Cantor Confusion
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From: Dik T. Winter on 30 Sep 2006 19:52 In article <451dddd9(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > Randy Poe wrote: > > Poker Joker wrote: .... > >> 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. > > > > Consider r = 0.00000000... > > > > 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. Uh? By constructing a number where each digit is both 0 and 4? Wouldn't -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 30 Sep 2006 19:58 In article <YDmTg.25600$QT.1073(a)tornado.rdc-kc.rr.com> "Poker Joker" <Poker(a)wi.rr.com> writes: > > "Virgil" <virgil(a)comcast.net> wrote in message > news:virgil-9C1609.21071129092006(a)comcast.dca.giganews.com... > > > It is in mathematics. Once a proof for any list is established, it > > covers every list. > > This list doesn't contain 4: > > 1 > 2 > 3 > > Proof: > > The 1st number isn't 4. > The 2nd number isn't 4. > The 3rd number isn't 4. > That list does't contain 4 > > Therefore, Virgil believes that in mathematics, no > list contains 4. No. You are confusing "some list" with "any list". Understandable if you do not know mathematical terminology. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 30 Sep 2006 20:03 In article <virgil-D50D47.00443230092006(a)comcast.dca.giganews.com> Virgil <virgil(a)comcast.net> writes: > In article <451dddd9(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > > > The original proof was regarding a complete language using > > at least two symbols, m and w, no? > > Not quite. Two disjoint sets of synbols. Also not quite. Exactly two symbols. > > That was later conflated to a proof about the reals. > > It was later shown that it could be modified to form a proof that the > set of all reals is uncountable. That was already pretty early. Zorn already did show in his annotation that it could be so modified, but he noted that there was a problem with the dual representation. He solved this by noting that the set of numbers with dual representations was countable, and subtracting a countable set from an uncountable set still left an uncountable set. Later somebody (and I would really like to know who) modified the proof to use non-binary notation so that the dual representations would not be a (potential) problem. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 30 Sep 2006 20:17 In article <1159611187.639749.195260(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Tonico schrieb: > > > > The argument seems extremely simple to me...so simple that some seem to > > feel it MUST be wrong, for some reason (perhaps the reason is that if > > something ain't complex then it can't be maths...): take ANY list of > > real numbers (or take ANY injective mapping > > f:N --> R). Then it can be shown with the genial and simple diagonal > > argument that Cantor came up, > > Unfortunately, his method fails in some cases which is fatal for an > impossibility proof. The list, for instance, > > 0.0 > 0.1 > 0.11 > 0.111 > ... This is not a list that conforms to what Cantor wrote. You forget the trailing 0's. > with the prescription that the diagonal digit 0 is replaced by 1, > delivers a number which is not different from any list number, except > in the last digit, which, however, does not exist. 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 30 Sep 2006 20:12
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. 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. 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |