Nothing as good as a warm bath to refresh the ideas :-)
I wrote:
>
> I didn't undestand your proof. I would prove it somehow
> like: let Acc(X) be the set of accumulation points of X.
>
> _a_ is in Acc(X) iff (every open set that contains _a_
> contains also an infinite number of elements of X)
>
> So, if Acc(X) is enumerable and X is bounded, then
> every open set that contains Acc(X) must include all
> but a finite number of elements of X [otherwise there
> would be an accumulation point not in Acc(X)]
>
> Then I thought about building a shrinking sequence
> of open sets containing Acc(X), so that X would be
> included in a countable union of countable sets, plus
> the intersection of the shrinking open sets - and I
> can't prove that this intersection _is_ Acc(X)
>
Duh, because any element _b_ that belongs to the intersection
of those shrinking open sets is as close as we wish to
any element of Acc(X) (take _a_), and since this element
_a_ is as close as we wish to any element of X (say, _x_), then
(using those epsilon / 2 + epsilon / 2 = epsilon stuff) it
results that _b_ is an accumulation point, etc
QED
Alberto Monteiro
