Why can no one in sci.math understand my simple point?
in [Logic]
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From: Peter Webb on 15 Jun 2010 20:39 > 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. Perhaps if you were to identify the step in Cantor's proof that you consider wrong, then we might gain some idea as to what you are actually objecting to? > Regards, WM
From: Virgil on 15 Jun 2010 20:40 In article <57db5bbc-d26a-4b66-a3e5-3d41f306570f(a)u7g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > A real number cannot be defined other than by a finite definition. All > are in the list. Given any such "list", one can construct from it, a la Cantor, an endless new list, and certainly one of any desired finite length, of numbers not listed in the original list.
From: Virgil on 15 Jun 2010 20:44 In article <ec965aaf-77ee-4190-846a-4acf52b68dbc(a)z8g2000yqz.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 15 Jun., 18:46, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > WM says... > > > > > > > > >On 15 Jun., 16:32, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > >> The proof does not make use of any property of infinite lists. > > >> The proof establishes: (If r_n is the list of reals, and > > >> d is the antidiagonal) > > > > >> forall n, d is not equal to r_n > > > > >As every n is finite, it belongs to a finite initial segment of the > > >infinite list. > > > > I'm not sure what you are saying. The fact is, we can prove > > that for every real r_n on the list, d is not equal to r_n. > > Of course. Every real r_n belongs to a finite initial segment of the > list. And an infinite terminal segment, neither of which is relevant. > That does not yield any result about the whole list It yields a result about each and every member of that list being different from d. > Look here: We can prove for any finite segment > {2, 4, 6, ..., 2n} > of the ordered set of all positive even numbers that its cardinal > number is surpassed by some elements of the set. What you cannot prove is that that statement is in any way relevant to the argument you are trying (and failing) to justify.
From: Virgil on 15 Jun 2010 20:52 In article <39a1fada-4fc1-4214-a858-df63a2444a7c(a)a30g2000yqn.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 15 Jun., 18:49, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > WM says... > > > > > > > > > > > > > > > > >On 15 Jun., 16:18, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > >> Peter Webb says... > > > > >> >"WM" <mueck...(a)rz.fh-augsburg.de> wrote in message > > >> >news:62ae795b-1d43-4e1f-8633-e5e2475851aa(a)x21g2000yqa.googlegroups.com.. > > >> >. > > >> >> On 15 Jun., 12:26, stevendaryl3...(a)yahoo.com (Daryl McCullough) > > >> >> wrote: > > > > >> >>> (B) There exists a real number r, > > >> >>> Forall computable reals r', > > >> >>> there exists a natural number n > > >> >>> such that r' and r disagree at the nth decimal place. > > > > >> >> In what form does r exist, unless it is computable too? > > > > >> >Of course its computable. > > > > >> No, it's computable *relative* to the list of all computable reals. > > >> But that list is not computable. > > > > >That is nonsense! > > > > >The list of all definitions is possible and obviously contains all > > >definitions of real numbers. > > > > I was talking about the list of all *computable* reals. There are > > definable reals that are not computable, and Cantor's proof shows > > how to define one. > > No, it does not show that because it is impossible to define a real by > an infinite definition. A Cantor-list without a finite definition > however is an infinite definition. > > Sensible definitions are finite. As far as I am aware, "sensible" has no mathematical meaning at all. There is nothing in mathematics which requires anything to be sensible. > So you explanation is nonsense in highest degree. Not even up to WM's usual level of nonsense. > > > > You can similarly get a list of all definable reals for a specific > > language L. > > Every real that is definable in a specific language is definable in > another specific language. The set does not grow or shrink or change > when the language is changed. That is a statement that requires proof, which WM, as usual, has not provided. There are, I believe, in both mathematics and logic, "Languages" which are proper "sub-languages" of other languages so that things expressible in a super-language are not always expressible in its sub-languages. Does WM still claim otherwise?
From: Mike Terry on 15 Jun 2010 21:05
"Virgil" <Virgil(a)home.esc> wrote in message news:Virgil-3B2F0B.16291815062010(a)bignews.usenetmonster.com... > In article <87sk4ohwbt.fsf(a)dialatheia.truth.invalid>, > Aatu Koskensilta <aatu.koskensilta(a)uta.fi> wrote: > > > Virgil <Virgil(a)home.esc> writes: > > > > > Note that it is possible to have an uncomputable number whose decimal > > > expansion has infinitely many known places, so long as it has at least > > > one unknown place. > > > > You need infinitely many unknown places. > > If the value of some decimal digit of a number depends on the truth of > an undecidable proposition, can such a number be computable? Yes - e.g. imagine just the first digit of the following number depends on an undecidable proposition: 0.x000000000... There are only 10 possibilities for the number, and in each case it is obviously computable... Mike. |