Cantor Confusion
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From: Poker Joker on 1 Oct 2006 09:08 "Peter Webb" <webbfamily-diespamdie(a)optusnet.com.au> wrote in message news:451e79ad$0$28952$afc38c87(a)news.optusnet.com.au... > > "Poker Joker" <Poker(a)wi.rr.com> wrote in message > news:3HmTg.25601$QT.205(a)tornado.rdc-kc.rr.com... >> I never tried to refute the uncountability of the reals. Too bad >> you've never been able to understand that. > No, but you introduced a specious point that seemed to support this > argument though. No I didn't. Showing a flaw in a proof is *FAR* from proving the opposite. > You said that for any real x exists y such that x/y=0. > > This statement is false, at least within standard arithmetic, as you point > out. Not if you neglect when x = 0. Then its true in general. > Cantor is different in that the statement is true, and easily proved. If you neglect x = 0. Then my statement is true and easily proved.
From: Poker Joker on 1 Oct 2006 09:13 "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.
From: mueckenh on 1 Oct 2006 09:43 Dik T. Winter schrieb: > In article <1159611066.767146.101490(a)e3g2000cwe.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > cbrown(a)cbrownsystems.com schrieb: > ... > > > Therefore, the assertion "M is a complete list of reals" is only true > > > if the assertion "M is complete, and M is not complete" is true. > > > > > > (A and ~A) = false. > > > > A system has the property W, if it can be proved that the reals can be > > well-ordered. A system has the property ~W if it can be proved that the > > reals cannot be well-ordered. A system is self-contradictive, if W and > > ~W can be proved. Therefore the system does not exist. > > The situation is slightly different. Neither W nor ~W can be proven, at > least, so mathematicians think. Zermelo was not a mathematician? He proved by what today is known as ZFC: Zermelo, E., "Beweis, daß jede Menge wohlgeordnet werden kann", Math. Ann. 59 (1904) 514 - 516 Zermelo, E., "Neuer Beweis für die Möglichkeit einer Wohlordnung", Math. Ann. 65 (1908) 107 - 128 > So either W or ~W can be taken as a new > axiom, leading to different branches of set theory. The case is similar > to the parallel postulate which can not be proven from the other postulates, > so either that postulate or its negation can be taken as an axiom, leading > to different branches of geometry. By forcing it can be proved that, even including AC, the reals cannot be well ordered. And if one had no proof of that, then experience would enforce this conclusion. > Morover, the negation can be taken in > two forms leading to elliptical and hyperbolic geometry. Similar with > ~W. You can take c = aleph_n for any n. The situation here is different. The axioms are the same in both cases, in particular AC is present. It is the same as if in *Euclidean* geometry the existence of two parallels through one point could not be be excluded. Regards, WM
From: mueckenh on 1 Oct 2006 09:55 Dik T. Winter schrieb: > Let's refrase it Cantor's way, please: > (m, m, m, m, m, ...) > (w, m, m, m, m, ...) > (w, w, m, m, m, ...) > (w, w, w, m, m, ...) > there is no element of the list that contains w's only. But the > diagonal constructed contains w's only. This is the typical one-eyed view of a set theorist. The same we have with Han's vase: Of course there is no ball which has not jumped out at noon. We cannot name any such number. But the other eye should see that there are more balls in than out at any time, including noon. The refore set theory is useless. One cannot calculate meaningfully with infinites! To come back to your argument: The diagonal differs from all the list numbers at most in one w. If there was not a list number begining with w for every w (except the first) of the diagonal, then the diagonal could not exist. So the diagonal differs in at most one w. That does not matter, however, because a last w can be shown to be not existing. Regards, WM
From: William Hughes on 1 Oct 2006 10:54
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. - William Hughes - William Hughes |