Re: Well Ordering the Reals

Virgil said:
> In article <MPG.1debd23d3457fa4098a736@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> Tony Orlow <aeo6@xxxxxxxxxxx> wrote:
>
> > William Hughes said:
>
> > > Suppose we start at 1 and the size of the set is a natural number
> > > within the set, when we start at 0, the set size is 1 more that a
> > > natural number within the set. In both cases the set size is
> > > finite.
> >
> > Yes, there is essentially no difference, except that with the current
> > construction of the axiom of infinity, we are led logically to
> > believe that the set size is larger than all elements of the set, so
> > when the set includes all finite numbers, we say the size is larger
> > than all finites, and is infinite.
>
>
> Right on!
>
> > That's a logical leap, when you
> > consider that at each point the set size is 1 more than the largest
> > element, so your set size larger than all finites exceeds the largest
> > by 1, and should also be considered finite, as opposed to the
> > standard consideration that calls it infinite.
>
> Given N = {1,2,3,...}, let S_n = {1,2,3,...,n} = {x in N: 1 <= x <= n}.
>
> For all n in N there is m in N\S_n,
> so for all n in N, size(N) > size(S_n) = n.
No, here is your mistake. size(N) >= size(S_n) = n. The size is a member of the
set.

>
> Thus the size of N cannot be a member of N.
> And the finiteness of members of N is no restriction on the size of N.
>

--
Smiles,

Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
.

Relevant Pages

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