CANTOR DISPROOF <<<<<<<<<<<<<<<<
in [Logic]
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From: |-|ercules on 28 Jun 2010 22:13 "Sylvia Else" <sylvia(a)not.here.invalid> wrote > On 29/06/2010 10:42 AM, |-|ercules wrote: >> "Sylvia Else" <sylvia(a)not.here.invalid> wrote >>> On 28/06/2010 11:15 PM, |-|ercules wrote: >>> >>>> You prove case n, by logical inspection, that the property holds for all >>>> elements in the list of size n. (a whole object) >>>> >>>> I prove case n, by logical inspection, that the property holds for >>>> prefixes of width n. (a subset of digits) >>>> >>>> You prove, by induction, that all (finite) sizes of lists the property >>>> holds. >>>> >>>> I prove, by induction, that the property holds for all digit widths. >>>> (all digits). >>> >>> You only prove that it holds for finite digit widths. Somehow you want >>> to be allowed to extrapolate from finite to infinite when it suits >>> you, but not for me to allowed to do the same thing when it doesn't >>> suit you. >>> >> >> Nope! Your prove it for increasing different objects. >> >> I sample larger and larger sizes of the one object. Different style of >> proof! >> > > <sigh> So we're sampling now, not constructing. Fine. I used the term sample a week ago, and that the computable reals are deterministic and I'm not constructing per se, etc. etc. You can say I construct subcomponents too, as opposed to your construction of individual objects. > > By sampling, I suppose you mean that you look through the list of > computables to find sequences with the required prefixes. > > I question whether your inductive reasoning even works with that > approach. It is certainly possible to find in the list of computables a > sequence that starts with any finite prefix, but that doesn't mean that > the inductive step is valid, merely that the result is true. > > An inductive step would need to take the form: > > --- > If a sequence with a specific prefix p is in the list of computables, > then prefixes constituting p followed by 0, p followed by 1, p followed > by 2, etc. are also in the list. Yes, you could have saved yourself a lot of time by reading the proof. -> there are 10 computable copies of the -> complete permutations of width w -> each ending in each of digits 0..9 (at position w+1) -> which generates a set larger than width w > > Then prove (by inspection presumably) that the prefixes 0 thru 9 exist. Yes, in your wording you start with a set of complete prefixes and add each of 0..9 to the end of each real, my wording is equivalent, but I use a "block append" operation 10 times.. > --- > > If the proof merely consists of observing that since prefixes one digit > longer than p are also of finite length, and therefore exist in the list > of computables, then that is not an inductive step (except in the very > limited sense that if n is finite, then so is n+1). your n is the digits that the property holds for, all you're doing is reminding everyone that digit positions (natural numbers) are finite. > > That aside, since the list of computables contains all finite prefixes, > then when you've sampled the list to find sequences starting with all > prefixes of length n digits, with the sample constituting a sublist[*] > of length 10^n, I can sample the list of computables to find a sequence > with a prefix of length 10^n which is not in your sublist. This remains > true for all finite n. > > [*] I use the term loosely. It's really a listing of the subset. > > Sylvia. So, it doesn't work for length n antidiagonals as all prefixes of length n digits are counted for, we've done enough analogies of bogus proofs. Herc
From: herbzet on 28 Jun 2010 22:18 Sylvia Else wrote: > The Raven wrote: > > > Sylvia and all the other followers of this thread, could you please keep it > > out of aus.tv and take it back to sci.math? It's off topic to aus.tv. > > > > Please (I'm asking politely) :-) > > I can't stop Herc posting them to aus.tv in the first place, and if I > remove groups from my replies, he just puts them back again. Oh, you poor dear! Oh, you poor, poor dear thing! Being *forced* by that wicked Herc to crosspost to ngs that have no interest in this topic at all! You poor, poor thing, being *forced* into this discourteous behaviour by that wicked Herc, who's not really wicked because he's so kuh-RAZY, the poor dear thing. What can a reasonable person do, other than suggest that dozens or perhaps hundreds of other people go to the trouble of filtering out *your* off-topic noise? Let me help you out, darling. 1) Set your own filter to kill any post with aus.tv in the "Newsgroups" field, and inform Herc that you won't see any such posts. 2) If your newsreader does not allow filtering on that field (I know mine certainly doesn't) just be your own filter -- announce firmly that you will not respond to posts with aus.tv in the Newsgroups field. Like this: I will not wittingly respond to posts about foundations of math that are posted to the aus.tv newsgroup, as that is just off-topic spam to them. I will not indulge Herc's narcissistic desire to keep his name in front of his countrymen's eyes. Herc may find me in sci.math if he wishes to converse with me. See how easy it is? All the people at aus.tv will not find it necessary to forward your posts to abuse(a)individual.net , prepending the sentence "Off-topic for aus.tv" to the body of the forwarded post, and otherwise following the guidelines at http://individual.net/faq.php#5.5 . You don't have to be an intercontinental pest if you don't want to be, darling. You really don't. You're welcome, dear. -- hz P.S. -- I won't wittingly be responding to posts with aus.tv in in the Newsgroups field any further.
From: herbzet on 28 Jun 2010 22:24 Jim Burns wrote: > George Greene wrote: > > On Jun 26, 6:05 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > >> It is perplexing if outputs of all computer programs are listed, > >> how do you find a program to compute the diagonal digit > >> at the position that is contradictory? > > > > You just write the program that says output(n) = 9 - L(n,n) > > AND YOU'RE DONE, VOILA, > > WHOOT, THERE IT IS! > > It is a VERY SIMPLE program. > > But there is no "contradictory" position. > > The number being computed simply IS NOT ON the list. > > Consider the list L Heck, consider a list L of sequences of the digits '5' and '7'. Your program is just If L(n,n) = 5 then output(n) = 7 else output(n) = 5. The number being computed simply is not on the list. You can't even get a list of all sequences of *two* digits, much less a list of all sequences of *ten* digits, of which the list of sequences of just *two* digits is a proper part. -- hz
From: herbzet on 28 Jun 2010 22:28 Ars�ne Lupin wrote: > > Why people bother replying? I really don't know, but it's very sad.
From: Sylvia Else on 28 Jun 2010 22:32
On 29/06/2010 12:13 PM, |-|ercules wrote: > "Sylvia Else" <sylvia(a)not.here.invalid> wrote >> On 29/06/2010 10:42 AM, |-|ercules wrote: >>> "Sylvia Else" <sylvia(a)not.here.invalid> wrote >>>> On 28/06/2010 11:15 PM, |-|ercules wrote: >>>> >>>>> You prove case n, by logical inspection, that the property holds >>>>> for all >>>>> elements in the list of size n. (a whole object) >>>>> >>>>> I prove case n, by logical inspection, that the property holds for >>>>> prefixes of width n. (a subset of digits) >>>>> >>>>> You prove, by induction, that all (finite) sizes of lists the property >>>>> holds. >>>>> >>>>> I prove, by induction, that the property holds for all digit widths. >>>>> (all digits). >>>> >>>> You only prove that it holds for finite digit widths. Somehow you want >>>> to be allowed to extrapolate from finite to infinite when it suits >>>> you, but not for me to allowed to do the same thing when it doesn't >>>> suit you. >>>> >>> >>> Nope! Your prove it for increasing different objects. >>> >>> I sample larger and larger sizes of the one object. Different style of >>> proof! >>> >> >> <sigh> So we're sampling now, not constructing. Fine. > > I used the term sample a week ago, and that the computable reals > are deterministic and I'm not constructing per se, etc. etc. > > You can say I construct subcomponents too, as opposed to your construction > of individual objects. > > >> >> By sampling, I suppose you mean that you look through the list of >> computables to find sequences with the required prefixes. >> >> I question whether your inductive reasoning even works with that >> approach. It is certainly possible to find in the list of computables >> a sequence that starts with any finite prefix, but that doesn't mean >> that the inductive step is valid, merely that the result is true. >> >> An inductive step would need to take the form: >> >> --- >> If a sequence with a specific prefix p is in the list of computables, >> then prefixes constituting p followed by 0, p followed by 1, p >> followed by 2, etc. are also in the list. > > > Yes, you could have saved yourself a lot of time by reading the proof. > > -> there are 10 computable copies of the > -> complete permutations of width w > -> each ending in each of digits 0..9 (at position w+1) > -> which generates a set larger than width w > > > >> >> Then prove (by inspection presumably) that the prefixes 0 thru 9 exist. > > > Yes, in your wording you start with a set of complete prefixes and add > each of 0..9 to the end of each real, > my wording is equivalent, but I use a "block append" operation 10 times.. That doesn't make the proof inductive, which was my point. You haven't proved any relation between the existence of prefixes of length n, and the existence of prefixes of length n+1. As it happens, we know from other considerations that they all exist in the list, but you haven't proved it by an inductive process. > > > >> --- >> >> If the proof merely consists of observing that since prefixes one >> digit longer than p are also of finite length, and therefore exist in >> the list of computables, then that is not an inductive step (except in >> the very limited sense that if n is finite, then so is n+1). > > your n is the digits that the property holds for, > all you're doing is reminding everyone that digit positions (natural > numbers) are finite. > > >> >> That aside, since the list of computables contains all finite >> prefixes, then when you've sampled the list to find sequences starting >> with all prefixes of length n digits, with the sample constituting a >> sublist[*] of length 10^n, I can sample the list of computables to >> find a sequence with a prefix of length 10^n which is not in your >> sublist. This remains true for all finite n. >> >> [*] I use the term loosely. It's really a listing of the subset. >> >> Sylvia. > > > So, it doesn't work for length n antidiagonals as all prefixes of length > n digits are counted for, > we've done enough analogies of bogus proofs. It doesn't have to work for length n antidiagonals, it merely has to work for some finite number, which it does. As long as your prefixes are of finite length, there is a sequence that is not in your sublist. You want that awkward sequence to vanish when you start dealing with infinite prefixes, but you've offered no reason to think it does. Sylvia. |