Cantor Confusion
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From: David Marcus on 27 Dec 2006 23:32 Virgil wrote: > In article <1167256020.615407.109810(a)42g2000cwt.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > Dik T. Winter schrieb: > > > > > > > Note that in each finite tree, the complete edge assigned to a path is > > > > > the last edge on the path. As there is no last edge in an infinite > > > > > path, this does not hold in the infinite tree. > > > > > > > > So there are different paths without different edges? > > > > > > No. I never did state that. > > > > So at least one path has its own edge? > > Deliberate obfuscation. Do you think it is deliberate? Or, is WM really as confused as he appears? -- David Marcus
From: David Marcus on 27 Dec 2006 23:34 mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > "...; it is also possible to show that the ZFC axioms are not sufficient to > > prove the existence of a definiable (by a formula) well-order of the > > reals. However it is consistent with ZFC that a definable well- > > ordering of the reals exists -- > > I have not read the proof (because I don't like to spend my time with > useless topics), but a real expert of set theory told me that it has > been proved that there is no definable well ordering in ZFC Did this real expert also point out that most of the things that you think ZFC proves it doesn't? -- David Marcus
From: David Marcus on 27 Dec 2006 23:38 mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > > This bijection between sets (initial segments {1,2,3,...n} and > > > {2,4,6,...2n}) is only valid for finite sets. > > > > Suppose I instead counter-claim that this bijection is obviously "not > > valid" for finite sets; only for infinite sets. > > > > Is that what you call a "correct mathematical argument"? Simply > > claiming something is valid or not valid? How is it different from your > > own argument? > > It would be as impossible to contradict as your other claims or as the > following: > In every finite initial segment of natural numbers, there are as many > even numbers as odd numbers (plus or minus one number). But in the > infinite set N there are 250 times more even numbers than odd numbers. Are you saying this is true? If so, why/how? > > Thus "|N|/|R| < 1" is nonsense; as I previously stated > > quite clearly. > > I did not calculate |N|/|R| but edges per path. This calculation yields > a real number for any finite path - and the infinite path is nothing > else but the union of all finite paths. And, as you've said many times, anything true for finite is also true for infinite. Oh, wait. There is no theorem that says that. So, how does the fact that the infinite path is the union of its finite segments tell you anything about the edges per infinite path? -- David Marcus
From: David Marcus on 27 Dec 2006 23:41 mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > As every "personalization" separates a set of infinitely many paths > > which contain that edge from a set of infinitely many which do not, such > > an edge is not "personal" to a single edge, but only to a tree which is > > , except for a finite initial sub-path, tree-isomorphic the the entire > > tree. > > If the path representing an irrational number does not exist as an > individuum in an infinite tree, then there is no individual existence > of an irrational number. That's what I have been preaching for some > years meanwhile. How is this related to what Virgil wrote? -- David Marcus
From: David Marcus on 27 Dec 2006 23:44
mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > You are extending conclusions valid for finite trees to the infinite tree > > without proof, it is just your intuition. > > Not at all. Forget about all finite trees. In the infinite tree there > is always an edge which continues a path. There is no continuation > without an edge. That is unavoidable and it is enough to prove that > there are not less edges than paths. Rather than say it is enough to prove, why not simply give the proof? And, why, when you are asked about one alleged proof, do you switch to a completely different proof? -- David Marcus |