Why can no one in sci.math understand my simple point?
in [Logic]
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From: Virgil on 15 Jun 2010 16:07 In article <0fec59c4-4347-42d5-9c05-2dc0c7b49804(a)k39g2000yqb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 15 Jun., 06:13, "|-|ercules" <radgray...(a)yahoo.com> wrote: > > Consider the list of increasing lengths of finite prefixes of pi > > > > 3 > > 31 > > 314 > > 3141 > > .... > > > > Everyone agrees that: > > this list contains every digit of pi � (1) > > There is no "every digit of pri". If it were, this must be proved by > showing the last digit. It is only in dreamworlds like WM's that infinite sequences do not exist. > > If you write every digit in the same line, no set theorist will > disagree. If you write the next digit always in the next line, they > will diagree. You can see that they have restricted capabilities of > thinking. Not nearly as restricted as WM's. An infinite sequence in mathematics is merely a function with domain N (or, more generally, with domain any ordered set order-isomorphic to N).
From: Virgil on 15 Jun 2010 16:11 In article <b9778106-8fb9-4363-b61c-582711cc49a4(a)u7g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 15 Jun., 12:39, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > > That's *all* that matters, for Cantor's theorem. The claim > > is that for every list of reals, there is another real > > that does not appear on the list. > > The claim is only proved for every finite subset of the list. On the contrary, a rule is provided which SIMULTANEOUSLY applies to every member of the list. > Then it > is extrapolated to an infinite list without more reason than to > extrapolate other laws from the finite to the infinite. When a rule can so easily seen (and proved, if one is too blind to see it without such proof) to apply to every member of a set, only fools like WM will claim that it does not apply to every member of that set.
From: Virgil on 15 Jun 2010 16:17 In article <26b02bef-c1c8-4e29-a29e-020b437f8721(a)c33g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 15 Jun., 10:28, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> > wrote: > > > > > > But no single member of the list contains all digits of pi. > > > > Pi doesn't appear anywhere on the list.- > > And wheres does it appear ,as an infinite sequence not obtaiend from a > finite formula? Limit statements are finite. And for a number to exist it is certainly sufficient (though not necessarily necessary) to be able to find it to within any given error, which is certainly the case for pi.
From: WM on 15 Jun 2010 16:22 On 15 Jun., 21:03, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > WM says... > > >On 15 Jun., 18:46, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > >> 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. > >That does not yield any result about the whole list > > On the contrary, the definition of "d is on the list" > is that "there exists a natural number n such that r_n = d". > We proved "forall n, r_n is not equal to d". So that > means "there does not exist a natural number n such that r_n = d", > so that means "d is not on the list". > > We have thus proved something about the whole list. > > >> That means that d is not on the list. There is no extrapolation > >> involved. > > >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. > > >Nevertheless this appears not be a proof that the cardinal number of > >the whole set is less than some elements of the set. > > So there we have an example of an illegitimate extrapolation. > If you prove Phi(n) for an arbitrary natural number n, then you are allowed > to conclude: > > forall natural numbers n, Phi(n). > > So you can conclude: > > forall natural numbers n > 0, the set of all even numbers less than or > equal to 2n has a cardinality less than 2n. > > That's true. That's a legitimate proof. On the other hand, it is not > legitimate to conclude: > > The set of all even numbers has a cardinality that is less than > some even number. > > That's an unwarranted extrapolation. > > So there are legitimate proofs, and there are bogus proofs. And Cantor's proof shows something for every natural number: Every line numerated with a natural number from 1 to n does not contain the diagonal. But to conclude that the whole set, i.e., the infinite list, does not conclude the diagonal is a bogus conclusion. Not a legitimate proof. Regards, WM
From: Virgil on 15 Jun 2010 16:24
In article <62ae795b-1d43-4e1f-8633-e5e2475851aa(a)x21g2000yqa.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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? > > But if r is computable, then this theorem shows that the computable > numbers are uncountable. Contradiction. Not if it is uncomputable. 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. > > And if r is not computable, then it is impossible to prove > disagreement with any r'. 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. > > Regards, WM |