Cantor Confusion
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From: Virgil on 3 Oct 2006 20:08 In article <AiCUg.42$LU2.22(a)tornado.rdc-kc.rr.com>, "Poker Joker" <Poker(a)wi.rr.com> wrote: > "William Hughes" <wpihughes(a)hotmail.com> wrote in message > news:1159714452.259786.215790(a)b28g2000cwb.googlegroups.com... > > > > Poker Joker wrote: > >> "Randy Poe" <poespam-trap(a)yahoo.com> wrote in message > >> news:1159578269.577169.76000(a)m73g2000cwd.googlegroups.com... > >> > >> >> 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. > >> > > >> > That doesn't make sense. You are saying that every digit of r > >> > both is equal to 4 and is equal to 5. > >> > >> So when it's put in extremely simple terms, then you understand > >> that the process doesn't always make sense. > > > > No the sentence above makes no sense. The sentence > > above has nothing to do with the proof. > > > > Diagonal process. > > > > We make a number > > d by specifying every one the digits > > in a decimal expansion. > > > > for every n > > > > find the real number r=f(n) > > > > express r in decimal notation (note this won't > > work if you express r in binary notation. Soluiton. > > Don't do that!) > > > > find the n'th decimal digit of r, If > > this is 5 set d_n to 4. Otherwise set > > d_n to 5. > > > > The only assumption made is that f(n) is a real number > > for every natural number n (i.e. that we have a list). > > Even if (contrary to fact) we assume the list contains all > > real numbers we can still find d. (Since we made an > > assumption contrary to fact this will lead to a contradiction. > > This is of course the whole point!) > > > > Assume that the list contains all real numbers. > > Make d as above (note there is no step you > > cannot do). Since the list contains all > > real numbers d must be in the list. Let d = f(m) > > Then the mth digit of d must be both 4 and 5. > > Contradiction. > > > > Note that we create d first and find the > > contradiction second. We do not start > > with a contradiction. > > f(m) is the mth real. What does d have to do with it? > d is the output of a seperate process, isn't it? > > R is the image of f(m). > d = PROCESS( f( n : n in N ) ) > so d = PROCESS(R). > if d in R then d is self-referential and therefore meaningless. How does 'd' being a member of R make anything "self-referential"? Is PJ saying that 'd' cannot be a member of R because it might be a member of R? PJ is still off in the clouds with no contact with reality.
From: Ross A. Finlayson on 3 Oct 2006 20:14 Virgil wrote: > In article <J4CUg.36$LU2.19(a)tornado.rdc-kc.rr.com>, > "Poker Joker" <Poker(a)wi.rr.com> wrote: > > > "Virgil" <virgil(a)comcast.net> wrote in message > > news:virgil-B9C4DD.12563101102006(a)comcast.dca.giganews.com... > > > In article <oaPTg.5137$3E2.3874(a)tornado.rdc-kc.rr.com>, > > > "Poker Joker" <Poker(a)wi.rr.com> wrote: > > > > > >> "Randy Poe" <poespam-trap(a)yahoo.com> wrote in message > > >> news:1159578269.577169.76000(a)m73g2000cwd.googlegroups.com... > > >> > > >> >> 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. > > >> > > > >> > That doesn't make sense. You are saying that every digit of r > > >> > both is equal to 4 and is equal to 5. > > >> > > >> So when it's put in extremely simple terms, then you understand > > >> that the process doesn't always make sense. > > > > > > Joker's processes do not make sense to anyone but Joker. > > > > > > Which is possibly why he calls himself "Joker". > > > > Virgil spends his days babbling. > > PJ has a very limited vocabulary. > Which matches his limited mathematical and logical comprehension. > Which is why he was so wrong. Virgil, ya pig, quit trying to drag people in the mud. You're getting mud on people. You do that to everybody, in extended discussion. I thought Poker Joker had some interesting things to say. I thought it was funny when he got back, saw what Virgil had written, and got really pissed. I actually disagree with Virgil's claims about your vocabulary, so I can make fun of his. Joker, what are you telling me about infinite sets? A poker deck has got jokers in it, you see, 54 cards, plus sometimes the instruction card. Joker, and I've told people this before, please try and ignore Virgil. Better, or worse? Better here, or better here? If you're to learn about Cantor, George Cantor, I suggest you go somewhere else. Obviously this discussion is not on-topic. Ross
From: MoeBlee on 3 Oct 2006 20:15 Poker Joker wrote: > "William Hughes" <wpihughes(a)hotmail.com> wrote in message > news:1159714452.259786.215790(a)b28g2000cwb.googlegroups.com... > > > > Poker Joker wrote: > >> "Randy Poe" <poespam-trap(a)yahoo.com> wrote in message > >> news:1159578269.577169.76000(a)m73g2000cwd.googlegroups.com... > >> > >> >> 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. > >> > > >> > That doesn't make sense. You are saying that every digit of r > >> > both is equal to 4 and is equal to 5. > >> > >> So when it's put in extremely simple terms, then you understand > >> that the process doesn't always make sense. > > > > No the sentence above makes no sense. The sentence > > above has nothing to do with the proof. > > > > Diagonal process. > > > > We make a number > > d by specifying every one the digits > > in a decimal expansion. > > > > for every n > > > > find the real number r=f(n) > > > > express r in decimal notation (note this won't > > work if you express r in binary notation. Soluiton. > > Don't do that!) > > > > find the n'th decimal digit of r, If > > this is 5 set d_n to 4. Otherwise set > > d_n to 5. > > > > The only assumption made is that f(n) is a real number > > for every natural number n (i.e. that we have a list). > > Even if (contrary to fact) we assume the list contains all > > real numbers we can still find d. (Since we made an > > assumption contrary to fact this will lead to a contradiction. > > This is of course the whole point!) > > > > Assume that the list contains all real numbers. > > Make d as above (note there is no step you > > cannot do). Since the list contains all > > real numbers d must be in the list. Let d = f(m) > > Then the mth digit of d must be both 4 and 5. > > Contradiction. > > > > Note that we create d first and find the > > contradiction second. We do not start > > with a contradiction. > > f(m) is the mth real. What does d have to do with it? > d is the output of a seperate process, isn't it? What d has to do with it is that d is the decimal expansion of a real number that is not in the range of f. As to "process", d is a function, which, yes, is not the function f. > R is the image of f(m). If by 'R' you mean the set of real numbers, then if the proof is by contradiction, then R is assumed to be the range of f; then a contradiction is derived, which proves that R is not the range of f. If not a proof by contradiction, then we don't assume that R is the range of f (and we don't assume that R is not the range of f; we simply make no assumption either way) but we still prove that R is not the range of f. > d = PROCESS( f( n : n in N ) ) Whatever notation that is, and whatver it is supposed to mean, it is not notation we use for the proof and it is not obvious what notation in the proof it corresponds to. What we actually have is: f: N -> {e | e: N -> {0 1 2 3 4 5 6 7 8 9}}. I.e., f is a function from the set of natural numbers into the set of denumerable sequences of members of {0 1 2 3 4 5 6 7 8 9}. d: N -> {0 1 2 3 4 5 6 7 8 9} defined by: d(n) = the function that is equal to f(n) except d(n)(n)= 4 if f(n)(n) = 5, and d(n)(n) = 5 otherwise. There is no "self reference". For each n, we defined the value of d(n) in terms of n and the already given function f. > so d = PROCESS(R). Only you can say mathematical theory and/or mathematical language that notation regards. > if d in R then d is self-referential and therefore meaningless. d is a decimal expansion of a member of R. The meaning of that is clear and there is no disqualifying self-referentiality, whatever that might be. Again, just first order logic applied to axioms of set theory. There's nothing else that the proof requires. And the steps in a fully formalized proof can be machine verified. The proof is just basic and correct mathematical reasoning applied to axioms. MoeBlee
From: Randy Poe on 3 Oct 2006 21:56 Poker Joker wrote: > "Randy Poe" <poespam-trap(a)yahoo.com> wrote in message > news:1159801355.575790.302710(a)i3g2000cwc.googlegroups.com... > > > > Poker Joker wrote: > >> "Randy Poe" <poespam-trap(a)yahoo.com> wrote in message > >> news:1159578269.577169.76000(a)m73g2000cwd.googlegroups.com... > >> > >> >> 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. > >> > > >> > That doesn't make sense. You are saying that every digit of r > >> > both is equal to 4 and is equal to 5. > >> > >> So when it's put in extremely simple terms, then you understand > >> that the process doesn't always make sense. > > > > No, I understand that you wrote something that doesn't > > make sense, but that also bears no resemblance to > > the proof you're confused about. > > If you don't understand, then you are confused. That's > why you have no real answer. > > > Nowhere does anybody sane say that there is a digit > > which is simultaneously required to have two different > > values, in any variant of the Cantor proof (except > > yours). > > Nobody "says" anything in any version. Nobody, including Cantor, has ever written a proof of the uncountability of the reals? Are you sure? Then what proof exactly are you taking issue with? - Randy
From: David R Tribble on 4 Oct 2006 01:12
Ross A. Finlayson wrote: > Joker, and I've told people this before, please try and ignore Virgil. Why not ignore Poker Joker? At least Virgil knows set theory. Virgil, you really ought to refrain from feeding the trolls. Their responses are only cluttering up the newsgroup. |