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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