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

**21.**

**Problem: **Let and be separated subsets of , suppose that and define

Let . a) Prove that are separated subsets of . b) Prove that there exists such that . c) Prove that every convex subset of is connected.

**Proof:**

**a) **Clearly we have that and so, if we assume that then we must have that . So, we may choose such that and so it easily follows that and thus so and thud which is a contradiction.

**b) **We must merely note that if then it would be a path from to which is impossible since they are separated.

**c)** We can prove this more generally. Let be a normed vector space and let be convex, then is path connected. This clearly follows since the straight line is a path from to .

**22.**

**Problem:** A metric space is called *separable* if it contains a countable dense subset. Show that is separable.

**Proof:** We prove the much, much deeper following theorem.

**Lemma:** Let be a countable class of non-empty separable topological spaces with corresponding countable dense subsets . Then, is separable with the product topology.

**Proof: **For each select an arbitrary but fixed and define where

Clearly we have that and thus countable (the finite product of countable sets is countable). So, let

Then , being the countable union of countable sets, is countable. We now claim that it is dense in . To see this let be arbitrary and any basic neighborhood of it. Then, there are only finitely many indices such that . So, let . Now, since for each we have that is a neighborhood of in there is some . So, doing this for we arrive at points . So, then

Which proves that is dense. The conclusion follows.

The above is kind of heavy duty (though it works since the product topology and usual topology coincide). For people who are more interested in analysis, the *real* way to do this is to note that is countable and note that for any you can choose corresponding such that and do the simple calculation.

**23.**

**Problem:** A collection of open subsets of ( a metric space) is said to be a *base* if for every and every neighborhood of it there is some such that . Prove that every separable metric space has a countable base (in other words that it’s second countable).

**Proof:** Let be the countable dense subset and let . Clearly is countable . To see this we merely note that is clearly a surjection. We shall now prove that (as it’s letter suggests) is a base. So, let be arbitrary and be given. Since is dense in there exists some . So, let and choose such that . then, and if we have that and thus . The conclusion follows.

**24.**

**Problem:** Let be a metric space in which every infinite subset of it has a limit point. Prove that is separable.

**Proof:** Consider the open cover and suppose that it did not have a finite subcover. Then, we may form a sequence such that , for otherwise if there existed some such that for no we have that then for every we have that

or in other words that contradicting the assumption that has no finite subcover. But, let and consider clearly contains at most one point of . To see this suppose that it contained more than one point, say then we may assume WLOG that but this contradicts the construction of the sequence since . It follows that has a finite subcover. So, let for each we may cover with finitely many open balls of the form . So, let be the finite set of points such that and let . Clearly is countable and given any and any neighborhood of it we may (by the Archimedean principle) find some such that . But, since every point of is less than in distance from some point of it follows there is some . The conclusion follows.

*Remark:* All we really showed above (the hard part anyways) is that every limit point compact metric space is totally bounded. As we prove in the next theorem limit point compactness is implied by compactness in metric spaces (they are in fact equivalent) and thus every compact metric space is totally bounded. It actually is true that any space such that every there is a *countable* -net is separable, this is done precisely as in the last part of the above.

**25.**

**Problem:** Prove that every compact metric space has a countable base.

**Proof: **It suffices to show that every infinite subset of has a limit point, from where it will follow from 24. that it is separable, and thus second countable. So, let have no limit point. Then, for each there exists some open ball such that . So, we may cover and so by assumption this open cover must admit a finite subcover . So,

The conclusion follows.

**26.**

**Problem:** Let be a metric space such that every infinite subset has a limit point. Show that is compact.

**Proof:**

**Lemma:** Let be a limit point compact metric space and a sequence of descending () of non-empty closed subsets of . Then,

**Proof:** For each choose some . We break this into two cases, namely whether the set of all these ‘s is finite or infinite.

**Case 1:** Suppose that is finite but . We first note that for each there is some such that . To see this assume not and let then otherwise and thus which was assumed to be impossible. So, for all and thus which contradicts our assumption that it was empty. So, for each there exists some for which the above idea is true. Let then which contradicts the construction of the set. It follows that the intersection is non-empty.

**Case 2:** Suppose that is infinite. By assumption it then has a limit point . So, we show that by showing that it is contained in a fixed but arbitrary . So, suppose that then which is open. But, since it follows that is finite contradicting that is a limit point of that set. It follows that .

So, now let be an open cover for . Since is second countable we know that admits a countable subcover . Now, suppose that for every then where is a sequence of decreasing, non-empty closed sets. But,

Which contradicts the lemma. The conclusion follows.

**27.**

**Problem:** Let be a metric space and call a *condensation point *of a set if every neighborhood of contains uncountably many points of . Now, suppose that is uncountable and let be the set of all condensation points of . a) Prove that is perfect and show that is countable.

**Proof:**

**a) **Let but suppose that then there exists some neighborhood of it such that for every there exists a neighborhood of it such that . So, we now need a quick lemma

**Lemma: **Let be a second countable topological space and any subset of . Then, if is a collection of open sets in such that then there exists some countable subcollection such that

**Proof:** For each there is some such that but if is the countable base there is some such that . Doing this for each clearly gives a collection such that . But, by design it is clear that is countable and so for each we may choose some (since this is how they were begotten) some such that . So, doing this for each we arrive at a countable collection such that

from where the conclusion follows.

So, note that since is second countable and we clearly have that is a collection of open sets whose union contains and so by the above lemma it follows that it must admit a countable subcover so then

which contradicts that . So, this proves that . Conversely, let then every neighborhood of contains a point of , but since is also a neighborhood of that point it follows that from where it follows by the arbitrariness of that .

**b) **Let . We claim that . To see this let then there is some such that and thus . Conversely, let then there exists some neighborhood of such that but there exists some such that . Thus, . So, we see that

The last part gotten from the fact that is countable and each also countable. The conclusion follows.

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