
Re: Who's up for a friendly round of debating CANTORS PROOF?
Posted:
Aug 17, 2011 7:00 PM


On Aug 17, 1:37 pm, WM <mueck...@rz.fhaugsburg.de> wrote: > On 17 Aug., 18:05, at1with0 <at1wi...@gmail.com> wrote: > > > Just to reiterate/elaborate. > > You can prove that there is no function from N onto R which implies R > > is uncountable. > > That would be the fact if R existed and the theory was free of > contradictions. > > Neither is true.
I'm not that interested in a mathematics without R.
Can you cite any semblance of a reason why Th(R) is inconsistent? Can you name two statements derivable from Th(R) which directly contradict each other?
> > > You do not need a diagonal trick to prove it because > > you can prove that no function exists from a set onto its powerset in > > general. > > No, not in general. Only in case of existing set and power set.
If not, you need to show me a set for which there is a function onto its powerset. So show me. And don't show me any bullshit pictures with trailing magical ellipses with wonderful properties. I want a set with a function onto its powerset.
Good luck with that.
> > > This is easy to prove in the finite case but the same proof > > works for the infinite case. > > This proof shows the inconsistency of set theory: > > The complete infinite binary tree contains all real numbers between 0 > and 1 as infinite paths i. e. infinite sequences { 0, 1 }^N of bits. > > 0. > / \ > 0 1 > / \ / \ > 0 1 0 1 > / > 0 ... > > The set { a_k  k in N } of nodes (bits) a_k of the tree is > countable. > > a0. > / \ > a1 a2 > / \ / \ > a3 a4 a5 a6 > / > a7 ... > > The complete infinite binary tree is the limit of the sequence of its > initial segments B_k: > ____________________ > B_0 = > > a0. > ____________________ > B_1 = > > a0. > / > a1 > ____________________ > B_2 = > > a0. > / \ > a1 a2 > ____________________ > ... > ____________________ > B_k = > > a0. > / \ > a1 a2 > / \ / \ > ... > ....a_k > ___________________ > ... > > The structure of the Binary Tree excludes that there are any two > initial segments, B_k and B_(k+1), such that B_(k+1) contains two > complete infinite paths both of which are not contained in B_k. > Nevertheless the limit of all B_k is the complete binary tree > including all (uncountably many) infinite paths. Contradiction. There > cannot exist more than countably many infinite paths.
The powerset of N is uncountable. What makes you so sure there are countably many infinite paths? Your path towards proving that would be to provide a function from the set of infinite paths into N (an injection). Or by providing a function from N onto the set of infinite paths. I hope you don't give me a circular argument, lol
> > Alternative consideration: Obviously every B_k is finite. None does > contain any infinite path. The infinite paths come into the play only > after all B_k with k in N (by some unknown mechanism). If that is > possible, however, this mechanism can also act in Cantor's diagonal > proof such that the antidiagonal enters the list only after all lines > at finite positions. > > Regards, WM
Irrelevant because, like I said, the proof that N < R (or N <  P(N), if you like) need not rely on Cantor's diagonal argument but, instead, can be done in purely settheoretical terms.
Of course you're going to deny Cantor's diagonal argument if you believe R does not exist.

