MTG 6316 HOMEWORK Spring 2017

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2 MTG 636 HOMEWORK Sprig Let {U k } k= be a fiite ope cover of X ad f k : U k! Y be cotiuous for each k =,...,. Show that if f k (x) = f j (x) for all x 2 U k \ U j, the the fuctio F : X! Y defied by F (x) =f k (x) for all x 2 U k is cotiuous. Proof. First of all, F is well defied o X. Now for ay fixed x 2 X, thereexists some U K, 6 K 6, such that x 2 U K. Note that f K is cotiuous over U K. The for ay eighborhood V Y of y = F (x) =f K (x), there is some eighborhood W U K of x such that f K (W ) V. Note that W is ope i U K ad that U K is ope i X, sow is ope i X, i.e.,w is a eighborhood of x i X. ButF (W )=f K (W ) V.Sicex ad V were arbitrary, we coclude that F is cotiuous.

3 Cosider a = 0,, 2,...,,,q 0,q,...,q 2 A ad let g = h(a). The 9k, where 0 apple k apple ad k k+ apple x<. I particular, g g (x) + apple g = g g k+ g k k+ k = g k+ g k + apple g k + k + f + f k x + g k x g f + f g < 3 5. The f(x) g(x) apple f(x) f + f g + g g(x) < = <. As x was arbitrary, (f,g) =sup{d(f(x),g(x)) : x 2 I} =sup{ f(x) g(x) : x 2 I} apple proves that D is dese i C(I,R). 2

4 MTG 636 HOMEWORK Sprig (a) (Sectio 3, #) Show that if X is regular, every pair of poits of X have eighborhoods whose closures are disjoit. Proof. Recall the followig: Defiitio. Suppose that oe-poit sets are closed i X. The X is said to be regular if for every pair of a poit x 2 X ad a closed set A X disjoit from x, there exist disjoit ope sets cotaiig x ad A, respectively. The space X is said to be ormal if for each pair A, B of disjoit closed sets of X, there exist disjoit ope sets cotaiig A ad B, respectively. Lemma 2. Let X be a topological space. Let oe-poit sets i X be closed. (i) X is regular if ad oly if give a poit x 2 X ad a eighborhood U of x, there is a eighborhood V of x such that Cl(V ) U. (ii) X is ormal if ad oly if give a closed set A ad a ope set U cotaiig A, there is a ope set V cotaiig A such that Cl(V ) U. The, for ay pair of poits x, y 2 X, by regularity of X, there are disjoit ope sets U, V cotaiig x, y, respectively. However, by (i) of Lemma 2, there are eighborhoods U 0 of x ad V 0 of y such that Cl(U 0 ) U ad Cl(V 0 ) V.SiceU, V are disjoit, so are Cl(U 0 ) ad Cl(V 0 ). (b) (Sectio 3, #2) Show that if X is ormal, every pair of disjoit closed sets have eighborhoods whose closures are disjoit. Proof. Let A, B X be disjoit closed sets. By ormality of X, there are disjoit ope sets U A ad V B. However, by (ii) of Lemma 2, there are ope sets U 0 A ad V 0 B such that Cl(U 0 ) U ad Cl(V 0 ) V.Sice U, V are disjoit, so are Cl(U 0 ) ad Cl(V 0 ).

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6 Let p X Y be a closed cotiuous surjective map. Show that if X is ormal, the so is Y. Let A ad B be disjoit closed sets i Y. By cotiuity, p (A) ad p (B) are disjoit closed i X. Sice X is ormal, there exist disjoit ope sets U ad U 2 i X such that p (A) L U ad p (B) L U 2. Now as U ad U 2 are ope, the X U ad X U 2 are closed. Ad as p is a closed map, the p(x U ) ad p(x U 2 ) are also closed. Thus V = Y p(x U ) ad V 2 = Y p(x U 2 ) are ope with A L V ad B L V 2. Now V = V 2 = (Y p(x U )) = (Y p(x U 2 )) = Y (p(x U ) < (X U 2 )). We claim ow that Y = p(x U )<(X U 2 ). Suppose y " Y.Thesicep is surjective, there is some x " X such that p(x) = y. SiceU = U 2 = o, thex " X U or x " X U 2.Thusy " p(x U ) < (X U 2 ), implyig Y = p(x U ) < (X U 2 ). Therefore, V = V 2 = o. ThusY is ormal.

7 Not officially assiged: 59. Show that a closed subspace of a ormal space is ormal. P: Let Y be a closed subspace of the ormal space X. The Y is Hausdorff by Thm 7.. Let A ad B be disjoit closed subspaces of Y. Sice A ad B are closed also i X, they ca be separated i X by disjoit ope sets U ad V. The U Y ad V Y are ope sets i Y separatig A ad B. Y is ormal.

8 Show that every locally compact Hausdor space is regular. Sice X is locally compact ad Hausdor, by theorem 29. there exists a compact Hausdor space Y such that X is a subspace of Y ad Y \ X is a sigle poit. Moreover, by theorem 32.3 Y is ormal sice it is compact ad Hausdor. But every ormal space is regular, thus Y is regular. Fially, by theorem 3.2 a subspace of a regular space is also regular. Thus the space X is regular, as desired.

9 MTG 636 HOMEWORK Sprig (Sectio 32, #4) Show that every regular Lidelöf space is ormal. Proof. Recall: Defiitio 3. A space for which every ope coverig cotais a coutable subcoverig is called a Lidelöf space. The idea of the proof is similar to provig Lemma 32. (o page 200). Let X be regular ad Lidelöf. Let A, B X be a arbitrary pair of disjoit closed sets. The, for every p =(x, y) 2 C = A B, there are disjoit ope sets U p,v p such that x 2 U p, y V p.sicea, B are closed, we may assume that each U p is disjoit from B ad that each V p is disjoit from A (if ot, subtract A or B from V p or U p, respectively, ad we still keep all the properties metioed above). Moreover, by part (a) of Lemma 3. (o page 96), we may assume that Cl(U p ) \ B =Cl(V p ) \ A = ;. () Let W p = U p [ V p. Note that {W p } p2c [ {X \ (A [ B)} forms a ope coverig of X. The by the Lidelöf coditio, there is a coutable subcoverig {W } =0 of X, wherew 0 = X \ (A [ B). Thus {W } = {W p} p2c is a coutable ope coverig of A[B. By earlier assumptio, we have that {U } = {U p} p2c covers A ad that {V } = {V p} p2c covers B. The rest of the proof is idetical to the proof of Lemma 32.. For every, defie [ [ U 0 = U Cl(V i ) ad V 0 = V Cl(U i ). (2) i= Note that by (), {U} 0 = ad {V } 0 = are ope coverigs of A ad B,respectively. Defie U 0 = [ U 0 A ad V 0 = = i= [ V 0 B, both are ope. Claim: U 0 \ V 0 = ;. Suppose ot, the there is z 2 U 0 \ V 0,so there are i, j 2 N such that z 2 Ui 0 \ V j 0. Without loss of geerality, assume that i 6 j. The by (2), we have z 2 U i but z/2 Cl(U i ) U i, which is a cotradictio, completig the proof of the claim. Therefore X is ormal by defiitio. = 2

10 Theorem (Strog form of the Urysoh Lemma). Let X be a ormal space. There is a cotiuous fuctio f : X! [0, ] such that f(x) =0forx 2 A, adf(x) =forx 2 B, ad 0 <f(x) < otherwise,ifadolyifa ad B are disjoit closed G sets i X. Defiitio A is a G set i X if A is the itersectio of a coutable collectio of ope sets of X. Proof. ()) Letusseethat G A := \ = apple f 0, = A. () Note that 0, are ope i [0, ] so by cotiuity of f, G A is a G set. Let x 2 G A ad suppose towards a cotradictio x/2 A. The f(x) =p>0. Let <p;butx 2 f ([0, )) is a cotradictio. Observig that A = f ({0}) G A proves G A = A. Similarly let us see that G B := \ f =, = B. (2) Note that, are ope i [0, ] so by cotiuity of f, G B is a G set. Let x 2 G B ad suppose towards a cotradictio x/2 B. The f(x) =p<. Let be such that p< ; but x 2 f ((, ]) is a cotradictio. Observig that B = f ({}) G B proves G B = B. I summary, G A = f (0) = A ad G B = f () = B are disjoit closed G sets i X. (() ByProblem4(MukresSectio33,page23),letf,g : X! [0, ] be a cotiuous fuctios so that 8 8 < 0, if x 2 A < 0, if x 2 B f(x) = g(x) = : f(x) > 0, x /2 A : g(x) > 0, x /2 B Let The h is cotiuous with h(x) = f(x) f(x)+g(x). 8 0, if x 2 A >< h(x) =, if x 2 B. >: 0 <h(x) < otherwise

11 67. 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 a ope subspace of a compact Hausdorff space. Every compact Hausdorff space is ormal (Thm 32.3). Z is the subspace of a ormal space. A ormal space is completely regular, by the Urysoh lemma (Textbook p. 2). 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.

12 70. Let X be a locally compact Hausdorff space. Is it true that if X has a coutable basis, the X is metrizable? Is it true that if X is metrizable, the X has a coutable basis? P: Let X be a locally compact Hausdorff space. X 2 d coutable X metrizable. : Sice a discrete ucoutable space is metrizable but ot 2 d coutable. : Every locally compact space is regular (Ex #6). Every 2 d coutable regular space is metrizable (Urysoh metrizatio Thm [34.])

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14 Show that cotractibility is a topological property, i.e. if X Y,theX is cotractible if ad oly if Y is cotractible. By assumptio, there exists a homeomorphism f : X! Y, so that f : Y! X is a homeomorphism as well. Assume that f is cotractible, therefore there exists a homotopy H : X I! X ad x 0 2 X such that H(x, 0) = x ad H(x, ) = x 0, 8x 2 X. We defie G(y, s) =f(h(f (y),s)), so that G is cotiuous sice it is the compositio of cotiuous fuctio ad G(y, 0) = f(h(f (y), 0)) = f(f (y)) = y ad G(y, ) = f(h(f (y), )) = f(x 0 ). Thus the idetity o Y is homotopic to the poit f(x 0 ) ad it follow that Y is cotractible. The other directio is doe the same way by switchig the roles of X ad Y. Thus, cotractibility is a topological property, as desired.

15 Let p : E! B be a coverig map, with E path coected. coected, the p is a homeomorphism. Show that if B is simply Proof. Let p(e 0 )=b 0 ad deote the liftig correspodece. Sice B is simply coected, (B,b 0 )istrivial.sicetheliftofb 0 that begis at e 0 is e 0, for ay [f] 2 (B,b 0 )withf the lift that begis at e 0, f() = ([f]) = ([b 0 ]) = e 0. Sice E is path coected, the liftig correspodece is surjective (Mukres, Theorem 54.4, page 345) hece p (b 0 )={e 0 }. This shows that p is ijective. Sice p is a coverig map, p is a ope map (Mukres, page 336). We coclude that p is a homeomorphism.

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