Date: Nov 13, 2012 2:22 PM
Subject: Re: Cantor's first proof in DETAILS

On Nov 13, 8:33 pm, "LudovicoVan" <> wrote:
> "Zuhair" <> wrote in message
> <snip>

> > Theorem 4. for all i. x_i =/= L
> > <...>
> > Let J=L
> > QED
> Same argument, same objection: as easily proven, the limit interval here
> must be degenerate, that is it is the singleton (in interval notation)
> [L;L]. So what you claim amounts to saying that the limit value L is not in
> (x_n), but that is just incorrect in that if you consider the limit value,
> then of course it does belong to (x_n), in the limit!  More formally L =
> lim_{n->oo} (a_n) = lim_{n->oo} (b_n) , then just consider an injection from
> N* instead of N and you can even talk meaningfully about that "last value".
> You should rather try and show the mistake in my objection instead of
> proposing the same argument again and again.  As I had put it there:
> << an omega-th end-point, a_oo, would necessarily be drawn from an omega-th
> entry of the sequence!  Formally, we have the following property:
>     A m : a_m e (x_n) & b_m e (x_n)
> That works not only for n and m in N, but also for n and m in N*.  >>
> Objection to Cantor's First Proof
> <>
> (Note that mine is against Cantor's First Proof of which yours remains a
> paraphrase, not a faithful reproduction.)
> -LV

Honestly I couldn't understand your objection because it is not well
phrased (to me at least).
For instance what is the domain of (x_n), (a_n) , (b_n) in your
account? The first part of the argument (BEFORE the COROLLARY) that
I've presented here is stipulated clearly to be about (x_n) being an
injection from N to R, and also clearly it follows from the definition
of the a's and b's that (a_n) as well as (b_n) would have their
domains being N also, and N is stipulated clearly to be the set of all
natural numbers. So in this presentation there is NO omega-th end
point in (x_n) nor there is in (a_n) nor in (b_n) because to have an
Omega-th end point you need the domain of those functions to have
Omega (or any limit of all naturals) as an element, and N clearly
doesn't have Omega as an element, nor does it have any element of it
that has an omega-th position.

So your objection is (as far as I can tell) is not applicable to the
first part of the proof presented here (i.e. the part before the
corollary). Do you agree with that? I mean do you agree that the FIRST
part of the proof which is about injections from N to R succeed in
proving the existence of a real that is not in the range of all such
injections. Remember again N is fixed to be the set of all natural
numbers! so do you agree to this part of the proof?

Now about the idea of using N* instead of N as the domain of (x_n),
(a_n) and (b_n), where N* as far as I understand is some countable
infinite well ordered set that has an omega-th place like for example
N Union {oo} where oo is any limit of all natural numbers, an explicit
example would be N*=Omega+1={0,1,2,...,Omega} as defined by Von

Of course it is crystal clear that if we use N* as the domain of
(x_n), (a_n) and (b_n) then of course it can "Temporarily" elude
Cantor's argument (for that setting, but not for the setting where N
is the domain) this is clear, much as the diagonal put on top of the
original set does temporarily cast such impression, but unfortunately
still Cantor's argument Catches it, see the Corollary that I've
presented, it tackles the case where N* is the domain of (x_n) (a_n)
and (b_n), you'll see that we can define a new sequence (x'_n) that
have the same range as (x_n) but with the domain being N, and since
the domain of (x'_n) is N, then we'll simply apply the first part of
the proof on (x'_n) and elucidate some real that is not in its range,
and thus not in the range (x_n) (since (x_n) and (x'_n) have exactly
the same range) and thus even with this case (x'_n) which has the
extended domain N*, still would be proved to be missing a real (See
Corollary of this proof).