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MATH 4181 Problem Set 6 Solutions 1 MATH 4181 001 Fall 1999 Problem Set 6 Solutions 1.Identify the two path components of the topologist's sine curve. Show that one of the components is closed and the other is not. The two path components of the topologist's sine curve are the component along they-axis,A=f(0;y)j1y1g, and the sin( =x) curve component: B= f(x;sin(=x))j0<x1g. We proved in class that there is no path joining any point ofAto any point of B. The component Ais closed, but Bdoes not contain its limit points, which consist of A, and hence Bis not closed. 2.Prove that path connectedness is a topological invariant. Can you prove that it is a continuous invariant? Letf:X!Ybe a homeomorphism and assume that Xis path connected. Let a;b2Ybe any two distinct points. Since fis onto, there are points a0;b02Xso that f(a0) =aandf(b0) =b. SinceXis path connected, there is a path, p:I!X, so that pis continuous, p(0) =a0andp(1) =b0. Letq=fp. Thenqis continuous, being the composition to two continuous functions; q(0) =f(a0) =aandq(1) =f(b0) =b. Thus, any two points of Yare connected by a path, meaning that Yis path connected. All we required in the above proof is that fwas onto and continuous. Thus, the continuous image of a path connected set is path connected, using the same proof. 3.Prove that local connectedness is a topological invariant, but that the continuous image of a locally connected set need not be locally connected. Letf:X!Ybe a homeomorphism and assume that Xis locally connected. Let p2Ycontained in UYan open set. We need to show that there is an open, connected set containing pand contained in U. Letx=f1(p). Thenx2f1(U) which is open in X. SinceXis locally connected, there is an open connected set, C, containing xand contained in f1(U). We then know that f(C) is connected and p=f(x)2f(C). Sincefis a homeomorphism, fis an open map and f(C) is open. Hence,Yis locally connected. Note that this time, we did need the fact that fis a homeomorphism to guarantee thatf(C) was open. Thus, we should not expect the continuous image of a locally connected set to be locally connected, because not every continuous map is open. As an example, take the identity function that maps the rationals with the discrete topology to the rationals with the usual topology. We showed in class that the rationals with the usual topology is not locally connected. However, any discrete space is locally connected, specically because each point is an open set. 4.Prove that every locally path connected space is locally connected. LetXbe a locally path connected space and let x2Xbe contained in an open set U. Now,Ucontains an open path connected set containing x, and any path connected c David C. Royster Introduction to Topology For Classroom Use Only MATH 4181 Problem Set 6 Solutions 2 set is connected. Thus, Ucontains an open connected set containing X. Hence,Xis locally connected. 5.Prove that every nite subset of a topological space is compact. LetA=fa1;a2;:::;angbe a nite set in X. LetObe an open cover of A. Eachai must be contained in at least one open set Oifrom the open cover O. Hence, we only need the setsfO1;O2;:::;Ongto coverA. This makes Acompact. c David C. Royster Introduction to Topology For Classroom Use Only