Virgil
Posts:
6,972
Registered:
1/6/11


Re: Mathematics in brief
Posted:
Dec 9, 2012 4:16 AM


In article <cf69cd13dc0d42af93204e52a2a503b9@u19g2000yqj.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 9 Dez., 05:55, Zuhair <zaljo...@gmail.com> wrote: > > > You need to prove that the set of all paths is countable, and so far > > you didn't present a proof of that. > > The set of all finite paths is countable. Therefore it is not possible > to define an infinite path by adding nodes to any finite path. That is both false and a nonsequitur. Given any finite sequence of left and right branchings, one can always in theory extend it in any of uncountably many ways to an infinite sequence. In practice one merely appends to it any infinite sequence of left and or right branchings.
> All > nodes to be added are already in finite paths. Therefore, by following > the nodes of a path, you never define an infinite path. It is > interesting that practically everybody not yet brainwashed can > understand that.
In actual mathematics, outside of WM's Wolkenmuekenheim,it is ONLY the brainwashed who "understand" that.
While the set of finite paths will have exactly one path ending at each node, the set of all paths will have uncountably many passing through each node. > > Hence, there remains only the possibility to define the infinite path > by a finite definition. But there are only countably many.
The set of all (possibly infinite) paths is determined only by whether an object is or is not a path, not by whether someone as far out as WM knows how to construct one. > > >I think using Cantor's argument > > one can prove that for any countably many paths of the Binary Tree > > there is a path that is not among them and thus establish a proof of > > uncountability of those paths. > > Then there is a contradiction.
You keep claiming that there is some such contradiction but have never produced a valid mathematical argument (that is valid outside Wolkenmuekenheim) supporting that claim.
> But you dismiss every contradiction?
What you claim to be contradictions are not provably contradictions outside of Wolkenmuekenheim.
> Therefore my proof mus be invalid???
You got it! > > But there is not even a contradiction. More precisely: There would be > a contradiction, if there wer uncountably many diagonal possible.
If you are talking about the Cantor "diagonal" proof, it does not claim uncountably many diagonals, it only claims, and proves, at least one more than can exist in any listing.
> But > in fact, no infinite sequence of paths defines a diagonalpath unless > the sequence has a finite definition such that every path is known.
Nonsense! In such a diagonal proof, the nth listed path need not be known at all beyond branch n+1.
> As > you know an infinite sequence cannot be defined by listing its terms. > You need a finite definition to define the sequence and its diagonal. > Again we reach the conclusion: There are only countably many finite > definitions.
Sets are not defined only by a list of definitions of their members. They are always defined by a rule distinguishing between their members and nonmembers, which may be, but need not be, list of those members. > > > > Math is discourse about "possible" form. Uncountability of the reals > > is PROVED in very weak fragments of ZFC, actually in PREDICATIVE > > fragments of second order arithemtic, which are PROVED to be > > consistent. This provide a discourse about form, thus uncountability > > of the reals is a possibility! thus it is mathematical, since reals > > can be interpreted as forms in the set hierarchy. > > See above. People have not taken into account that only a *defined* > sequence gives a defined diagonal. There are only countably many > finite definitions.
Again: none of the paths in such a list of paths need be defined beyond a finite number of places in order to define an antidiagonal provably not in that list.
This should be obvious even to someone as habitually unable to grasp the obvious as WM. > > Regards, WM 

