At 12:23 AM 5/21/01 +0000, Alberto Monteiro wrote:
>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.
In reality, of course, the set of numbers that can be constructed on a
calculator is _finite_. For example, if the calculator can represent
numbers to 12-place accuracy and, like a HP calculator can deal with
exponents in the range +/- 499, then there are about 10^12 * 1000 * 2
(positive or negative) = 2 x 10^15 different numbers that can be
represented on the calculator. (Actually, somewhat fewer, since that would
include meaningless "numbers" such as -0.00000000000. [That's a negative
zero, if the line breaks in your e-mail program mess it up.]) In MIRCALC,
the calculator program I was playing with when I started this thread,
numbers can be represented with up to 10,000 digits of precision, so there
are about 2 x 10^10,000 possible numbers (larger but still
finite). _Mathematica_ says its precision is limited only by the available
memory (about 2 decimal digits per byte), so one could in principle have
much longer numbers, but the set of them would still be finite.
Just to be picky . . .
-- Ronn! :)
