Could you, who enjoys Math, give your opinion about this proof, please?

Question

Some months ago I asked the question at http://answers.yahoo.com/question/index;_ylt=AoZQQ796ABG1xfCSMoOOT9pIzKIX;_ylv=3?qid=20070326151241AAxToH1
about bilateral condensation points of R. We should prove that, if S is an uncountable subset of R, then the set of it's bilateral condensation points, as well as the set of the bilateral condensation points of S that are in S, is uncountable. Then, I came up with a proof cited in the additional details to the original question. A bit messy, but I think it's right. I would like opinions of people who, like you, enjoy such problems.
Thank you

Definitions: If S is a subset of a topological space, we say x is a condensation point of S if every neighborhood of x contains uncountably many elements of S. We can prove that, if S has a countable topological base and S is uncountable, then the set of condensation points of S is uncountable. In the case of R, there are 2 kinds of condensation points:

Bilateral, if, for every eps >0, both the intervals (x-eps, x) and (x, x +eps) contain uncountably many elements of S. Like 0 in S= [-1, 1].

Unilateral, if only one of the intervals (x-eps, x) and (x, x +eps), but not both, contains uncountably many elements of S. Like -1 and 1 in S = [-1, 1]

Answer 1

I read the proof and it is interesting.
There are some things that I don't get:
2.For each n, (a_n, b_n) contains no condensation point of S and,
therefore, (a_n, b_n) Inter S is countable

Why?

The rest of proof is correct for me.
cheers

Answer 2

I star it for you steiner, so someone more knowledgeable than me takes notice.

Answer 3

me 2 fan of maths, but cant understand ya!!!