Alberto Monteiro wrote
>
> Robert Shaw wrote:
>
> >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).
Yes, in any interval between neighbouring accumulation points.
That is if there is no accumulation point between a and d and a<d
I'm considering all the intervals [b,c] with a<b<c<d.
Then (a,d), the interval I'm really interested in, is the union of all
the intervals [b,c]
If (a,d) is the union of a countable family of such intervals the
set of constructable points in it is countable, being the union
of countable number of finite sets.
>
> >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.
>
That doesn't matter. We can list the points as being first closest
to (a+d)/2, second closest, etc which gives a mapping to the integers.
Comsider a monotonic mapping between (a,d) and the real line.
Any subinterval of (a,d) maps to an interval of the reals. The image
of the constructable set's intersection with (a,d) under this mapping
is a set such that any finite interval contains only a finite number of
points,
since any subinterval of (a,d) does and any such set of reals is clearly
countable (with the same technique) so, since the mapping is one-to-one,
their can only be a countable number of points in the interval (a,d)
>
> >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)
>
> 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
To me, this looks like the same proof more abstractly stated.
In any case, this means the accumulation points of an uncountable
set of reals must themselves be uncountable.
The set of all infinite sequence of calculator operations is uncountable
so, unless there's at least one number which can be made in uncountably
many different ways by these operations, the set of all numbers
constructable
by an infinite sequence of such operations must be uncountable and hence
have an uncountable number of accumulation points the minimum distance
between which is zero, as is the minimum distance between points in this
set of constructable numbers.
That is not enough to make the set dense, as the cantor set demonstrates,
but it is more than can be stated for countable sets.
In fact we know that the set of calculator constructable numbers is dense
since it includes the rationals. It would be interesting to know if this
is typical for sets constructed in this manner.
--
Robert
