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*. ... 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! (Hermann Weyl)
> > 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.