In article <firstname.lastname@example.org>, "Peter Webb" <webbfamilyDIEspamDie@optusnet.com.au> wrote:
> Shmuel (Seymour J.) Metz wrote: > > > In <email@example.com>, on 11/03/2012 > > at 02:35 AM, "Peter Webb" <webbfamilyDIEspamDie@optusnet.com.au> > > said: > > > > > Obviously they are. Somebody constructs a list of purported Reals, > > > > Nobody constructs anything. There is a universal quantifier, that's > > all. > > > > > Sorry, but the proof commences with a purported list of all Reals. Look > at it.
Cantor's own proof definitely does not start by assuming that any list is complete, or even that it is a list of reals. He merely shows that if one has any list of binary sequences of m's and w's at all, of any sort, and then that list is not complete. > > > > > Cantor showed that the Reals are not recursively > > > enumerable, not that they are uncountable. > > > > No, he proved that they are uncountable. Theories like ZFC may not be > > to your taste, > > Ohh no, that's definitely false.
Cantor's first proof showed that any list of reals omitted some reals, i.e., that any listing of reals was necessarily incomplete.
His second, the diagonal proof, did not even mention reals, but was later modified by someone else to apply to lists of reals s well. > > > > but Cantor's proof is valid in them and nobody has ever > > found an inconsistency in them. > > > > The proof we are discussing is not formulated in the language of ZF. > The equivalent proof in ZF is correct. > > > > > > The proof actually says that such a list cannot be constructed. > > > > No, it proves that such a list does not exist, a stronger result. > > By showing any purported list is missing a Real.
Actually by showing that any list of binary sequences is missing some binary sequences. > > > > > > > It starts with a hypothetical list that has already been > > > constructed. > > > > It uses a shorthand notation for a longer expression involving > > universal quantifiers. > > That's the source of some of the problem in the proof - it uses a > shorthand which isn't quite correct and isn't properly explained. > > I have no problem with the very similar proof in ZF. --