Virgil
Posts:
6,972
Registered:
1/6/11


Re: ZFC and God
Posted:
Jan 22, 2013 4:58 PM


In article <980a4e98fdcd4d999cfbeba365a22e82@m35g2000vbn.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> 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 > finite place*.
It does if one phrases it as "For all n, d differs from b_n at place n."
> > > > > 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.
It is not evidence of an increase in cardinality, but of the magnitude of cardinality. > > > > > 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.
Every decimal representation is effectively an infinite infinite one,, but terminal zeroes may be omitted. > > > > > 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. But there are infinitely many FIS's none of which represents 1/3, thus the list of all of them also does not contian 1/3.
> 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.
Every fraction whose denominator in lowest terms is not of form 2^m*5^n has a necessarily infinite decimal representation, though it is finitely representable as ending with infinitely many repetitions of some finite sequence.
As for irrationals, most of them cannot be exactly expressed decimally, but that does not prevent them from existing. 

