"Virgil" <Virgil@home.esc> wrote in message news:Virgil-966F99.20493515062010@bignews.usenetmonster.com... > In article <4c181d87$0$17178$afc38c87@news.optusnet.com.au>, > "Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote: > >> > Nevertheless your "definition" belongs to a countable set, hence it is >> > no example to save Cantors "proof". >> > >> > Either all entries of the lines of the list are defined and the >> > diagonal is defined (in the same language) too. >> >> Yes. If you provide a list of Reals, then the diagonal is computable and >> does not appear on the list. >> >> > Then the proof shows >> > that the countable set of defined reals is uncountable. >> >> No, it shows that all "definable" (computable) Reals cannot be explicitly >> listed. This is *not* the same as being uncountable. >> >> >> > Or it does not >> > show anything at all. >> > >> >> It shows that all "definable" (computable) Reals cannot be explicitly >> listed. This is a well known proof in set theory. This is *not* the same >> as >> being uncountable. >> >> >> > To switch "languages" is the most lame argument one could think of. >> > The diagonal argument does not switch languages. And it cannot be >> > applied at all because the list of all finite defiitions does not >> > contain infinite entries. Those however are required for the diagonal >> > argument. >> > >> >> No, that paragraph above is close to gibberish. Cantor said and proved >> that >> any purported list of all Reals cannot contain all Reals. His proof is >> simple and clear, provides an explicit construction of at least one >> missing >> Real, and does contain or require any concepts of uncomputable numbers, >> or >> use of the Axiom of Choice. > > Isn't there a missing "not" in that last sentence between "does" and > "contain"?
