From: "Alberto Monteiro"
>
> Robert Shaw wrote:
> >
> >>> Generally, composing just two functions will give an uncountable
> >> >
> >> Countable :-)
> >>
> >Uncountable, if you allow infinite sequences.
> >If you've got functions f and g you can do the
> >diagonal argument on any list of sequences.
> >
> But I was arguing against your presumed assertion that
> "composing two functions a finite number of times..."
> because this was the a priori rule that generated the
> countable set that must get as close as possible to the
> target set:
>
Composing a finite number of times gives a countable
infinity but the limit points of this set are given by
composing infinitely often. The finite sequences can get
arbitary close to the infinite sequences by making them long
enough.
> >>> If you start with 3 you can only generate a countable infinity
> >>> of numbers. Take the factorial n times, then the square root
> >>> m times. Because of the sparsity of the factorials I'd suspect
> >>> this set isn't dense.
> >>>
> >>> Generally, composing just two functions will give an uncountable
> >>> infinity of numbers which I think means the set of numbers
> >>> must be dense over a nonzero interval.
>
> >
> >I'm not clear what you mean here.
> >Can you give a uncountable set of numbers whose
> >closure doesn't include any connected sets?
> >
> Ok. I interpreted your assertion above [>>>] as:
>
> Every infinite set restricted to a finite interval is dense
> in the interval
>
> When it should be:
>
> Every infinite set restricted to a finite interval accumulates
> around at least one point
>
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.
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
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.
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.
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.
--
Robert
