Just For Fun(Rudin’s Topology Section) Part II
Problem: Form a compact subset of the real numbers (with the usual topology) which has countably many limit points.
Proof: Let which is tediously, but easily verified. Sorry folks, I’m not interested enough to Tex it all out. Comment me asking for a full solution if you so wish.
Problem: Given an open cover (with the usual topology) which has no finite subcover.
Proof: How about . So, we must first show that this in fact covers . So, let be arbitrary and let . Then, we have that and so by the Archimedean principle there exists some such that and thus . Now, to see that no finite subcover works, let be any finite subclass of of . Then, let . Now, it clearly follows that and so we mus merely show that there is some . But, we merely need realize that and so the conclusion follows.
Problem: Show that the intersection of arbitrarily many closed subsets of a metric space with the finite intersection property may be empty if they are just closed, or just bounded.
Proof: Just take for example. Then, the class is a class of closed subsets of with the FIP but their intersection is empty. To see this we merely note that for each we have that
Also, if we just have bounded then the claim is also wildly false in . Take . This is a class of bounded subsets of with the FIP, but clearly since choosing we have that for some (by the Archimedean principle) and thus .
Problem: Regard as a metric space with . Let . Prove that a) is closed in , b) it is bounded in , but it is not compact in .
a) We first prove a nice little lemma (which is almost immediate from theorem 2.30)
Lemma: Let be a metric space and a subspace (in the metric space sense, don’t get too ahead of yourself with topological terms…we haven’t yet proven they are equivalent), then is closed in it if and only if for some set which is closed in .
Proof: First suppose that is open, then is open and thus by theorem 2.30 we have that for some set closed in . A little set theoretic self-convincing then shows that .
Conversely, suppose that for some set closed in . Then, (once again with a little set theoretic magic…be sure to be mindful of what set theoretic difference means here!) and thus by theorem 2.30 again we see that this means that is open and thus closed.
From this it immediately follows that is closed in since evidently and is closed in .
b) It is evidently bounded.
c) We prove that is not compact in by a roundabout but instructive way.
Lemma: Let be a topological space and a subspace. Then, is a compact subspace of if and only if it’s a compact subspace of .
Proof: First suppose that is a compact subspace of and let be an open cover as a subspace of where is open in . Then, since is a subspace of it evidently follows that is open in for every . Thus, it is evident that is an open cover of as a subspace of . So, by assumption it admits a finite subcover and so it readily follows that is a finite subcover of
Conversely, suppose that is compact as a subspace of and is an open cover for as a subspace of . Then, since is a subspace of we have that for some open in . Thus, it is evident that is an open cover for as a subspace of and so by assumption it admits a finite subcover from where a quick check shows then that is the desired finite subcover of .
From the above lemma we see that will be compact as a subspace of if and only if it is compact as a subspace of . But, it is not since evidently is a limit point of but not a point of , and thus is not closed and thus not compact.
Problem: Let ; a) Is countable? b) Is dense in ? c) Is dense in ? d) Is perfect in ?
a) No, is not countable. It is evident that . The only one that may be slightly hazy is that but it is pretty routine to check that is a bijection. (this is just the correspondence between subsets and indicator functions).
b) Of course it isn’t dense. It is clear that .
c) It suffices from prior comment to show that it’s compact as a subspace, and by virtue of the Heine-Borel theorem we must merely show that is closed and bounded. Boundedness is clear though since . So, suppose that but . Then, we have that where for some . But, note then that for any we have that and thus which of course is a contradiction. Thus, the conclusion follows.
d) It suffices to prove (since we proved in c) that is closed) that . To do this let and let be given. Then, by the Archimedean principle there exists some such that and so if then we have that
Where obviously we choose to be the opposite of what is. Then, and . It follows that .
Problem: Is there a non-empty perfect subset of (with the usual topology) which contains no rational values?
Proof: The answer is yes, but to say it right takes some measure theory. We know that (the usual measure) and so there exists countably many open intervals such that and . So, we have that is closed, non-empty (since it is non-zero measure). So, once again since it has positive measure it must be uncountable and by problem 28 it follows that the set of all condensation points of is perfect, and since each condensation point is a limit point it must be a point of and thus not a rational number.
Problem: a) If are disjoint closed sets in a metric space then they are separated. b) If are disjoint and open they are separated. c) Fix and define and , and prove that are separated. d) Prove that every connected metric space with more than one point is uncountable.
a) Clearly are separated since .
b) Suppose that then it must follow that there is some . But, since every neighborhood of intersects and thus intersects and thus which is a contradiction. An identical argument works for the other cases.
c) Merely note that is continuous (since it’s the normal metric restricted to ). And, that which are disjoint and open. The conclusion follows from b)
d) Suppose that were countable and has more than one point, then so is and then so is . So, choose to be distinct and let . But, since is uncountable and countable there exists some . So, let and . Then, are non-empty ( and disjoint and since their union is equal to . It follows that is not connected.
Problem: Are closures and interiors of connected sets always connected?
Proof: Closures are, we reprove this fact for general topological spaces.
Lemma: Let be a topological space, a connected subspace and then is a connected subspace.
Proof: Evidently . So, let (where is the two-point discrete space) be continuous. Then since is connected must have that is constant, but since is dense in , continuous and Hausdorff it follows that is constant. The conclusion follows.
The interior of connected subspaces is not always connected though. Consider the “dumbbell” shape . This is clearly connected (it is the union of three connected sets with non-empty intersection) but it’s interior is which is as classically disconnected as one can get.
Remark: It is interesting to note that the result is true for , since the only connected subsets are interiors which (as can easily be checked by case) the interiors of which are intervals.