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Re: Objection to Cantor's First Proof
Posted:
Aug 1, 2012 1:45 PM
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On 8/1/2012 9:16 AM, LudovicoVan wrote:
> "Jim Burns" <burns.87@osu.edu> wrote in message
> news:501918E4.3030709@osu.edu...
>> The impression I draw from your posts is that you accept that
>> there _are no_ functions from N _onto_ R
>
> No, I don't. Let's make a quick recap: I have *proven* (no correction,
> no mistake) that if we are given an *enumeration* of the reals, the
> limit interval is degenerate (it is a singleton: I had wrongly called
> this interval "improper"). In a standard setting, with sequences as
> functions N->R, I claim that that is enough to prove that Cantor's First
> Theorem does not hold in general.
Since an enumeration of the reals would be a counter-example to
Cantor's First Theorem, it is to be expected that that would be
enough to prove it does not hold. If I assume that I have an
example where 1+1=3, that would be enough to prove that it is
not a theorem of PA that 1+1=2. But I don't.
Do you have an example of a complete enumeration of the reals?
> It is to the subsequent objection
> that a_oo (the limit end-point) itself would not be in the given
> enumeration that we reply: no, that needs an extended setting (the
> sequence would be extended, i.e. a function N*->R), then nothing is
> missing anyway.
In this new enumeration, a_oo would not be missing. That is not the
same as showing nothing is missing.
Do you agree that there exist bijections between N and N* ?
Do you agree that, if g: N -> N* is a bijection, and
f: N* -> R, f': N-> R, with f'(n) = f(g(n)), then
either f and f' are both onto or neither are?
The original argument still applies to the new enumeration,
so there is some real not in the enumeration, not a_oo but
some other real.
Also, remember that the original enumeration was assumed to be
complete. Despite this, a_oo was proven to not be in that enumeration.
Would you not agree that this makes the assumption of completeness
problematic?
>> That is to say, they are the same problem, with slightly
>> different decorations.
>
> It is not as simple as that:
All I say here is that every enumeration f: N* -> R corresponds to
an enumeration f': N -> R and that either f and f' are both onto
or neither onto. It really is very nearly that simple. I doubt it
it takes more than a couple lines to prove it.
> without at least compactification some
> problems are simply mis-modeled (and the standard definitions of
> limit inf and sup are just broken). Also, speaking in general, with
> transfinite ordinals the structures we get are much finer.
I don't see how compactification, lim sup, lim inf and transfinite
ordinals affect this point. All I see that is needed is the definition
of function, the definition of bijection, the definition of surjection,
and just a bit of logic.
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