Re: infinity

Virgil said:
> In article <MPG.1d87c7758667715298a1f5@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
>
> > Virgil said:
>
> > > TO has specifically claimed that for m and n in the set of finite
> > > naturals N,
> > >
> > > (a) for each n there is an m such that m > n
> > >
> > > implies
> > >
> > > (b) there is an m such that for each n, m > n
> > >
> > > despite that obvious fact that (a) is true and (b) is false.
> > >
> > > Proof that (a) is true: for each n choose m = n+1, and m > n.
> > >
> > > Proof that (b) is false: for any M choose n = m+1, and not (m > n).
> > >
>
>
> > Look, if I say m is aleph_0,
>
> Note that each m and n are specifically required to be members of the
> set, N, of finite naturals.
>
> Thus TO is saying that aleph_0 is a member of the set of finite
> naturals, in particular, TO is saying that aleph_0 is finite.
>
> But then there is certainly no need for any of those infinite numbers,
> that TO keeps trying to bring in.
>
If the size of a set of all consecutive whole numbers starting at 1 is always a
member of that set, and you claim your set of naturals defined this way has a
size of aleph_0, then aleph_0 must be a member of your set. Is it finite?
--
Smiles,

Tony
.

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