Re: Cantor and the binary tree

In article <1120285537.541672.105210@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
> *** T. Winter wrote:
> > There is a
> > distinction between "the set is not infinite" and "none of the elements
> > is infinite".
>
> And just this distinction is wrong for ordered sets where every element
> has an initial segment.

Eh? I wonder.

> What does it mean, that every number is finite?
> It means that the belonging initial segment is finite, for *every*
> number.
> The limit n --> omega is not taken in N as 0 is not taken by any 1/n.

What do you mean with "the limit is not taken"? By convention the
limit of 1/n is 0. But there is no n such that 1/n = 0. By the
same convention, the limit of {1,...,n} is N. But there is no n such
that {1,...,n} = N.

> Sets like N are potentially infinite but never actually infinite.

So 1/n is potentially 0, but never actually 0. What is your problem
with that? The limit (which is just a convention) gets you at 0 and N.

> {1,2,3,...,omega} would be actually infinite, but it would not be
> countable, because we cannot count beyond n e N.

You are using a non-standard meaning of countable. There is a
bijection from N to {1,2,3,...,omega} so, by definition, that set is
countable. Countable does *not* mean that you can count to every
element. Countable means that there is a bijection between N and the
set.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
.