Date: Jan 22, 2013 4:47 AM
Subject: Re: ZFC and God
On 22 Jan., 09:53, Virgil <vir...@ligriv.com> wrote:
> > Consider, for instance, all terminating binary fractions b_n
> > 0.0
> > 0.1
> > 0.00
> > 0.01
> > 0.10
> > 0.11
> > 0.000
> > where some numbers are represented twice (in fact each one appears
> > infinitely often). Constructing the diagonal d we find that d differs
> > from *every* b_n *at a finite place*.
> Quite so!
Yes, but this does not imply that d differs from *all* b_n *at a
... one afterwards mistook that logic for something above and prior to
all mathematics, and finally applied it, without justification, to the
mathematics of infinite sets. This is the Fall and Original sin of set
theory even if no paradoxes result from it. Not that contradictions
showed up is surprising, but that they showed up at such a late stage
of the game!
> > Since the above list is complete, which is possible because all
> > terminating fractions, as a subset of all fractions, are countable, it
> > is impossible that the diagonal differs from all entries b_n at a
> > finite place.
> It does not have to "differ from all entries b_n at a finite place",
> but it does differ from each entry b_n at a finite place.
> Which is an entirely different thing.
Yes, but in order to take it as evidence for an increase in
cardinality, d must differ from all entried b_n.
> > If this was possible, the list would have a gap, namely
> > a finite initial segment of d. That means, the diagonal up to every
> > bit can be found in the list. And after every finite place there is
> > nothing that could distinguish two numbers.
> > Therefore the diagonal does not increase the cardinal number of the
> > listed entries b_n.
> The diagonal is not present in the list, if for no other reason than
> every member of WM's list has a last digit but the diagaonl does not.
Every entry of the list has a last digit, but there is no last digit
in the list. Therefore every digit of the diagonal can be compared
with entries of the list.
> > The diagonal may be infinitely long. But what does that mean? Every
> > given number of bits is surpassed. But the same holds for the entries
> > of the list.
> Every member of the list has a last entry but the diagonal does not.
Decimal representations of numbers are not identified by digits they
do not have, but by digits they have.
> > The only difference could be a bit of the diagonal that
> > has no finite index.
> Outside of WMytheology, one can have infinitely many indices without
> having an infinite index.
If the decimal representation of a number consists of nothing but its
finite initial segments FISs, then a list containing every FIS
contains that number. If the decimal representation of a number
consists of more than its FISs, then it should be expressible in
mathematical discourse what this "more" consists of.