Cantor Confusion
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From: Virgil on 10 Nov 2006 16:51 In article <1163154101.329683.20480(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > If it is not a member, then A' has one member less than the otherwise > > > same set A where it is a member. Therefore A and A' are different, > > > aren't they? > > > > > > > Completely irrelevent. The bijection is not between one set without > > supremum > > and this set plus supremem. The fact that there are aleph_0 terms in > > the first column, does not mean there is an aleph_0 th line. > > What makes up the difference to the fact that there are not aleph_0 > terms in any line? > Or is there no difference between aleph_0 terms and less than aleph_0 > terms? Not if the "difference" is finite. > > > Remember, > > the fact that there are infinitely many lines does not mean > > that ther is a line with index aleph_0. > > The column has infinitely many indexes. No line has infinitely many > indexes. No difference? Not significant in comparing the number of lines, when the line lengths are unbounded to the number of columns. > Lines with finitely many indexes cannot exhaust the column with its > infinitely many. We learn: Infinity cannot be exausted. Infinite sets CAN be exhausted, but only by infinite processes. WM's problem is that he does not, perhaps cannot, understand that there is a difference in the way infinite sets and infinite processes work from the way that finite sets and finite processes work. Getting over than hurdle is one of the major problems that all beginning calculus students must face and conquer. WM has not yet done so.
From: Virgil on 10 Nov 2006 16:52 In article <1163154585.218606.42810(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > The length of the diagonal will be longer than every > > line if and only if there is no line with maximum length. > > Therefore also the diagonal cannot have maximum lengh omega > n. > > > As there is no last line, there is no line with maximum length. > > Exactly. And therefore there is no "number" omega. Except that there is an ordinal omega.
From: Virgil on 10 Nov 2006 16:53 In article <1163154922.989872.54560(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > Okay, so for you there is no point to draw about your quote other than > > understanding Cantor's own writings. And, I'll add that in this > > particular instance, Cantor, as you translated, is not in conflict with > > the theorem of current set theory that omega is a limit ordinal and the > > first oridinal that is greater than all natural numbers. > > That is what I said. WRONG!
From: Virgil on 10 Nov 2006 16:57 In article <1163155421.780070.100420(a)e3g2000cwe.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > > > This number does not exist. The diagonal has not omega positions. > > > > So I guess the answer to my question is that you are not prepared to > > show that my proof has anything more than first order logic applied to > > Z set theory nor are you prepared to demonstrate a contradiction in Z > > set theory. > > Your proof shows either a contradiction in Z or it shows that Z is > false. Claimed but not proven, and, in fact, false. > > Z contains a contradiction to your proof, if one can show in Z that a > diagonal of a matrix has not more elements than every line. If every line is finite but of length at least equal to its line number then the diagonal for such a list IS of "greater length" than EVERY line. I.e., omega is greater that every member of omega. Z is false, > if one cannot show that. So, why further bother about Z? > > Regards, WM
From: William Hughes on 10 Nov 2006 17:01
Franziska Neugebauer wrote: > William Hughes wrote: > > > The length of the set of natural numbers is the supremum of the > > lengths of the initial segments. > > Since when do sets have a length? Substitute size or cardinality if you do not like the term length in this context. - William Hughes |