Why can no one in sci.math understand my simple point?
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From: Virgil on 15 Jun 2010 22:49 In article <4c181d87$0$17178$afc38c87(a)news.optusnet.com.au>, "Peter Webb" <webbfamily(a)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"?
From: Peter Webb on 15 Jun 2010 23:14 "Virgil" <Virgil(a)home.esc> wrote in message news:Virgil-966F99.20493515062010(a)bignews.usenetmonster.com... > In article <4c181d87$0$17178$afc38c87(a)news.optusnet.com.au>, > "Peter Webb" <webbfamily(a)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"? Oops.
From: Aatu Koskensilta on 15 Jun 2010 23:37 Virgil <Virgil(a)home.esc> writes: > But until you can determine which of those 10 cases, how can you > compute the number? You can't. The number is computable nonetheless, in the sense that there exists an effective procedure for churning out its decimal expansion. As noted, computability is a purely extensional notion. Recall the classical recursion theory exercise, which we find, in some form or other, in pretty much any text on the subject: Let f : N --> N be a function such that f(x) = 0 if Goldbach's conjecture is true, and 1 otherwise. Is f computable? -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Virgil on 16 Jun 2010 01:33 In article <4c1841c6$0$17174$afc38c87(a)news.optusnet.com.au>, "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > "Virgil" <Virgil(a)home.esc> wrote in message > news:Virgil-966F99.20493515062010(a)bignews.usenetmonster.com... > > In article <4c181d87$0$17178$afc38c87(a)news.optusnet.com.au>, > > "Peter Webb" <webbfamily(a)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"? > > Oops. When I was younger, one excused such thing by saying "Even Homer nods", but the young today don't seem to know about the Iliad and Odyssey.
From: Virgil on 16 Jun 2010 01:34
In article <874oh3ir6m.fsf(a)dialatheia.truth.invalid>, Aatu Koskensilta <aatu.koskensilta(a)uta.fi> wrote: > Virgil <Virgil(a)home.esc> writes: > > > But until you can determine which of those 10 cases, how can you > > compute the number? > > You can't. The number is computable nonetheless, in the sense that there > exists an effective procedure for churning out its decimal expansion. Except for that one digit. > > As noted, computability is a purely extensional notion. Recall the > classical recursion theory exercise, which we find, in some form or > other, in pretty much any text on the subject: > > Let f : N --> N be a function such that > > f(x) = 0 if Goldbach's conjecture is true, and 1 otherwise. > > Is f computable? Not yet! |