
Re: Cantor's first proof in DETAILS
Posted:
Nov 13, 2012 2:26 PM


On Nov 13, 10:22 pm, Zuhair <zaljo...@gmail.com> wrote: > 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
sorry for the last line, I meant,.... thus even with this case (x_n) which has the extended domain N*,.....
Zuhair

