## Miscellaneous Rudin Like Topology questions

**Problem: **Let be a second countable topological space and be such that . Then,

**Proof:** Suppose that is countable then is uncountable. But, by definition for each there exists some neighborhood of it such that . So, clearly is an open cover for and since is second countable Lindelof’s theorem guarantees that it must have a countable subcover . Thus,

But this contradicts that is uncountable. The conclusion follows.

**Corollary:** If and is the set of all isolated singularities then

**Corollary: **Let be a separable metric space and uncountable, then there exists a sequence in such that .

**Problem: **Let be a metric space and perfect and open. Then, is perfect.

**Proof:** The fact that follows from the fact that it’s closed. Now, let and let be any neighborhood of it. Then, by definition we have that and so there exists some . Now, since and is a neighborhood of it it follows that there are infinitely many values of in it. In other words is infinite and in particular there exists some such that . Thus, . The conclusion follows.

**Corollary:** If is a metric space and is perfect we can choose and see that and we have a perfect subset of which is also bounded.

**Problem:** Let be a compact metric space. Prove that is connected if and only if it cannot be written as a union with

**Proof: **It is tacitly assumed that . So, in any metric space suppose that with . Remember that and thus if we have that and thus . Thus, (the reverse direction gotten using the exact same logic) from where it follows that is disconnected.

Before we continue to prove the converse we prove a nice little lemma

**Lemma:** Let be a compact metric space and non-empty disjoint closed subspaces of . Then, .

**Proof:** Since is compact and closed subspaces it follows that they themselves are compact and thus so is . So, notice that is continuous since it is the distance function (trivially continuous) restricted to . Thus, by the extreme value theorem it follows that for some and since we see that .

So, now assume that were not connected. Then, where are non-empty disjoint closed subsets of . But, by the lemma this contradicts the assumption that cannot be written as with .

**Problem:** Let be a separable metric space, then for any subspace we have that is separable.

**Proof:** We prove something stronger. But, first we prove a small lemma

**Lemma:** Let be a second countable topological space, then any subspace is second countable.

**Proof:** Let be the countable base for and let . Clearly is a countable collection of open subsets of . Now, let be arbitrary and let be any neighborhood of it. Then, by definition for some open set in . So, we may find some such that and thus and since the conclusion follows.

So, if is a separable metric space, then as proven earlier it is second countable and thus so is (by the lemma) but since every second countable space is trivially separable the conclusion follows.

**Problem: **Let be a compact metric space and an isometry. Then, is surjective.

**Proof:** Suppose and consider . Since is a compact metric space this must have a convergent subsequence . In particular, that sequence must be Cauchy. So, noting that are disjoint non-empty compact subspaces of a compact space it follows by an earlier lemma that 4. So, since is Cauchy there exists such that . So,

which is a contradiction.

**Problem:** Let be a compact metric space and be a function such that

for all distinct . Prove that has precisely one fixed point.

**Proof:** Clearly if there is a fixed point it’s unique because if then . Now, to see that it must have a fixed point we note that

is continuous since evidently the diagram

commutes. So, since is continuous and it follows that obtains a minimum at some . Now, assume that then contradicting that achieved it’s minimum at . The conclusion follows.

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