Re: infinity

David Kastrup said:
> Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> writes:
>
> > William Hughes said:
>
> >> If you want you can define "complete initial segment" to mean a set
> >> of the form 1 through n, In this case your proof applies to
> >> complete initial segments. However, in this case the set of all
> >> finite natural numbers is not a complete initial segment.
> >
> > For every finite n in N there is defined a complete initial segment
> > which is finite.
>
> Correct. Aleph_0 is not a finite n in N, so your reasoning does not
> make N a "complete initial segment".
N is the complete initial segment of itself, like every set.
>
> > No initial segment of the finite naturals has an infinite number of
> > elements.
>
> No _complete_ initial segment. And N is not a "complete initial
> segment" of N according to your definition. You admitted yourself
> that there is no largest finite natural, and that would be required to
> define N as a complete initianl segment ending with it.
N is the initial segment which includes all initial segments. Your argument is
vacuous.
>
> >> You can't have it both ways. Either your proof does not apply to
> >> complete initial seqments, or the set of all finite natural numbers
> >> is not a complete initial seqment.]
>
> > It obviously applies to all complete initial segments defined by ANY
> > finite natural number.
>
> Fine. But N is not such a segment.
It applies to very n in N. No finite n has an infinite number of predecessors.
>
> > There is no finite n for which there are an infinite number of
> > predecessors.
>
> Correct. But there also is no finite n for which there is not an
> infinite number of successors.
What drugs are you on?? Did you eat a can of stupid for breakfast? How many
predecessors does 1 have? None? Is that infinite? How about 2? 1 predecessor?
That's finite isn't it? What a dumb statement. Try thinking, before you make
blindly contradictory statements.
>
>

--
Smiles,

Tony
.

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