Date: Nov 13, 2012 2:26 PM
Author: Zaljohar@gmail.com
Subject: Re: Cantor's first proof in DETAILS

On Nov 13, 10:22 pm, Zuhair <zaljo...@gmail.com> wrote:
> On Nov 13, 8:33 pm, "LudovicoVan" <ju...@diegidio.name> wrote:
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> > "Zuhair" <zaljo...@gmail.com> wrote in message
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> >news:86a85cce-2a84-4c9f-b860-527958274b50@o8g2000yqh.googlegroups.com...
> > <snip>

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> > > Theorem 4. for all i. x_i =/= L
>
> > > <...>
>
> > > Let J=L
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> > > 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:

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> >     A m : a_m e (x_n) & b_m e (x_n)
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> > That works not only for n and m in N, but also for n and m in N*.  >>
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> > Objection to Cantor's First Proof
> > <https://groups.google.com/d/msg/sci.math/T2V4Jh7zzD8/wDM_wsyQZ0QJ>

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> > (Note that mine is against Cantor's First Proof of which yours remains a
> > paraphrase, not a faithful reproduction.)

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> > -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
> Neumann.
>
> 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).
>
> Zuhair


sorry for the last line, I meant,.... thus even with this case (x_n)
which
has the extended domain N*,.....

Zuhair