Can you give me your opinion abou this proof, please?

Question

Some months ago I posted the question at

http://answers.yahoo.com/question/index;_ylt=AiIJp41Ma4ni_sOh.AnJaR7J6gt.;_ylv=3?qid=20070326151241AAxToH1

involving condensation points of R. In the additional details of this question I gave my proof to the posted assertion, correcting some mistakes. I'd like you, if you into these things, to give me your opinion about my proof, please.

In addition, I propose an interesting problem: Show that, if A is an uncountable subset of R, then A contains a subset S with the in-between property, that is, for every s1 and s2 in S, there is s3 in S between s1 and s2.

In order to avoid those vacuoulsly true statements, S must contain at least 2 elements (otherwise, any subset {s} would satisfy the condition).

Thank you

Ok, let's see a proof. Let M be a separable metric space, A an uncountable subset od A and C the set of all condensation points of A (each a
in A has a neighborhood which contains uncountably many elements of A). Then, since M is separable, M has a countable topological basis {B1, B2...B_n...}. Put W = {intersection of all B_n that contains only countably many elements of A}

Lemma: W = C'

Proof:
If w is in W, then w is in some basic neighborhood B_m that intersects A according to only countably many elements. It follows w is not a condensation point of A and, therefore, w is in C'. Hence, W is subset of C'

If w is in C', then w is not a condensation point of A and, therefore, lies in some neighborhood - hence in a basic neighborhood B_m - whose intersection with A is countable. From the definition of W, it follows w is in W. Hence, C' is subset of W, completing the proof of the Lemma.
(to be continued)

(continued)
Now, we observe that W inter A is given by Union (B_n_k Inter A), where B_n_k is a subcollection of the basic neighborhoods. So, W inter A = C' Inter A is given by a countable union of countable sets, which implies it's countable. Since A is uncountable, an immediate consequence is that C inter A is uncountable. That is, the set of condensation points of A that belongs to A is uncountable. Of course C is uncountable, too.

In the case of separable metric spaces, an easy way to see C is closed is observing that, being given by an union of open sets, C' is open. So, C is closed.

But this is not restricted to separable metric spaces. In any topological space, if c is in C' then c has a neighborhood V that contains ns only countably elements of A. Since V is a neighborhood of each of its elements, this is true for every v of V, so that V is in C'. This shows every element of C' is an interior point, so that C' is open and C is closed.

In the case of R, B is uncountable and U is countable. B doesn't have to be open and U doesn't have to be closed. But, as we've seen in the general case, C = B union U is closed.

A TYPO: W is the union, not the intersection, of course, of the B_n whose intersection with A is countable. Sorry.

CLARiFYING:

If T is any topological space and A is a subset of T, we say x is a condensation point of A if every neighborhood of x contains uncountaby many elements of A. Like in the case of accumulation points, x doesn't need to be in A.

In the case of R, an ordered field, we also have unilateral and bilateral condensation points.

Examples.

1 is condensation pont of (0,1) and of [0,1]. In both cases, an unilateral condensation point.

1/2 is a bilateral condensation point of (0,1) and of course of [0,1]. It's also a bilateral condensation point of [0, 1/2) U (1/2 , 1].

0 is an accumulation point, but not a condensation point, of {1, 1/2, 1/3,....1/n....}

If a topological space T has a countable topological base (which includes the case of separable metric or metrizable spaces), A is an uncountable subset of A and C is the set of condensation points of A, the C is uncountable.

Answer 1

@Steiner:
It is generally bad form to mix definitions as you did in your proof. M being separable has nothing to do with your proof; it is the fact that M is second countable that matters. Yes, being separable and second-countable are equivalent for metric spaces, but since your proof is topological in nature, there is no reason to make extraneous assumptions. Simply say "M is second-countable".

Steve