the principle of mathematical induction, any finite product of locally compact spaces is locally compact.�

4-18Let M1 and M2 be n−manifolds. For i = 1, 2, let Bi ⊆ Mi be regular coordinate balls, and let M ′i = Mi \ Bi. Choose ahomeomorphism f : ∂M ′2 → ∂M ′1 (such a homeomorphism exists by problem 4-17. ) Let M1#M2 (called a connected sumof M1 and M2) be the adjunction space M ′1 ∪f M ′2 (Fig.4.13).(a) Show that M1#M2 is an n−manifold (without boundary).(b) Show that if M1 and M2 are connected and n > 1, then M1#M2 is connected.(c) Show that if M1 and M2 are compact, then M1#M2 is compact.

Proof. .

(a)Note that M ′i = Mi \Bi for i = 1, 2 are n−manifold with nonempty boundary since Bi are regular coordinate ball ∃B′i ⊆Mi

such that Bi ⊆ B′ and ∂Bi ⊆ ∂M ′i so M ′i has nonempty boundary. Then, by Theorem 3.79 (Attaching Manifold along TheirBoundary),

M1#M2 = M ′1 ∪f M ′2is a n−manifold without boundary.

(b)By the same theorem, there are topological embeddings ei : M ′i →M1#M2 for i = 1, 2 such that

e1(M ′1) ∪ e2(M ′2) = M1#M2

e1(M ′1) ∩ e2(M ′2) = e1(∂M ′1) = e2(∂M ′2).

Note that ∂M ′i 6= ∅ for i = 1, 2, so e1(M ′1)∩ e2(M ′2) 6= ∅. Also, note that each M ′i is connected( ∵ Mi and Bi are connected.)so is ei(M

′i) for i = 1, 2. Therefore, e1(M ′1) ∪ e2(M ′2) = M1#M2 is connected.

(c)Suppose that M1 and M2 are compact. Note that M ′i = Mi \ Bi is closed subset of compact set Mi for each i = 1, 2. Sincethe mapping e1 and e2 are continuous, e1(M ′1) and e2(M ′2) are compact. Since M1#M2 is a union of two compact sets, it iscompact.

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4-23Let X be a locally compact Hausdorff space. The one-point compactification of X is the topological space X∗ definedas follows. Let ∞ be some object not in X, and let X∗ = X

∐{∞} with the following topology:

T = { open subsets of X} ∪ {U ⊆ X∗ : X∗ \ U is a compact subset of X}.(a) Show that T is a topology.(b) Show that X∗ is a compact Hausdorff space.(c) Show that a sequence of points in X diverges to infinity if and only if it converges to ∞ in X∗. (See p.118)(d) Show that X is open in X∗ and has the subspace topology.(e) Show that X is dense in X∗ if and only if X is noncompact.

Proof. (a)Note that ∅ is open in X, so ∅ ∈ T . Also, since X∗ \X∗ = ∅ is compact, X∗ ∈ T .Let, for an arbitrary index set A, {Uα}alpha ⊆ T be given. If ∞ 6∈ Uα ∀α ∈ A, observe that⋃

α∈AUα ∈ T ,

since it is open in X. If the union contains ∞,

X∗ \

(⋃α∈A

Uα

)=⋂α∈A

X∗ \ Uα ⊂ X∗ \ Uβ for some β ∈ A

where ∞ ∈ Uβ , so X∗ \ Uβ = K is compact. Since the intersection above is closed subset of a compact set K, it is compactin X, so the union is open.Let {Uk}nk=1 is a finite family of open sets in X∗. If ∞ ∈ Uk for all k ∈ {1, 2, . . . , n}, X∗ \ Uk is compact, so

∞ ∈n⋂k=1

Uk and X∗ \

(n⋂k=1

Uk

)=

n⋃k=1

(X∗ \ Uk) is compact =⇒n⋂k=1

Uk ∈ T

If ∞ 6∈ Ui for some i, then, since Uk \ {∞} is open in X for all k, observe the below.n⋂k=1

Uk =

n⋂k=1

(Uk \ {∞}) ∈ T

Therefore, T is a topology on X∗.(b)N (X∗, T ) is compact.(→) Let O = {Uα}α∈A be an given open cover of X∗. Then there exists γ ∈ A such that ∞ ∈ Uγ . Note that C =let

{Uα}α∈A\{γ} is an open cover of U cγ and U cγ is compact in X, so there exists finite subcover S ⊂ C. Therefore, S ∪ {Uγ} ⊂ Ois the finite sub open cover of X∗. So X∗ is compact.N (X∗, T ) is a Hausdorff space.(→) If x, y ∈ X with x 6= y, then clearly there exists disjoint open sets Ux(3 x), Uy(3 y) ∈ T since X is a Hausdorff space.Thus, it suffice to prove that for any x ∈ X, there exists disjoint open sets Ux(3 x), U∞(3 ∞) ∈ T . Note that x ∈ U∞and U∞ is compact. Thus, there exists a precompact neighborhood K ⊆ U c∞ of x since X is locally compact Hausdorff.Therefore, the space is Hausdorff.(c)Suppose that (xi) ⊆ X diverges to infinity. Let U ∈ T with ∞ ∈ U be given. Since X∗ \ U is compact, xi ∈ X∗ \ U at mostfinitely many i ∈ Z+ which implies that ∃N ∈ N such that xi ∈ U ∀i ≥ N . Therefore, (xi) converges to ∞.Conversely suppose that (xi) converges to∞ in X∗. Then ∀U ∈ T with∞ ∈ U , ∃N ∈ N such that xi ∈ U ∀i ≥ N . It impliesthat xi ∈ X∗ \ U at most N of i’s. Therefore, for any compact set K ⊆ X, xi ∈ K at most finitely many i. Therefore, (xi)diverges to infinity.

(d)Clearly, X ∈ T since X is an open subset of itself. Therefore, X is open in X∗. Let the original topology of X be T1 andT2 = {X ∩U : U ∈ T }. Clearly, T1 ⊆ T2 since T1 ⊆ T . In order to show the reverse inclusion, it suffices to show that for anyU ∈ T with ∞ ∈ U , X ∩U ∈ T1. Let such U ∈ T be given. Then note that X∗ \U = X \U = X \ (X ∩U) is compact, so itclosed since X is Hausdorff. Therefore, X ∩ U is open in X so T∈ ⊆ T1. Therefore, T1 = T2 so X has subspace topology ofX∗.(e)(Contrapositive ) Suppose that X is compact. Then {∞} is open in X∗, so X∗ \ {∞} = X is closed and it means that∞ 6∈ X, so X 6= X∗ i.e., X is not dense in X∗.Conversely, suppose that X is not dense, then X ⊆ X ( X∗. It implies that {∞} is open in X∗. Then X = X∗ \{∞} shouldbe compact by definition of T . �


4-29Let X be a complete metric space or a locally compact Hausdorff space. Show that every nonempty countable closed subsetof X contains at least one isolated point. [Hint : use the Baire category theorem.]

Proof. .

Let X be either complete metric space or locally compact Hausdorff space. Note that, in either cases, X is a Hausdorffspace.Let A = {an}∞n=1 be a countable. Then, since A is closed, A is locally compact Hausdorff space or a complete metric spacedepending on the kind of space X.Assume that A does not have any isolated point and let Un = A\{an} ∀n ∈ N. Note that Un is closed since X is Hausdorff soevery singleton set is closed. And let V ⊆ A be any nonempty open set. If V ∩ Un = ∅ for some n, then V = {an}. But it isimpossible since then an is an isolated point. Therefore, V ∩Un 6= ∅ ∀n ∈ N. It implies that Un is open dense subset for every n.

Then, by Baire category theorem( we can apply this theorem in either cases of X), ∩n∈NUn = ∅ is dense in A. However,∅ is a closed set and A is nonemptyset, so ∅ cannot be dense in A. Then it is a contradiction. Therefore, A should have atleast on isolated point.

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4-32Prove that every closed subspace of a paracompact space is paracompact.

Proof. .

Suppose that A is a closed subspace of a paracompact space X. Let A = {Uα} be an open cover of A. Then, there is acollection of open sets {Vα} in X such that A ∩ Vα = Uα for each α. Note that A′ = {Vα} ∪ {X \A} is an open cover of X.Since X is paracompact, we have locally finite open refinement S ′ of A′. Then clearly, S = {A ∩ S : S ∈ S ′} is a collectionof open subsets of A.Let x ∈ A be given. Then it has a neighbor hood U ⊂ X such that U intersects only finitely many of S ∈ S ′, say {S1, . . . , Sn}.So A ∩ U is a neighborhood of x and it intersects with at most {S1 ∩A, . . . , Sn ∩A} ⊆ S. Thus, S is locally finite.Let S ∩ A ∈ S be given. Since S ′ is a refinement, ∃Vα ∈ A′ such that S ⊆ Vα. Then S ∩ A ⊆ A ∩ Vα = Uα ∈ A. Therefore,S is an open refinement of A.Therefore, A is paracompact.

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math 671 : topology homework solutions

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