CANTOR DISPROOF <<<<<<<<<<<<<<<<
in [Logic]
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From: Jim Burns on 27 Jun 2010 10:23 Sylvia Else wrote: > On 27/06/2010 4:34 PM, |-|ercules wrote: >> "Sylvia Else" <sylvia(a)not.here.invalid> wrote >>>> If you think my induction only works on finite prefixes >>>> and not over entire infinite expansions >>>> then YOU prove that assertion. >>> >>> Consider the following proposition: >>> >>> For any *finite* list of infinite digit sequences, >>> one can use the anti-diagonal method to produce >>> a sequence that is not in the list. >>> >>> Do you have any difficulty with that? >> >> No. This is precisely my point. >> >> 123 >> 456 >> 789 >> >> Diag = 159 >> Anti-Diag = 260 >> >> 260 is a NEW DIGIT SEQUENCE. > > OK, so you accept that it is true for any finite list. > That is, that for any finite list there is a sequence > that is not in the list, or to put it another way, > all finite lists omit at least one sequence. > Note that the requirement that the length of the list > be finite doesn't impose any maximum on the length. Please pardon my interruption. I would like to point out that you don't need to extrapolate to the infinite case here, because every real on the list is at a finite position. If we assume that the anti-diagonal is on the infinite list, then it must be on the list for some finite position n*. If we then truncate the list after n*, we have a finite list which (by the same argument) /we have agreed/ cannot have the anti-diagonal anywhere, including position n*. Except that it does. Contradiction. And so the anti-diagonal cannot be on the infinite list, either. Jim Burns > By analogy with your argument that extrapolates from a list that > contains all finite permutations to a list that contains all infinite > sequences, I'll argue that by extroplating from the fact that all finite > lists omit at least one sequence one can conclude that an infinite list > omits at least one sequence. > > The latter of course contradicts your thesis, but either extrapolating > from the finite to the infinite is valid, or it isn't. Without some > demonstration that the circumstances are materially different, you can't > argue that the extrapolation is valid in one case, and invalid in the > other.
From: |-|ercules on 27 Jun 2010 16:55 "Jim Burns" <burns.87(a)osu.edu> wrote ... > Sylvia Else wrote: >> On 27/06/2010 4:34 PM, |-|ercules wrote: >>> "Sylvia Else" <sylvia(a)not.here.invalid> wrote >>>>> If you think my induction only works on finite prefixes >>>>> and not over entire infinite expansions >>>>> then YOU prove that assertion. >>>> >>>> Consider the following proposition: >>>> >>>> For any *finite* list of infinite digit sequences, >>>> one can use the anti-diagonal method to produce >>>> a sequence that is not in the list. >>>> >>>> Do you have any difficulty with that? >>> >>> No. This is precisely my point. >>> >>> 123 >>> 456 >>> 789 >>> >>> Diag = 159 >>> Anti-Diag = 260 >>> >>> 260 is a NEW DIGIT SEQUENCE. >> >> OK, so you accept that it is true for any finite list. >> That is, that for any finite list there is a sequence >> that is not in the list, or to put it another way, >> all finite lists omit at least one sequence. >> Note that the requirement that the length of the list >> be finite doesn't impose any maximum on the length. > > Please pardon my interruption. > > I would like to point out that you don't need to extrapolate > to the infinite case here, because every real on the list is at > a finite position. > > If we assume that the anti-diagonal is on the infinite list, > then it must be on the list for some finite position n*. > If we then truncate the list after n*, we have a > finite list which (by the same argument) /we have agreed/ > cannot have the anti-diagonal anywhere, including position n*. > Except that it does. Contradiction. > > And so the anti-diagonal cannot be on the infinite list, either. > > Jim Burns > Isn't Sylvia making a NON correct proof to illustrate a point about MY proof? Works on all finite cases --/--> Works on infinite case There is another "mid-way" example where you form a finite diagonal on the infinite list, more akin to my induction of prefixes on an infinite string, which would not work in your case. Herc
From: Sylvia Else on 27 Jun 2010 20:14 On 28/06/2010 6:55 AM, |-|ercules wrote: > "Jim Burns" <burns.87(a)osu.edu> wrote ... >> Sylvia Else wrote: >>> On 27/06/2010 4:34 PM, |-|ercules wrote: >>>> "Sylvia Else" <sylvia(a)not.here.invalid> wrote >>>>>> If you think my induction only works on finite prefixes >>>>>> and not over entire infinite expansions >>>>>> then YOU prove that assertion. >>>>> >>>>> Consider the following proposition: >>>>> >>>>> For any *finite* list of infinite digit sequences, >>>>> one can use the anti-diagonal method to produce >>>>> a sequence that is not in the list. >>>>> >>>>> Do you have any difficulty with that? >>>> >>>> No. This is precisely my point. >>>> >>>> 123 >>>> 456 >>>> 789 >>>> >>>> Diag = 159 >>>> Anti-Diag = 260 >>>> >>>> 260 is a NEW DIGIT SEQUENCE. >>> >>> OK, so you accept that it is true for any finite list. >>> That is, that for any finite list there is a sequence >>> that is not in the list, or to put it another way, >>> all finite lists omit at least one sequence. >>> Note that the requirement that the length of the list >>> be finite doesn't impose any maximum on the length. >> >> Please pardon my interruption. >> >> I would like to point out that you don't need to extrapolate >> to the infinite case here, because every real on the list is at >> a finite position. >> >> If we assume that the anti-diagonal is on the infinite list, >> then it must be on the list for some finite position n*. >> If we then truncate the list after n*, we have a >> finite list which (by the same argument) /we have agreed/ >> cannot have the anti-diagonal anywhere, including position n*. >> Except that it does. Contradiction. >> >> And so the anti-diagonal cannot be on the infinite list, either. >> >> Jim Burns >> > > > Isn't Sylvia making a NON correct proof to illustrate a point about MY > proof? > Works on all finite cases --/--> Works on infinite case Yes. > > There is another "mid-way" example where you form a finite diagonal on > the infinite list, > more akin to my induction of prefixes on an infinite string, which would > not work in your case. I cannot for the life of me see how you regard that as more akin to your induction. It looks to me as if you recognised the issue I raised, and looked for a non-analagous failing case in an attempt to save your position. Sylvia.
From: |-|ercules on 27 Jun 2010 21:24 OK extending WM's example. You come across a fork in a road with two options, left and right. Next to the fork is a sign that reads Each path left or right leads to a new fork with two more options, left or right. Beside the sign is a shop, INFINITE PATH MAPS. The shop contains infinite maps, each numbered 1, 2.. Each map is an infinite sequence of lefts and rights. The shop is infinitely long, and each row has a sign that reads All maps in this section contain all paths X levels deep (plus the remaining infinite paths of Lefts and Rights in each map) The first row X=1, the second row X=2 and so on. So the millionth row, contains well over 2^1,000,000 infinite maps that cover every path option on the first million branches. Considering if you have a map for a certain path you can always find your way back, and you legs are solar powered bionics that never wear, and your cyborg body is suitably matched to your legs, the sun contains infinite hydrogen fuel etc. etc. would you get lost? Herc
From: Transfer Principle on 28 Jun 2010 17:00
On Jun 27, 6:24 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > OK extending WM's example. > You come across a fork in a road with two options, left and right. Ah yes, WM's infinite binary tree example. > Considering if you have a map for a certain path you can always find your way back, > and you legs are solar powered bionics that never wear, and your cyborg body is suitably matched to your legs, the sun contains > infinite hydrogen fuel etc. etc. > would you get lost? As I'm already familiar with WM's examples, I assume the point of this is to show that according to the standard theory, there are only countably many maps, yet there are uncountably many infinitely paths. Thus anyone who believes in the uncountability of the paths would be forced to answer "yes" to the question "Would you get lost?" since there aren't enough maps for all of the paths, and anyone who answers "no" would be admitting that there are only countably many paths, in accordance with Herc's and WM's beliefs. I assume that the correct answer for someone who uses ZFC to give would be something like, since each map is located only a finite distance from the starting point (even though each path extends away from the start infinitely), one can only prove that there are only countably many initial _finite_ segments of each path, not that there are countably many infinite paths. Still, I see nothing wrong with having a theory which does satisfy either Herc's or WM's intuitions. |