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


On Nov 13, 8:33 pm, "LudovicoVan" <ju...@diegidio.name> wrote: > "Zuhair" <zaljo...@gmail.com> wrote in message > > news:86a85cce2a844c9fb860527958274b50@o8g2000yqh.googlegroups.com... > <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 omegath endpoint, a_oo, would necessarily be drawn from an omegath > 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 > <https://groups.google.com/d/msg/sci.math/T2V4Jh7zzD8/wDM_wsyQZ0QJ> > > (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 omegath end point in (x_n) nor there is in (a_n) nor in (b_n) because to have an Omegath 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 omegath 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 omegath 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

