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".

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> > 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