Math 131: Homework 4 Solutions

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1 Math 3: Homework 4 Solutons Greg Parker, Wyatt Mackey, Chrstan Carrck October 6, 05 Problem (Munkres 3.) Let {A n } be a sequence of connected subspaces of X such that A n \ A n+ 6= ; for all n. Then S nn A n s connected. Proof. Let B n = S n = A n. By nducton these are connected for each n. For n =, B = A s connected by assumpton. Assumng B n s connected then B n+ = B n [ A n+ but A n B n and A n \ A n+ 6=, hence B n \ A n+ 6= and by Theorem 3.3 each B n s connected. Notce, however, that [ nn A n = [ nn Yet B n contans A for each n, hence B n have a pont n common, so agan by Theorem 3.3 the ntersecton s connected. Problem (Munkres 3.9) Let A X and B Y be proper subsets. Suppose X, Y are connected. Then s connected. Proof. Fx x 0 / A and y 0 / B. Let B n S =(X Y ) (A B) S 0 = X {y 0 }[{x 0 }[Y I clam that S 0 s connected. We know both peces X {y 0 } and {x 0 } Y are connected, as they are canoncally somorphc to X and Y, whch are connected by assumpton. Theorem 3.3 therefore mples that S 0 s connected. Lkewse, for each x X A, and y Y B we can defne S xy = X {y}[{x} Y. Ths s connected for each x, y by the same argument as for S 0. Now notce we have wrte [ (X Y ) (A B) =S 0 [ S xy (0.) x/a or Y/B As each pont (x, y) / A B has ether x/ A or Y / B, hences xy s contaned n the above. Moreover, every pont n the above has ether x/ A or y/ B, and so s contaned n the left hand sde. I clam that the above s connected. Note: Ths DOES NOT follow from Theorem 3.3. Ths theorem requres a pont of mutual ntersecton of all the sets n the unon, whch the above does not have. Many students made ths mstake. Instead, we must copy the method of proof used n that theorem. Suppose the above were dsconnected, hence could be wrtten U t V. For U, V open and non-empty. By Lemma 3. we have such a separaton, a connected subspace must le n ether one or the other. In partcular, ths s true for S 0. Wthout loss of generalty, say that S 0 U.

2 Now consder S xy for a d erent (x, y). It s connected, hence t s completely contaned n U or V.Yet t contans the ponts n common, wth S 0, namely (x 0,y) and (y 0,x). Therefore we must have S xy U as well, because t cannot be completely contaned n V. Therefore the entre unon s contaned n U, as each S xy has ponts n common wth S 0. Ths s a contradcton to V beng non-empty. Hence we must have that the unon (0.) s connected. Problem 3 (Munkres 3.0) Let {X } be an ndexed faml of connected spaces. Defne X := Y JX and let a =(a ) be a fxed pont n X. (a) Gven any fnte subset K J, denote X K by the subset of ponts so that x = a f / K. Then X K s connected. There s a homeomorphsm Q n k X k! X K va ( x f K (x,...,x n ) 7! otherwse and the fnte product s connected by Theorem 3.6. (b) Now let a Y := [ K J Ths s also connected by Theorem 3.3 as each X K contans the pont a. X K (c) X s the closure of Y, hence connected (by Theorem 3.4) Proof. Choose a pont (x ) X. We show that any neghborhood of (x ) contans a pont of Y. Choose a neghborhood U of (x ). Recall we are under the product topology, so ths neghborhood s of the form U = U U... U N [ 6= X So let K = {,..., N }. Then we can fnd a pont (y ) X K \ U by settng y = a for / K and smply choosng y U for K. Therefore we have found a pont n the neghborhood U (n fact we found nfntely many unless the X are fnte). Problem 4 Let f :[0, ]! [0, ] be a contnuous functon. Then there s a fxed pont such that f(x) =x. Ths s false for [0, ) and (0, ). Proof. At 0, we must have f(0) > 0 else 0 tself s a fxed pont. Lkewse we must have f() < safxed pont. Therefore the functon g(x) :=f(x) x has g(0) > 0 and g() < 0. Hence by the Intermedate Value Theorem, there s a pont c [0, ] so g(c) = 0 but by defnton of g, ths means f(c) =c. However, f we consder [0, ) or (, ) ths no longer holds. Any functon that les completely above the graph of y = x wthout touchng t wll have no fxed pont. In partcular, the functon f(x) = s always strctly greater than x so g(x) > 0 everywhere, and there s no fxed pont.

3 4.9 Assume R s uncountable. Show that f A s a countable collecton of ponts n R,then R A s path connected. Take two ponts x, y R A, x 6= y, andwewanttoshowthatthereexstsapathfrom x to y. Let ` be the perpendcular bsector of the segment from x to y. Correspondng to each pont on ths lne, there s a path from x to y gven by gong on a straght lne from x to the gven pont, and then a straght lne from the gven pont to y. Note that no par of these paths share any common ponts other than x and y. Thuswehaveuncountablemany dsjont lnes, hence only countably many can have nonzero ntersecton wth A. Thusthere must exst uncountably many paths n R A from x to y, thoughofcourse,twouldhave su ced to show only one. 4.0 Show that f U s an open connected subspace of R,thenU s path connected. Fx x 0 U, andleta denote the subset of U contang those ponts x U such that there s a path n U from x 0 to x. Snce x A, wehavethata s not the empty set. Thus t su ces to show that A s both open and closed, snce the only clopen subsets of a connected space are the empty set and the entre space. Frst, we must show t s open. Gven any pont x A, snceu s open, there s a ball of some radus such that B(x, ) U. ThentheresastraghtlnepathnU from x to any pont n ths ball, hence B(x, ) A, hencea s open. Next, we must show t s closed. Consder y U A. Snce U s open, there s a ball of radus around y such that B(y, ) U. Thentheresastraghtlnepathfromy to every pont n B(y, ). In partcular, f any of these were elements of A, thenbyconcatenatng these paths we d get a path from y to x 0, contradcton. Hence B(y, ) U A, sou A s open. Thus A s closed, and we can conclude the problem by above. 5.5 Let X denote the set of ratonal ponts of [0, ] {0} R. Let T denote the set of lne segments jonng the pont (0, ) to the ponts of X. (a) Show that T s path connected, but locally connected only at (0, ). (b) Fnd a subset of R that s path connected locally connected at none of ths ponts. We begn wth (a). T s evdently path connected by gong from any pont on a gven segment, back to (0, ), and then down the approprate segment to our second chosen pont. On the other hand, gven any open set U 3 x, wecantakeaballb small enough that B does not contan (0, ), and B U. Then t s clear that B s not connected, by splttng t between lnes wth slope (sum rratonal number such that the approprate lne ntersects the ball). Local connectedness at (0, ) follows from (0, ) havng to be n both open sets, f they are to be nonempty. For (b), we need only slghtly modfy the gven example. Let Y denote the set of lne segments jonng the ponts (0,n)toX for every n Z. Ths s nowhere locally connected by a smlar argument to (a), but s connected smply by travelng frst to X, thenuptothe appropate (0,n), and then back down to our second pont. 6. Show: (a) In the fnte complement topology on R, everysubspacescompact. (b) If R has a consstng of all sets A such that R A s ether countable or all of R, s [0, ] a compact subspace? (a) Take any subspace X, andanyopencover{u a } a of X. Taknganarbtraryelement of ths cover as our frst element, we have all but fntely many elements of X. Then

4 Math 3 Solutons snce t s a cover, we can just pck one element of the open cover to go over each of these mssng ponts, gvng us a fnte subcover. (b) No note that [0, ] has nfntely many ponts. Let B = {b } be a countable subset of [0, ]. Then consder the collecton of open sets (X B) [ {b }. These clearly form an open cover [0, ], but any fnte collecton of them wll contan only fntely many of the b, hence not all of [0, ].

5 (Pg 7 Problem 5) Let A and B be dsjont compact subspaces of the Hausdorff space X. Show that there exst dsjont open sets U and V contanng A and B respectvely. Fx a A, then 8b B, we can fnd dsjont neghborhoods U b(a) and V b(a) of a and b respectvely by Hausdorffness. The V b(a) cover B, so we may fnd a fnte subcover V b(a),...,v b(a)n(a,b) of B by compactness, defne W a := T Vb(a) U b(a), then W a \ =;. The W a cover A, so we can fnd a fnte subcover W a,...,w an T T by compactness of A. But we have W a j V b(a j ) =;snce, for all j, W a j \ V b(a j ) =;, and j a j T T V b(a j ) æ V b(a j ). Furthermore, V b(a j ) s open as the ntersecton of fntely many open a j a j sets, and hence s an open set contanng B, as S V b(a j ) contans B for all j, hence the ntersecton does. (Pg 78 Problem 5) Let X be a compact Hausdorff space. Show that the unon of a countable collecton of closed sets {A n }, all of whch have empty nteror, has empty nteror. It suffces to show that for an open set W Ω X, we can fnd a pont x W wth x S A n. I clam we can fnd a sequence of nonempty open sets W such that W Ω W A. By Hausdorffness, as n theorem 7.7, If A Ω X s closed, W 0 := W Ω X s non-empty and open, and W 6Ω A (ths s satsfed snce A has empty nteror), then there s a non-empty open set W Ω X such that W Ω W 0 A. We see by nducton that we have the desred sequence. But by compactness T W 6= ;, and, for x T W, x W Ω W, but x W S A n, as desred. 3 (Taubes Problem) Vew the crcle S as the set of pars (x, y) n R obeyng x + y =. Prove the followng: Let f denote any gven contnuous map from S! R. Then there exsts a pont (x, y) S such that f (x, y) = f ( x, y). (Note that ths proves that S s not homeomorphc to the real lne R. There are contnuous maps, njectve from R to S, for example the map that sends t R to the pont p, t +t Consder the functon g : [0,º)! R by g (µ) = f (cos(µ),sn(µ)) f ( cos(µ), sn(µ)), g s contnuous snce f, the parametrzaton µ 7! (cos(µ),sn(µ)), and the antpodal map (x, y) 7! ( x, y) are all contnuous. Fx µ [0,º) and assume g (µ) 6= 0 (assume WLOG g (µ) < 0), then g (µ + º (mod º)) = g (µ) and hence g (µ) > 0, so by the Intermedate Value theorem 0 g ([0,º)), thus there exsts µ 0 [0,º) wth g (µ) = 0 =) f (cos(µ),sn(µ)) = f ( cos(µ), sn(µ)). p +t