Robert Shaw wrote:
>
>In any case, this means the accumulation points of an uncountable
>set of reals must themselves be uncountable.
>
Yep.
Now back to our problem: which real numbers are the accumulation
points of the union of sets, where the initial set is finite and
the elements of the (n+1)-th set are the elements of the n-th
set operated by a finite set of functions.
>The set of all infinite sequence of calculator operations is uncountable
>
Unless there's some normal-group-like stuff in the process; for
example, if we consider the sequence generated by the two
functions x -> x + 1 and x -> x - 10
>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.
>
I don't get this. What do you mean - exactly - by "the minimum distance
between points" in a set?
>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.
>
The problem with the calculator functions is that most of them have
a natural accumulation point, namely, _error_. For example,
for x <= 0, log(x) = error. This is the problem with x!, for which
only (integer)! gives a non-error number.
Alberto Monteiro
