Math 535 - General TopologyFall 2012

Homework 8 Solutions

Problem 1. (Willard Exercise 19B.1) Show that the one-point compactification of Rn ishomeomorphic to the n-dimensional sphere Sn.

Solution. Note that Sn is compact, and a punctured sphere Sn\{p} is dense in Sn, with com-plement consisting of a single point {p}. Therefore it suffices to show that Rn is homeomorphicto Sn \ {p}.Consider Rn ∼= Rn × {0} ⊂ Rn+1 as the hyperplane xn+1 = 0 inside Rn+1. Take the “Northpole” p = (0, . . . , 0, 1) ∈ Sn ⊂ Rn+1. We will produce a homeomorphism f : Rn → Sn \ {p}using “stereographic projection”. Define the maps

f : Rn → Sn \ {p}

where f(x) is the unique point of Sn \ {p} which lies on the straight line through x and p and

g : Sn \ {p} → Rn

where g(y) is the unique point of Rn which lies on the straight line through y and p.

For all x ∈ Rn, g(f(x)) is the unique point of Rn which lies on the straight line through f(x)and p. By definition of f , x is a point of Rn which lies on the straight line through f(x) and p.This proves g(f(x)) = x.

For all y ∈ Sn \ {p}, f(g(y)) is the unique point of Sn \ {p} which lies on the straight linethrough g(y) and p. By definition of g, y is a point of Sn \ {p} which lies on the straight linethrough g(y) and p. This proves f(g(y)) = y.

It remains to show that f and g are both continuous.

A straightforward calculation yields

f(x) =1

‖x‖2 + 1

[2x+ (‖x‖2 − 1)p

]which is a continuous function of x. Indeed, the norm ‖x‖ is continuous in x, and the denomi-nator satisfies ‖x‖2 + 1 > 0 for all x ∈ Rn.

Another straightforward calculation yields

g(y) =1

1− yn+1

[y − yn+1p]

which is a continuous function of y. Indeed, the projection Rn+1 � R onto each coordinate iscontinuous (so that yn+1 is a continuous function of y), and the denominator satisfies 1−yn+1 > 0for all y ∈ Sn \ {p}.

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Problem 2. (Munkres Exercise 38.4) Let X be any topological space and f : X ↪→ Y acompactification of X, where moreover Y is Hausdorff. Show that there is a unique continuousclosed surjective map g : βX → Y that extends f , meaning g ◦ e = f , where e : X → βX is thecanonical evaluation map:

X

f

e// βX

g

��

Y.

Hint: You do not need anything about a specific construction of βX, only its universal property.

Solution. Since Y is compact Hausdorff, there is a unique continuous map g : βX → Ysatisfying g ◦ e = f (by the universal property of the Stone-C̆ech compactification).

Since βX is compact and Y is Hausdorff, the continuous map g : βX → Y is closed.

Note that g has dense image in Y :

g(βX) ⊇ g(e(X)) = f(X) = Y.

However, g is closed and thus its image is closed:

g(βX) = g(βX) = Y

which proves that g is surjective.

Remark. If X admits a Hausdorff compactification, we know that X must be Tychonoff (a.k.a.T3 1

2), so that e : X → βX is in fact an embedding and βX is therefore a compactification of X.

Problem 2 shows in particular that any Hausdorff compactification of X is a quotient of theStone-C̆ech compactification βX.

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Problem 3. Let f : X → Y be a continuous map between topological spaces. Show that f isclosed if and only if for all y ∈ Y and any open subset U ⊆ X satisfying f−1(y) ⊆ U , there isan open neighborhood V of y satisfying f−1(V ) ⊆ U .

Solution. Consider the equivalent statements:

f is closed, i.e. for all C ⊆ X closed, f(C) is closed in Y .

⇔ For all C ⊆ X closed and y /∈ f(C), there is an open neighborhood V of y satisfyingV ∩ f(C) = ∅.⇔ For all C ⊆ X closed and y ∈ Y satisfying f−1(y) ∩ C = ∅, there is an open neighborhoodV of y satisfying f−1(V ) ∩ C = ∅.⇔ (Taking U := X \C) For all U ⊆ X open and y ∈ Y satisfying f−1(y) ⊆ U , there is an openneighborhood V of y satisfying f−1(V ) ⊆ U .

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Problem 4. Let f : X → Y be a continuous map between topological spaces.

a. Assume that f : X → Y is proper. Let V ⊆ Y be an open subset. Show that the restrictionf |f−1(V ) : f−1(V )→ V is proper.

Solution. Let K ⊆ V be a compact subspace. Then the preimage

(f |f−1(V ))−1(K) = f−1(K) ⊆ f−1(V ) ⊆ X

is compact since f : X → Y is proper.

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b. Assume that Y is Hausdorff, and that for all y ∈ Y , there is an open neighborhood V of ysuch that the restriction f |f−1(V ) : f−1(V )→ V is proper. Show that f : X → Y is proper.

Solution.

Lemma. Let K be a compact Hausdorff space and {Vα}α∈A an open cover of K. Then thereare compact subspaces K1, . . . , Kn ⊆ K satisfying K = K1 ∪ . . . ∪Kn and each Ki is includedin some Vαi.

Proof. Since K is compact, there is a finite subcover {Vα1 , . . . Vαn}. Thus it suffices to showthe result for any finite open cover {V1, . . . Vn} of K.

Writing K as the union of two open subsets

K = V1 ∪ . . . ∪ Vn = V1 ∪ (V2 ∪ . . . ∪ Vn) ,

it suffices to show the result for an open cover consisting for two open subsets, and then applythe argument n− 1 times.

Let {V1, V2} be an open cover of K. Since K is compact Hausdorff, it is normal. The equalityK = V1 ∪ V2 implies V c

2 ⊆ V1, and note that V c2 is closed and V1 is open. By normality, there

is an open subset U ⊆ X satisfying

V c2 ⊆ U ⊆ U ⊆ V1.

Take K1 := U ⊆ V1 and K2 := U c ⊆ (V c2 )c = V2. Both K1 and K2 are closed in K, hence

compact, and they cover K:

K1 ∪K2 = U ∪ U c ⊇ U ∪ U c = K.

Now we solve the problem. Let K ⊆ Y be compact. By assumption, there exists an open cover{Vα}α∈A of Y such that the restrictions

f |f−1(Vα) : f−1(Vα)→ Vα

are proper.

By the lemma, K can be written as a finite union

K = K1 ∪ . . . ∪Kn

where each Ki is compact and satisfies Ki ⊆ Vαi ∩K ⊆ Vαi . This implies that the preimage

f−1(Ki) =(f |f−1(Vαi )

)−1(Ki) ⊆ f−1(Vαi) ⊆ X

is compact because f |f−1(Vαi )is proper. The preimage

f−1(K) = f−1(K1 ∪ . . . ∪Kn)

= f−1(K1) ∪ . . . ∪ f−1(Kn)

is a finite union of compact spaces, hence compact.

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Alternate solution. For each point y ∈ K, let Vy be a neighborhood of y such that therestriction f |f−1(Vy) is proper. Note that Vy ∩K is a K-neighborhood of y.

Since K is compact Hausdorff (in particular locally compact Hausdorff), for each y ∈ K thereis a compact K-neighborhood Ky of y inside Vy ∩K.

The cover {Ky}y∈K of K can be refined to an open cover of K. Hence, by compactness of K,there is a finite subcover

K = Ky1 ∪ . . . ∪Kyn

where each of these satisfies Kyi ⊆ Vyi ∩K ⊆ Vyi .

Conclude as above.

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Recall that a totally ordered set is a partially ordered set (X,≤) where any two elementsare comparable: for all x, y ∈ X, either x ≤ y or y ≤ x.

Problem 5. Let (X,≤) be a totally ordered set. The order topology on X is the topologygenerated by “open rays”

(a,∞) := {x ∈ X | x > a}

(−∞, a) := {x ∈ X | x < a}

for any a ∈ X.

a. Show that the order topology on a totally ordered set is always T1.

Solution. Because X is a totally ordered set, given any a ∈ X, every element x ∈ X satisfiesexactly one of the three conditions

x < a

x = a

x > a

so that the “closed rays”

[a,∞) := {x ∈ X | x ≥ a} = X \ (−∞, a)

(−∞, a] := {x ∈ X | x ≤ a} = X \ (a,∞)

are indeed closed in X. Their intersection

(−∞, a] ∩ [a,∞) = {x ∈ X | x ≤ a and x ≥ a} = {a}

is therefore closed in X, so that all singletons of X are closed.

Alternate solution. Let x, y ∈ X be distinct points of X. Since X is totally ordered, x andy are (strictly) comparable, WLOG x < y. Then the open ray (−∞, y) contains x but not y,while the open ray (x,∞) contains y but not x.

b. Show that the order topology on R with its usual order ≤ is the standard (metric) topologyon R.

Solution. (Tord ≤ Tmet) For any a ∈ R, the “open rays” (a,∞) and (−∞, a) are metricallyopen.

(Tmet ≤ Tord) The metric topology on R is generated by intervals (a, b) for any a < b. But theseare order-open since they are the finite intersection

(a, b) = (−∞, b) ∩ (a,∞)

of “open rays”.

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c. An interval in a partially ordered set (X,≤) is a subset I ⊆ X such that for all x, y ∈ I,the condition x ≤ z ≤ y implies z ∈ I.

Let (X,≤) be a totally ordered set endowed with the order topology. Show that every connectedsubspace A ⊆ X is an interval in X.

Solution. We will show the contrapositive: Any subset A ⊆ X which is not an interval is notconnected.

The fact that A is not an interval means that there are elements x, y ∈ A and an element z ∈ Xsatisfying x ≤ z ≤ y but z /∈ A. This implies

A ⊆ X \ {z}

= (−∞, z) t (z,∞) since X is totally ordered

so that A can be written as the disjoint union

A = (A ∩ (−∞, z)) t (A ∩ (z,∞))

of subsets that are both open in A. Both subsets are non-empty, by the conditions x ∈A ∩ (−∞, z) and y ∈ A ∩ (z,∞). Therefore A is not connected.

d. Find an example of totally ordered set (X,≤), endowed with the order topology, and aninterval A ⊆ X which is not a connected subspace.

Solution. Consider X = {0, 1}, and the subset X ⊆ X, which is trivially an interval in X.

Since X is finite and T1 (by part a), it is discrete. Since X contains at least two elements, it isnot connected.

Alternate solution. Consider N with its usual order ≤. Then N is discrete, because everysingleton

{n} = (−∞, n+ 1) ∩ (n− 1,∞)

is open. Since N is discrete and contains at least two elements, it is not connected.

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Problem 6. (Munkres Exercise 23.5) A space is totally disconnected if its only connectedsubspaces are singletons {x}.

a. Show that every discrete space is totally disconnected.

Solution. Every subspace A ⊆ X of a discrete space X is itself discrete. If A contains atleast two points, it is therefore not connected.

b. Find an example of totally disconnected space which is not discrete.

Solution. Take X = {0} ∪ { 1n| n ∈ N} ⊂ R viewed as a subspace of R. Then X is not

discrete, since the singleton {0} is not open in X. Indeed, any open ball around 0 containsinfinitely many points of the form 1

n.

However, X is totally disconnected. If A ⊆ X contains at least two points x, y ∈ A, then atleast one of them (WLOG y) is of the form y = 1

n6= 0. We can write A as the disjoint union

A = {y} t (A \ {y})

of two subsets that are open in A, and non-empty since x ∈ A \ {y}. Therefore A is notconnected.

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