Cantor Confusion
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From: Tony Orlow on 29 Sep 2006 23:02 William Hughes wrote: > Poker Joker wrote: >> "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message >> news:J6CsBJ.Jys(a)cwi.nl... >>> In article <070Tg.14143$8_5.3402(a)tornado.rdc-kc.rr.com> "Poker Joker" >>> <Poker(a)wi.rr.com> writes: >>>> "Randy Poe" <poespam-trap(a)yahoo.com> wrote in message >>>> news:1159494111.724651.95600(a)i3g2000cwc.googlegroups.com... >>>> >>>>> That's incorrect. You don't have to assume none map onto R in order to >>>>> prove none map onto R. >>>>> >>>>> The direct argument starts this way: Let f be any such function, from >>>>> naturals to reals. >>>> Certainly we should assume that f *MIGHT* have R as its image, right? >>> You may assume that, but that assumption is not needed. >> Certainly not for ostriches. >> >>>>> Now, are you saying that somehow that misses some possible functions >>>>> from naturals to reals? How so? >>>> No, but we haven't proven that the image of f can't be R in step #1, >>>> right? >>>> So step #2 isn't valid, right? >>> Remember: >>>> 1. Assume there is a list containing all the reals. >>>> 2. Show that a real can be defined/constructed from that list. >>>> 3. Show why the real from step 2 is not on the list. >>>> 4. Conclude that the premise is wrong because of the contradiction. >>> Why is step 2 invalid? >> Do you always accept steps that have questionable validity? >> >>>> Under the most general assumption, we can't count out that >>>> R is f's image, so defining a real in terms of the image of >>>> f *MIGHT* be self-referential, and it certainly is if the image >>>> of f is R. >>> What is the problem here? >> I assume you accept this proof that there are no complete lists >> of reals: >> >> Let r be a real number between 0 and 1. Let r_n denote the nth digit >> in r's decimal expansion. Let r_n = 5 if r_n = 4, otherwise let r_n = 4. >> r isn't on any list of reals. Therefore there isn't a complete list of >> reals. > > Hardly, but this "proof" does not reflect what is being done. > > You start out with a set of real numbers A, with a certain > property, there is a surjective function from the natural > numbers to A. In other words A is a list of real numbers. > > You define a process D(A) which gives you a real number. > D depends only on the fact that a surjective function > from the natural numbers exists, it does not depend on > any other property of A. Thus D can be applied to any > list . In particular, if we assume there is a list > containing all the real numbers, D can be applied to this > list. That this application will lead to a contradiction > does not change the fact that D can be applied to the > list. So step 2 is valid. > > Of course, we need not make this assumption. In this case > the proof goes > > 1 Let A be a list > 2 Use D to contruct a real number r > 3 Show that r in not an element of A > 4 Conclude that A does not contain all the > real numbers > > Again, since D can be applied to any list > step 2 is valid. > > - William Hughes > This list must be in representation using 2 or more symbols, yes? You cannot perform this proof in unary without the obvious result.
From: Virgil on 29 Sep 2006 23:07 In article <vOjTg.25590$QT.1504(a)tornado.rdc-kc.rr.com>, "Poker Joker" <Poker(a)wi.rr.com> wrote: > "Alan Morgan" <amorgan(a)xenon.Stanford.EDU> wrote in message > news:efkegr$6d9$1(a)xenon.Stanford.EDU... > > >>if its true for ANY list, then it must be > >>true for a specific list. So if considering a single specific list > >>shows a flaw, then looking at ANY (ALL of them) list doesn't > >>help. > > > > But if it's true for ANY list then it must be true for a specific > > list. So if considering a single specific list shows a flaw then > > perhaps that list doesn't really exist. > > That's true, but that's not the entire story. It is in mathematics. Once a proof for any list is established, it covers every list. > > Suppose I claim that I have a list that contains all the reals. Such claims, without supporting proofs, are to worth the electrons Joker used to post them. > You claim you can take that list and construct a real not > on the list. You procede to show the construction. I would > claim that your construction is flawed Again, such claims, as that of having a list of all reals, are worthless without evidence, and Joker has shown that his word is damned poor evidence.
From: Virgil on 29 Sep 2006 23:09 In article <qWjTg.25591$QT.23912(a)tornado.rdc-kc.rr.com>, "Poker Joker" <Poker(a)wi.rr.com> wrote: > "MoeBlee" <jazzmobe(a)hotmail.com> wrote in message > news:1159579135.514717.296590(a)m7g2000cwm.googlegroups.com... > > > Look, suppose x is an arbitrary even number and I prove that therefore > > x has property P. I then conclude that all even numbers have property > > P. > > The property could be say, that x is prime I wait with unbated breath for Joker to prove that an arbitrary even number has to be prime.
From: Virgil on 29 Sep 2006 23:10 In article <_YjTg.25592$QT.21259(a)tornado.rdc-kc.rr.com>, "Poker Joker" <Poker(a)wi.rr.com> wrote: > I understand everything you understand. I just understand more, and > that confuses you. Joker seems to have delusions of competency.
From: MoeBlee on 29 Sep 2006 23:11
Tony Orlow wrote: > > So you're saying the first digit of r is 4 because the first digit of > > r isn't 4? What the hell are you talking about? > > Duh. Sounds like PJ's constructing an anti-diagonal. No, he's not. There is no diagonal in what he's doing. What he's doing just amounts to stipulating the existence of a sequence that does not exist. > >> r isn't on any list of reals. Therefore there isn't a complete list of > >> reals. > > > > That bears no resemblance at all to a proof. > It bears much resemblance to Cantor's second regarding uncountability > of...a set. The original proof was regarding a complete language using > at least two symbols, m and w, no? That was later conflated to a proof > about the reals. Numbers are representations of quantity. It's related > to powerset, via N=S^L. Whatever that is all about, like Randy said, Poker Joker's arguments have no bearing on the uncountability proofs. MoeBlee > > Tony |