From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Topo Question
Date: 8 Oct 1996 14:09:08 GMT

In article <Pine.Sola.3.91.961004125958.25955A-100000@symcom.math.uiuc.edu>,
Phichet Chaoha  <chaoha@math.uiuc.edu> wrote:

>"Let A be a set and A' the set of accumulation points of A,
>	A and A' are disjoint ===> A is countable ?"

Depends on whether this is for a topology class or an analysis class.

In the former case, "set" might mean "subset of a topological space".
In that case, the above implication is false. Take A' to be any
(uncountable) set and  A ={a}, a singleton. Give  X = A'\union A the
topology whose only open sets are the empty set,  X,  and {a}. 

In the latter case, "set" might mean "subset of the real line". In
that setting, if  A'  and  A  are disjoint, then for each  a  in  A
there must be an  e(a)>0  so that  a  is the only point of  A  in
the interval  (a - e(a), a + e(a) ). Adding up the lengths  2e(a)  as
appropriate, we see there can be only a countable number of  a's in
each bounded interval  [-N, N], since an uncountable sum of positive
terms must diverge. Taking the union of those subsets, we see  A  is
a countable union of countable sets, hence countable.

So it looks like you'll have to indicate the setting in which you
want the above result to hold.

dave