CANTOR DISPROOF <<<<<<<<<<<<<<<<
in [Logic]
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From: herbzet on 28 Jun 2010 22:39 Alan Smaill wrote: > Transfer Principle writes: > > > Still, I see nothing wrong with having a theory which does > > satisfy either Herc's or WM's intuitions. > > Of course you're right; > there's nothing wrong with having such theories. You'll have to talk louder -- TP has problems hearing people make this sort of assertion. > Is there anyone who has said anything different in sci.logic? > > Do you think that Herc or WM has actually proposed > any such theory? Please don't confuse TP by asking two distinct questions. He has trouble enough answering one at a time. -- hz
From: |-|ercules on 28 Jun 2010 22:42 "herbzet" <herbzet(a)gmail.com> wrote > > > Ars�ne Lupin wrote: >> >> Why people bother replying? > > I really don't know, but it's very sad. Go herby! Gather strength in numbers, WHOOA! Herc
From: |-|ercules on 28 Jun 2010 22:46 "herbzet" <herbzet(a)gmail.com> wrote ... > > > 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 ....just say that, you have to PROVE IT. Here you go. Stick a forall(n) in front. forall n, > If L(n,n) = 5 then output(n) = 7 else output(n) = 5. There you go, now you've CONSTRUCTED and PROVED it! Herc
From: |-|ercules on 28 Jun 2010 22:54 "Sylvia Else" <sylvia(a)not.here.invalid> wrote... > 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. There's a very clear P(n) -> P(n+1) In your wording and mine. > >> >> >> >>> --- >>> >>> 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. The sequences' width grows slower than the list length, your example proves nothing, you could just as well get a 10 length list and a 20 digit antidiagonal and use that to prove transfiniteness. > > You want that awkward sequence to vanish when you start dealing with > infinite prefixes, but you've offered no reason to think it does. This is sidetracking that the proof has been fully justified, please make your objections to steps 1, 2, 3, or 4, or better still read how I justified the induction for all digit positions. You haven't commented on the difference between induction over a single data structure and induction over numerous data structures. Herc
From: Sylvia Else on 28 Jun 2010 23:27
On 29/06/2010 12:54 PM, |-|ercules wrote: > "Sylvia Else" <sylvia(a)not.here.invalid> wrote... >> 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. > > > There's a very clear P(n) -> P(n+1) > > In your wording and mine. > > > >> >>> >>> >>> >>>> --- >>>> >>>> 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. > > > The sequences' width grows slower than the list length, your example > proves nothing, > you could just as well get a 10 length list and a 20 digit antidiagonal > and use that > to prove transfiniteness. I'm not proving, nor seeking to prove, anything about infinities. I'm just pointing out that your proof is flawed, because to get anywhere you have to explain how the anti-diagonal that's present at every stage, disappears when the prefix length goes to infinity. But your reference to the different finite lengths makes me wonder - are you suggesting that infinite list's length is smaller than the anti-diagonal sequence's infinite length? > > >> >> You want that awkward sequence to vanish when you start dealing with >> infinite prefixes, but you've offered no reason to think it does. > > > This is sidetracking that the proof has been fully justified, please > make your objections > to steps 1, 2, 3, or 4, or better still read how I justified the > induction for all digit positions. > > You haven't commented on the difference between induction over a single > data structure > and induction over numerous data structures. > I have questioned whether what you're doing is induction. P(n) -> P(n+1) is the result of an inductive proof, not the proof itself. Sylvia. |