Re: Accumulation points in metric space

Adam wrote:

Suppose (X,d) is a metric space such that for every
uncountable subset A of X there exists an accumulation
point in X. Prove that (X,d) is separable.

The contrapositive is: If (X,d) is not separable, then
there exists an uncountable subset having no accumulation
point in X. In fact, I believe even more is true: If (X,d)
is not separable, then there exists an uncountable subset
that is uniformly isolated (i.e. there exists epsilon > 0
such that the distance between each pair of distinct points
in the uncountable subset is at least epsilon). If this is
important for something you're working on (and not just
a homework problem), let me know and I'll look it up when
I get home this evening. (It's sketched out in my copy of
Willard's topology text, which I don't have with me right now,
and I don't feel like trying to work it out either.)

Dave L. Renfro

.

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