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MATH 4181 Problem Set 4 Solutions 1 MATH 4181 001 Fall 1999 Problem Set 4 Solutions 1. (Exercise 4, page 74) Let(X;T)be a topological space. Prove that ;andXare closed sets, that a nite union of closed sets is a closed set, and that an arbitrary intersection of closed sets is a closed set. Let (X;T) be a topological space. To show that a set is closed we must show that its compliment is open. Since XnX=;is open,Xmust be closed. Likewise, since Xn;=Xis open, the empty set is closed. LetfFign i=1be a nite collection of closed sets. To show that their union is closed, we must show that the complement of the union is open. Xn n[ i=1Fi! =n\ i=1(XnFi) and each of the sets XnFiis open since Fiis closed. Since the nite intersection of open sets is open, we have then that Xn(Sn i=1Fi) is open, hence makingSn i=1Fi closed. LetfFj2Igbe a collection of closed sets. To show that the arbitrary intersection is closed, we must show that its complement is open. Xn \ 2IF! =\ 2I(XnF) and each of the sets XnFis open since Fis closed. Since the arbitrary union of open sets is open, we have then that XnTn 2IF
is open, hence makingTn 2IF closed. 2. (Exercise 6, page 75) Prove that in a discrete topological space, each subset is simulta- neously open and closed. We already know that each subset is open, by the denition of the discrete topology. LetAX. ThenXnAis a subset of Xand hence open. Thus, Ais closed. Therefore, every subset of a discrete topological space is simultaneously open and closed. 3.Show that a topological space (X;T)is discrete if and only if each set consisting of only one point is open. If (X;T) is discrete then every subset is open. Hence the singleton sets are open. If (X;T) has the property that the singleton sets are open, then each subset of X being a union of its elements is a union of open sets and is thus open. Thus, every subset is open, making ( X;T) a discrete topological space. c David C. Royster Introduction to Topology For Classroom Use Only MATH 4181 Problem Set 4 Solutions 2 4.LetXbe a set and let T0the nite complement topology for X. (a)Show that (X;T0)is discrete if and only if Xis a nite set. IfXis a nite set, then the complement of a singleton set is nite. Thus, the sin- gleton sets are open. By the previous problem that gives Xthe discrete topology. If the nite complement topology and the discrete topology coincide on the space X, then the singleton set fagmust be open (by the discrete topology). Since it is open, its complement must be nite (by the nite complement topology). But X=fxg[(Xnfxg. Each of the two sets on the right are nite, making Xa nite set. (b)Show that if Ais an innite subset of X, then every point of Xis a limit point ofA. LetAbe an innite set and let x2X. IfUis any open set containing x, then XnUis a nite set. If AXnU, then we would have a nite set with an innite subset | which is impossible. Thus, A\U6=;. This means that xis a limit point ofA. Sincex2Xwas arbitrary, each point of Xis a limit point of A. 5.LetXbe a set. The countable complement topology, or co-countable topology, T00for Xconsists ofX,;and all subsets OofXfor whichXnOis a countable set. (a)Show that T00is a topology on X. Clearly,;andXare inT00, by denition. LetfOj2Igbe a collection of open sets. Then, XnOis countable in X. To see thatS 2IOis open, we need look at XnS 2IO. Xn[ 2IO=\ 2I(XnO)XnO: Since each complement is countable, and the subset of a countable set is countable, we have that XnS 2IOis countable andS 2IOis open. LetfOign i=1be a nite collection of open sets. Then, XnOiis countable in X. To see thatTn i=1Oiis open, we need look at XnTn i=1Oi. Xnn\ i=1Oi=n\ i=1(XnOi)XnOn: Since each complement is countable, and the union of a nite number of countable sets is countable, we have that XnTn i=1Oiis countable andTn i=1Oiis open. (b)For the space (X;T00), show that a countable set AofXhas a derived set A0=; and that an uncountable set BhasB0=X. LetAbe countable and let x2X. Ifx62A, thenXnAis an open set containing xandA\XnA=;. Ifx2A, then (XnA)[fxgis an open set containing aand (Anfxg)\(XnA[fxg= ;. c David C. Royster Introduction to Topology For Classroom Use Only MATH 4181 Problem Set 4 Solutions 3 In neither case is xa limit point of A. Hence, no points are limit points of A, or A0=;. IfBis uncountable, let x2Xbe any point and UXbe any open set with x2U. SinceXnUis countable, we cannot have BXnU. This means that U\B6=;. Thus, each point is a limit point for BandB0=X. (c)Show that the intersection of any countable family of members of T00is a member ofT00. LetfOig1 i=1be a countable collection of open sets. Then, XnOiis countable in X. To see thatT1 i=1Oiis open, we need look at XnT1 i=1Oi. Xn1\ i=1Oi=1\ i=1(XnOi)XnO1: Since each complement is countable, and the union of a countable number of countable sets is countable, we have that XnT1 i=1Oiis countable andT1 i=1Oiis open. 6.LetX=fa;bgbe a two-element set and let T=f;;fag;fa;bgg. Show that Tis a topology on Xand identify the limit points of each subset of X. (This space is called Sierpenski space.) Clearly, Tcontains the empty set and the whole space. It is easy to see that it is closed under unions and intersections. Thus, it is a topology on X. For the limit points we have Limit Points ;; fag; fbg; fa;bgfbg c David C. Royster Introduction to Topology For Classroom Use Only MATH 4181 Problem Set 4 Solutions 4 7.How many dierent topologies are there for a set with three members? Since there are 8 subsets of this space, there are at most 28possibilities for topologies. Fortunately, there are not that many. It is probably best just to list the collections that form a topology on this set X= fa;b;cg. T1=f;;Xg T2=f;;fag;Xg T3=f;;fbg;Xg T4=f;;fcg;Xg T5=f;;fa;bg;Xg T6=f;;fa;cg;Xg T7=f;;fb;cg;Xg T8=f;;fag;fa;bg;Xg T9=f;;fag;fa;cg;Xg T10=f;;fag;fb;cg;Xg T11=f;;fbg;fa;bg;Xg T12=f;;fbg;fa;cg;Xg T13=f;;fbg;fb;cg;Xg T14=f;;fcg;fa;bg;Xg T15=f;;fcg;fa;cg;Xg T16=f;;fcg;fb;cg;Xg T17=f;;fag;fbg;fa;bg;Xg T18=f;;fag;fcg;fa;cg;Xg T19=f;;fbg;fcg;fb;cg;Xg T20=f;;fag;fa;bg;fa;cg;Xg T21=f;;fbg;fa;bg;fb;cg;Xg T22=f;;fcg;fa;cg;fb;cg;Xg T23=f;;fag;fbg;fa;bg;fb;cg;XgT24=f;;fag;fbg;fa;bg;fa;cg;Xg T25=f;;fag;fcg;fa;bg;fa;cg;XgT26=f;;fag;fcg;fa;cg;fb;cg;Xg T27=f;;fbg;fcg;fa;bg;fb;cg;XgT28=f;;fbg;fcg;fa;cg;fb;cg;Xg T29=f;;fag;fbg;fcg;fa;bg;fa;cg;fb;cg;Xg c David C. Royster Introduction to Topology For Classroom Use Only