|
|
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.fh-augsburg.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 anti-diagonal 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 set-theoretical terms.
Of course you're going to deny Cantor's diagonal argument if you
believe R does not exist.
|
|
