Cantor Confusion
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From: William Hughes on 29 Sep 2006 21:09 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
From: MoeBlee on 29 Sep 2006 21:10 Poker Joker wrote: > >> By analogy, what you're saying is: > >> > >> For ANY x > >> there is a procedure to find a y such that x/y = 1. > > > > I said no such thing by analogy or otherwise. > > I did. Right, so when you said "By analogy, What YOU're saying is", [all caps added], the word 'your'e' does not belong there. > >> Because we are using the verbage "ANY", we don't > >> have to worry about special cases like when x = 0. > >> That's how mathematicians work? > > > > No, you've got it reversed. > > No, I don't. You neglected to point out the difference in > the analogy. I did. There's no analogy with any signficant feature of what I said, and the reason I say that is just as I posted to which you just responded by saying I don't understand the analogy. My points stand. You've not refuted uncountability proofs nor Arturo's point to which you originally responded. If you reject first order logic, then fine. No one says you have to accept first order logic. Then maybe ou can tell us what system of logic for mathematics you use instead. MoeBlee
From: MoeBlee on 29 Sep 2006 21:18 Poker Joker wrote: > > The problem here is very basic: You don't understand what a > > mathematician means when he argues such as this: "Let x be an arbitrary > > even number. [Then he proves some property P about x.] So, since x is > > arbitrary, property P holds for all even numbers." > > There also needs to be assurance that the property P actually holds > for all x. The property holds for all even numbers x since it hold of an ARBITARY even number x. That is reasoning that is as valid now as it always has been. > Just because some people can't come to grips with the > fact that there are special cases to consider, doesn't mean they > aren't there. If there are special cases that disqualify the generalization, then the generalization will NOT be permitted by the rule of universal generalization. That is the REASON the rule is drawn up the way it is. 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. Then you say, "What if x is prime? What if x is greater than 100? What if x is a successor of a prime? What if x is square? What if any number of special cases hold?" Then I say, as any mathematician would say, "If any of those special cases disqualified some even number from having property P, then I would not have been able to prove that P holds for an ARBITRARY even number." You REALLY don't understand that? MoeBlee
From: Poker Joker on 29 Sep 2006 21:32 "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. Suppose I claim that I have a list that contains all the reals. 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 because it is self-referential, which it must be if I truly gave you a list of all the reals. So in that *SPECIAL CASE*, unlike the general case, your construction isn't valid. The only way to eliminate that special case is to use what the conclusion of the proof would be if you neglected the special case. > It's as if, after hearing the proof that sqrt(2) is irrational, > you reply "But what if I give you a and b, both integers, such > that a/b = sqrt(2)? Your argument is demolished!". Yes, if > you provide what the proof shows can't exist then the proof is > wrong. But you can't just assume that that thing exists. No its not. Its like if you give me a proof that zero isn't a real number: For ANY x, there is procedure to construct a y, such that x = 1/y. When x = 0, there is no y. Therefore x is not a real number. If everybody neglects the fact that the construction isn't valid for x=0, then the proof is flawless.
From: Poker Joker on 29 Sep 2006 21:40
"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 (assuming you neglect the special case of 2 and nobody notices like you've done below). > Then you say, "What if x is prime? What if x is greater than 100? What > if x is a successor of a prime? What if x is square? What if any number > of special cases hold?" > > Then I say, as any mathematician would say, "If any of those special > cases disqualified some even number from having property P, then I > would not have been able to prove that P holds for an ARBITRARY even > number." > > You REALLY don't understand that? Yes, I understand. I also understand other things that you don't. I guess we'll leave it at that. |