## Just For Fun(Rudin’s Topology Section) Part IV

**28.**

**Problem:** Prove that ever closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is a set which is countable.

**Proof:** It is easy to see that the above proof (proof 27.) holds in any space which is second countable, in particular for separable metric spaces. Thus, if is closed we surely have that . Thus, but since and we have then that which is the union of a perfect and countable set respectively.

**29.**

**Problem:** Prove that every open set may be written as the disjoint union of countably many open intervals.

**Proof:** We need to prove a quick lemma

**Lemma:** Let be a topological space and let be a class of connected subspace of such that then is a connected subspace of .

**Proof:** Suppose that is a separation of . We may assume WLOG that for some . So, now we see that otherwise would be non-empty disjoint subsets of whose union is contradicting that is connected. Thus, it easily follows that for any we have that so that and thus by a similar reasoning we see that . Thus, since was arbitrary it follows that contradicting that .

So, now for each define

And let and finally we prove that

is a countable class of disjoint open intervals whose union is . The fact that each is an open interval follows since it is the union of open connected spaces with non-empty intersection (each element of contains ) it is also an open connected subspace of (note that each element of is open in but since is open it is also open in . But, it was proven in the book the only connected subspace of are intervals and thus is an interval for each .

Now, to see that they are disjoint we show that if then from where the conclusion will follow. So, to see this we first note that if is non-empty then is an open connected subspace of containing both and and thus and but this implies that and respectively from where the conclusion follows.

Now, to see that is countable we notice that for each we have that and so if we let be a fixed but arbitrary then

is an injection since the elements of are pairwise disjoint. The fact that is countable follows immediately.

Thus, is a countable collection of open intervals and

Thus, the conclusion follows.

**30.**

**Problem:** Prove that if where each is a closed subset of then at least one has non-empty interior.

**Proof:**

**Lemma: **Let be a complete metric space and a descending sequence of non-empty closed subsets of such that . Then, contains one point.

**Proof:** Clearly does not contain more than one point since . So, now for each choose some and let . Now, if were somehow finite a similar argument to that used in the first case of problem 26. is applicable, so assume that is infinite. But, since it is evident that is a Cauchy sequence and thus by assumption it converges to some point . We claim that . To see this we note similarly to problem 26. that since is infinite it is easy to see that is a limit point of and so if for some then is a neighborhood of containing only finitely many points of which clearly contradicts that it is a limit point. The conclusion follows.

So, now suppose each had empty interior (i.e. nowhere dense) . Then, since is open and is nowhere dense there exists an open ball of radius less than one such that . Let be the concentric closed ball of whose radius is half that of . Since is nowhere dense contains an open ball of radius less than one-half which is disjoint from . Let be the concentric closed ball of whose radius is one-half that of . Since is nowhere dense we have that contains an open ball of radius less than one-fourth which is disjoint from . Let be the concentric closed ball of whose radius is half that of . Continuing in this we get a descending sequence of non-empty closed subsets for which . Thus, by our lemma we have that there is some . This point is clearly not in any of the ‘s from where the conclusion follows.

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