MTG 6316 HOMEWORK Spring 2017

153. Let U n be a finite open cover of X and f : U Y be continuous for each { k}k=1 k k ! k =1,...,n. Show that if f (x)=f (x) for all x U U , then the function k j 2 k \ j F : X Y defined by ! F (x)=f (x) for all x U k 2 k is continuous.

Proof. First of all, F is well defined on X. Now for any fixed x X,thereexists 2 some U ,1 K n, such that x U . Note that f is continuous over K 6 6 2 K K U . Then for any neighborhood V Y of y = F (x)=f (x), there is some K ⇢ K neighborhood W U of x such that f (W ) V . Note that W is open in U ⇢ K K ⇢ K and that UK is open in X,soW is open in X,i.e.,W is a neighborhood of x in X.ButF (W )=f (W ) V .Sincex and V were arbitrary, we conclude that F K ⇢ is continuous.

1 1 2 n 1 Consider a = 0, , ,..., , 1,q ,q ,...,q A and let g = h(a). Then k, where n n n 0 1 n 2 9 0 k n 1and k x< k+1 . In particular,   n  n k g g (x) n ✓ ◆ k+1 k k g n g n k k = g k+1 k x + g n n n ✓ ◆ n n ✓ ◆ ✓ ◆ k+1 k g n g n k = 1 x n n ✓ ◆ k +1 k g g  n n ✓ ◆ ✓ ◆ k +1 k +1 k +1 k k k 3✏ g f + f f + f g < .  n n n n n n 5 ✓ ◆ ✓ ◆ ✓ ◆ ✓ ◆ ✓ ◆ ✓ ◆ Then k k k k ✏ ✏ 3✏ f(x) g(x) f(x) f + f g + g g(x) < + + = ✏ < 1. | |  n n n n 5 5 5 ✓ ◆ ✓ ◆ ✓ ◆ ✓ ◆ As x was arbitrary,

⇢(f,g)=sup d(f(x),g(x)) : x I =sup f(x) g(x) : x I ✏ { 2 } {| | 2 }  proves that D is dense in (I,R). C

2 MTG 6316 HOMEWORK Spring 2017

154. (a) (Section 31, #1) Show that if X is regular, every pair of points of X have neigh- borhoods whose closures are disjoint.

Proof. Recall the following:

Definition 1. Suppose that one-point sets are closed in X. Then X is said to be regular if for every pair of a point x X and a closed set A X disjoint from x, 2 ⇢ there exist disjoint open sets containing x and A, respectively. The space X is said to be normal if for each pair A, B of disjoint closed sets of X, there exist disjoint open sets containing A and B, respectively.

Lemma 2. Let X be a topological space. Let one-point sets in X be closed. (i) X is regular if and only if given a point x X and a neighborhood U of x, 2 there is a neighborhood V of x such that Cl(V ) U. ⇢ (ii) X is normal if and only if given a closed set A and an open set U containing A, there is an open set V containing A such that Cl(V ) U. ⇢ Then, for any pair of points x, y X, by regularity of X, there are disjoint 2 open sets U, V containing x, y, respectively. However, by (i) of Lemma 2, there are neighborhoods U of x and V of y such that Cl(U ) U and 0 0 0 ⇢ Cl(V ) V .SinceU, V are disjoint, so are Cl(U ) and Cl(V ). 0 ⇢ 0 0

(b) (Section 31, #2) Show that if X is normal, every pair of disjoint closed sets have neighborhoods whose closures are disjoint.

Proof. Let A, B X be disjoint closed sets. By normality of X, there are ⇢ disjoint open sets U A and V B. However, by (ii) of Lemma 2, there are open sets U A and V B such that Cl(U ) U and Cl(V ) V .Since 0 0 0 ⇢ 0 ⇢ U, V are disjoint, so are Cl(U 0) and Cl(V 0).

1

Let p ⇥ X Y be a closed continuous surjective map. Show that if X is normal, then so is Y .

Let A and B be disjoint closed sets in Y . By continuity, p1 A and p1 B are disjoint closed in X.Since 1 1 X is normal, there exist disjoint open sets U1 and U2 in X such that p A L U1 and p B L U2. Now as U1 and U2 are open, then X U1 and X U2 are closed.( And) as p is( a closed) map, then p X U1 and p X U2 are also closed. Thus V1 = Y p X U1 and V2 = Y p X( )U2 are open with( )A L V1 and B L V2. Now ( ) ( ) ( ) ( ) V1 = V2 = Y p X U1 = Y p X U2 = Y p X U1 < X U2 .

We claim now that Y = p( X U(1 < X ))U2 (. Suppose( y " Y)).Thensince( ( p is surjective,) ( there)) is some x " X such that p x = y.SinceU1 = U2 = o,thenx " X U1 or x " X U2.Thusy " p X U1 < X U2 , implying Y = p X U1 (< X )U2(. Therefore,) V1 = V2 = o.ThusY is normal. ( ) ( ) ( ) ( ) ( )

1 Not officially assigned: 159. Show that a closed subspace of a normal space is normal. P: Let Y be a closed subspace of the normal space X. Then Y is Hausdorff by Thm 17.11. Let A and B be disjoint closed subspaces of Y. Since A and B are closed also in X, they can be separated in X by disjoint open sets U and V. Then U ∩Y and V ∩ Y are open sets in Y separating A and B. Y is normal. ∎

Show that every locally compact Hausdor↵ space is regular.

Since X is locally compact and Hausdor↵, by theorem 29.1 there exists a compact Hausdor↵ space Y such that X is a subspace of Y and Y X is a single point. Moreover, by theorem 32.3 \ Y is normal since it is compact and Hausdor↵. But every normal space is regular, thus Y is regular. Finally, by theorem 31.2 a subspace of a regular space is also regular. Thus the space X is regular, as desired. MTG 6316 HOMEWORK Spring 2017

162. (Section 32, #4) Show that every regular Lindel¨ofspace is normal.

Proof. Recall:

Definition 3. A space for which every open covering contains a countable subcovering is called a Lindel¨of space.

The idea of the proof is similar to proving Lemma 32.1 (on page 200). Let X be regular and Lindel¨of. Let A, B X be an arbitrary pair of disjoint closed sets. ⇢ Then, for every p =(x, y) C = A B, there are disjoint open sets U ,V such 2 ⇥ p p that x U , y V .SinceA, B are closed, we may assume that each U is disjoint 2 p ⇢ p p from B and that each Vp is disjoint from A (if not, subtract A or B from Vp or

Up, respectively, and we still keep all the properties mentioned above). Moreover, by part (a) of Lemma 31.1 (on page 196), we may assume that

Cl(U ) B =Cl(V ) A = . (1) p \ p \ ;

Let Wp = Up Vp. Note that Wp p C X (A B) forms an open covering [ { } 2 [ { \ [ } of X. Then by the Lindel¨ofcondition, there is a countable subcovering W { n}n1=0 of X,whereW0 = X (A B). Thus Wn n1=1 Wp p C is a countable open \ [ { } ⇢ { } 2 covering of A B. By earlier assumption, we have that Un n1=1 Up p C covers [ { } ⇢ { } 2 A and that Vn n1=1 Vp p C covers B. The rest of the proof is identical to the { } ⇢ { } 2 proof of Lemma 32.1. For every n,define

n n U 0 = U Cl(V ) and V 0 = V Cl(U ). (2) n n i n n i i[=1 i[=1 Note that by (1), U and V are open coverings of A and B,respectively. { n0 }n1=1 { n0}n1=1 Define 1 1 U 0 = U 0 A and V 0 = V 0 B, n n n=1 n=1 [ [ both are open. Claim: U V = . Suppose not, then there is z U V ,so 0 \ 0 ; 2 0 \ 0 there are i, j N such that z U 0 V 0. Without loss of generality, assume that 2 2 i \ j i j. Then by (2), we have z U but z/Cl(U ) U , which is a contradiction, 6 2 i 2 i i completing the proof of the claim. Therefore X is normal by definition.

2 Theorem (Strong form of the Urysohn Lemma). Let X be a normal space. There is a continuous function f : X [0, 1] such that f(x)=0forx A,andf(x)=1forx B, ! 2 2 and 0

Definition A is a G set in X if A is the intersection of a countable collection of open sets of X.

Proof. ( )Letusseethat ) 1 1 1 G := f 0, = A. (1) A n n=1 \ ✓ ◆◆ Note that 0, 1 are open in [0, 1] so by continuity of f, G is a G set. Let x G and n A 2 A 1 1 1 suppose towards a contradiction x/A. Then f(x)=p>0. Let

( )ByProblem4(MunkresSection33,page213),letf,g : X [0, 1] be a continuous ( ! functions so that

0, if x A 0, if x B f(x)= 2 g(x)= 2 8f(x) > 0, x / A 8g(x) > 0, x / B < 2 < 2 Let : : f(x) h(x)= . f(x)+g(x) Then h is continuous with

0, if x A 2 h(x)=81, if x B . > > 2 <0 :>

1 167. Show that every locally compact Hausdorff space is completely regular. P: Let Z be a locally compact Hausdorff space. By Corollary 29.4, Z is homeomorphic to an open subspace of a compact Hausdorff space. Every compact Hausdorff space is normal (Thm 32.3). Z is the subspace of a normal space. “A normal space is completely regular, by the Urysohn lemma…” (Textbook p. 211). Z is the subspace of a completely regular space. A subspace of a completely regular space is completely regular (Thm 33.2) Z is completely regular. ∎

170. Let X be a locally compact Hausdorff space. Is it true that if X has a countable basis, then X is metrizable? Is it true that if X is metrizable, then X has a countable basis? P: Let X be a locally compact Hausdorff space. X 2nd countable ⇒ X metrizable. ⇍: Since a discrete uncountable space is metrizable but not 2nd countable. ⇒: Every locally compact space is regular (Ex #161). Every 2nd countable regular space is metrizable (Urysohn metrization Thm [34.1])∎

Show that contractibility is a topological property, i.e. if X Y ,thenX is contractible if and ⇡ only if Y is contractible.

By assumption, there exists an homeomorphism f : X Y , so that f 1 : Y X is an ! ! homeomorphism as well. Assume that f is contractible, therefore there exists an homotopy H : X I X and x X such that ⇥ ! 0 2 H(x, 0) = x and H(x, 1) = x , x X. 0 8 2 1 We define G(y, s)=f(H(f (y),s)), so that G is continuous since it is the composition of continuous function and

1 1 1 G(y, 0) = f(H(f (y), 0)) = f(f (y)) = y and G(y, 1) = f(H(f (y), 1)) = f(x0).

Thus the identity on Y is homotopic to the point f(x0) and it follow that Y is contractible. The other direction is done the same way by switching the roles of X and Y . Thus, contractibility is a topological property, as desired. Let p : E B be a covering map, with E path connected. Show that if B is simply ! connected, then p is a homeomorphism.

Proof. Let p(e0)=b0 and denote the lifting correspondence. Since B is simply connected, ⇡ (B,b )istrivial.Sincetheliftofb that begins at e is e , for any [f] ⇡ (B,b )withf 1 0 0 0 0 2 1 0 the lift that begins at e0,

f(1) = ([f]) = ([b0]) = e0.

Since E is path connected, the lifting correspondence is surjective (Munkres, Theorem 54.4, 1 page 345) hence p (b )= e . This shows that p is injective. Since p is a covering map, p 0 { 0} is an open map (Munkres, page 336). We conclude that p is a homeomorphism.

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