Re: Well Ordering the Reals

In article <MPG.1debfc2c8aac108b98a73c@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:

> 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.


But if for all n , size(N) >= size(S_(n+1)) > size(S_n)
and there exists any n for which size(N) = size(S_n)
then for that n, size(N) < size(S_(n+1))

Failure to accept this would be only one more proof of TO's quantifier
dyslexia.



> > 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.
.

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