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MATH 4181 Problem Set 5 Solutions 1 MATH 4181 001 Fall 1999 Problem Set 5 Solutions 1.Letf:X!Ybe a function. Then fis an open mapping if for each open set OX,f(O)is open inY. (a)Give an example of a mapping that is continuous, but not open. LetX=Y=R. LetTXdenote the discrete topology on XandTYdenote the usual, metric topology on Y. Denef: (X;TX)!(Y;TY) byf(x) =x. Then,f is a continuous mapping, because the inverse image of any set is open in ( X;TX). However,fis not open since f(f0g) =f0gwhich is not open in ( Y;TY). (b)Give an example of a mapping that is open, but not continuous. Use the same notation as in (a) and dene g: (Y;TY)!(X;TX) byg(x) =x. Then,gis open, since the image of any set is open in ( X;TX), butgis not continuous since, for example, the inverse image of the open set f0gis not open in (Y;TY). (c)Prove that a one-to-one, onto mapping f:X!Yis a homeomorphism if and only iffandf1are open mappings. Letfbe a homeomorphism. Then we know that fandf1are both continuous. We also know that f= (f1)1, so that the statement that fis open is equivalent to saying that f1is continuous. Likewise, the statement that f1is open is equivalent to saying that fis continuous. Thus, fandf1are open mappings. If we know that fis one-to-one and onto, then assuming that fandf1are open mappings makes fandf1both continuous. Thus, fis a homeomorphism. 2.Prove that a nite subset Aof a Hausdor space Xhas no limit points. Conclude that Amust be closed. LetA=fx1;:::;xngand letpbe a limit point of A. Then, for every open set, U, containingpwe must have that U\Anfpg6=;. SinceXis Hausdor, there are open setsUicontainingpandVicontainingxiso thatUi\Vi=;. LetU=\n i=1Uiand let V=[n i=1Vi. Then, (i)p2U; (ii)AV; (iii)U\V=;. Thus, we found a neighborhood of pwhich does not intersect A. Hence,pcannot be a limit point of AandAhas no limit points. SinceA0=;, we have that A=A[A0=A, makingAclosed. c David C. Royster Introduction to Topology For Classroom Use Only MATH 4181 Problem Set 5 Solutions 2 3.IfXis a space which is homeomorphic to a subspace Aof a spaceY, thenXis said to beembedded inY. Give an example of spaces AandBfor whichAcan be embedded inBandBcan be embedded in A, butAandBare not homeomorphic. (Simple examples can be found in R.) LetA= (0;1) and letB= [0;1]. Clearly, Acan be embedded in Busing the identity function,f(x) =x. Dene a map g:B!Abyg(x) =1 2x+1 4. This maps Bonto the interval [1 4;3 4] embedding it into A. Thus, we can embed AintoBandBintoA. However, we know that the two intervals (0;1) and [0;1] are not homeomorphic. 4.Prove that every countable subset of Ris totally disconnected. We shall prove in Problem 7a that the property of being totally disconnected is a topological invariant. Since we have shown that Qis totally disconnected. Since Qis countable it is homeomorphic to every countable subset of R. Hence, every countable subset ofRis totally disconnected. 5.Give examples of subsets AandBinR2to illustrate each of the following. A drawing is sucient. (a)AandBare connected, but A\Bis disconnected. LetAbe the segment on the x-axis,A=f(x;0)j1x1g. LetBdenote the upper hemisphere of the unit circle: B=f(x;y)jx2+y2= 1 andy1g. Then each of AandBis connected but the intersection is the two disjoint points (1;0) and (1;0) which are disconnected. (b)AandBare connected, but AnBis disconnected. LetAbe the rectangle in the plane with vertices at (2 ;1), (2;1), (2;1) and (2;1). LetBbe the square in the plane with vertices (1 ;1), (1;1), (1;1) and (1;1). ThenAnBis two disconnected squares. (c)AandBare disconnected, but A[Bis connected. TakeAto be the two vertical sides of the unit square: A=f(1;t)j1t 1g[f 1;t)j1t1g. TakeBto be the two horizontal sides of the unit square:B=ft;1)j1t1g[ft;1)j1t1g. Then, each of AandB is disconnected, but A[Bis connected. (d)AandBare connected and A\B6=;, butA[Bis disconnected. LetA= (0;1) andB= (1;2). ThenAandBare connected, A\B= [0;1]\[1;2] =f1g6=;; butA[B= (0;1)[(1;2) is disconnected. c David C. Royster Introduction to Topology For Classroom Use Only MATH 4181 Problem Set 5 Solutions 3 6.Denition: A Hausdor space Xis0-dimensional ifXhas a basis Bof sets which are simultaneously open and closed. Prove that every 0-dimensional space is totally disconnected. LetXbe a 0-dimensional space. Let Cbe a component of Xcontaining the point x2X. Lety2C. SinceXis Hausdor, there are disjoint open sets, UandV, containing xandy, respectively. Now, each of these open sets consists of a union of basic open sets, thus, we can separate xandyby disjoint basic open sets. Thus, there are setsUxandVy, disjoint and open and closed, since Xis 0-dimensional. Then, C contains sets which are both open and closed, making Cnot connected. The only way in which this can be prevented is for C=fxg. Thus, each component consists of a single point and Xis totally disconnected. 7.Prove: (a)The property of being totally disconnected is a topological invariant but not a continuous invariant. Letf:X!Ybe a homeomorphism and assume that Xis totally disconnected. LetCbe a connected component of Y. Thenf1(C) is a connected subset of X, being the continuous image of a connected set. If x2f1(C), thenf1(C) lies in the connected component of Xcontainingx. SinceXis totally disconnected, this component is x. Thus,f1(C) is a single point. Since fis a homeomorphism, C consists of a single point. Thus, Yis totally disconnected. Let (X;T) denote the reals with the discrete topology. Then ( X;T) is totally disconnected. Let ( Y;S) denote the reals with the usual topology. Dene f:X! Ybyf(x) =x. This function is continuous, since the domain has the discrete topology, but the image space is connected, not totally disconnected. (b)The property of being totally disconnected is hereditary. LetXbe totally disconnected and let AX. LetCAbe the component of Acontainingx2A. Lety2C. Now,yis disconnected from xinX, sinceXis totally disconnected. The same disconnection in Xwill disconnect xandyinA. Hence,Ccannot contain more than one point, and Ais totally disconnected. c David C. Royster Introduction to Topology For Classroom Use Only