Robert Shaw wrote:
>
>That's true but not quite what I'm saying. For uncountably
>infinite sets I think there's a stronger statement: any
>uncountable set must accumulate at set of points whose
>minimum distance from each other is zero.
>
I guess this is true, but the proof isn't trivial.
>In any interval between accumulation points whose endpoints
>aren't accumulation points there can only be a finite set of
>points or there'd be a third accumulation point in the interval
>
Uh? I think you mean that "in any interval that does not
contain any accumulation points there can only be a finite
set of points". The counter-example for what you say
is just a countable set with accumulation points { -2, 0, 2},
and the interval (-1, 1).
>That means there can only be a countable infinity of points between
>any two accumulation points since they can be counted off in, say,
>order of their distance from the centre of the interval.
>
This is not a counting, because those distances might be irrational
numbers.
>If the minimum distance between any accumulation point
>is non-zero there can only be a countable infinity of such points
>in the reals and hence (omega^2=omega) only a countable
>infinity of points in the set.
>
omega^2 is not omega; just aleph-0^2 that is aleph-0
>That implies that for uncountable sets the accumulation points
>must be arbitarily close together, i.e they must have their own
>accumulation point. That isn't quite the same as being dense,
>which I wrongly thought, but it's a stronger statement than can be
>made for countably infinte sets.
>
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)
Alberto Monteiro
