Virgil
Posts:
4,480
Registered:
1/6/11
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Re: Objection to Cantor's First Proof
Posted:
Jul 26, 2012 4:53 PM
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In article <juqvng$euj$1@speranza.aioe.org>,
"LudovicoVan" <julio@diegidio.name> wrote:
> "Virgil" <virgil@ligriv.com> wrote in message
> news:virgil-95F48D.22544224072012@bignews.usenetmonster.com...
>
> > A PROOF OF THE UNCOUNTABILITY OF THE REALS
> > (A variation on Cantor's FIRST proof)
>
> > The intersection of such a nested sequence of closed intervals is not
> > empty, but the natural associated with any of its members is necessarily
> > larger than all of the infinitely many natural numbers associated with
> > those infinitely many endpoints.
>
> Wrong: by completeness (which you do assume), it must be a_oo == b_oo,
Even if that were true, it is clear that neither a_oo nor b_oo can be a
members of the original sequence from which the a_n and b_n were
derived. Thus EVERY sequence of reals STILL omits at least one real, and
Cared(N) < Card(R) as claimed.
Here is my proof again, so anyone can check that LV is wrong:
*********************************************
A PROOF OF THE UNCOUNTABILITY OF THE REALS
(A variation on Cantor's FIRST proof)
ASSUMPTIONS:
(1) the intersection of a strictly nested sequence of closed real
intervals (the endpoints of each interval being interior points of the
previous interval) is not empty.
(2) A strictly increasing sequence of naturals does not have a natural
as its limit,
(3a) A strictly increasing but bounded sequence of reals has a real
number as a limit, its least upper bound, different from every member of
the sequence.
(3b) A strictly decreasing but bounded sequence of reals has a real
number as a limit, its greatest lower bound, different from every member
of the sequence.
Proof:
If the reals are countable then we may assume each real can be and has
been paired with a natural so that different reals are paired with
different naturals with none of either left out.
Assuming this has been done, take the two reals corresponding to the
lowest naturals as endpoints of a real interval.
It is clear that all the interior points of this real interval must be
paired with naturals larger that paired with its endpoints.
Now take the two reals with the lowest naturals INTERIOR to the previous
interval to be the endpoints of a subinterval of that interval.
It is clear that the interior points of this real interval must be
paired with naturals larger than the naturals paired with its endpoints.
By repeating this process one generates a decreasing, but never empty,
sequence of closed real intervals each of which contains only points
with higher attached naturals than its endpoints have.
The intersection of such a nested sequence of closed intervals is not
empty, but the natural associated with any of its members is necessarily
larger than all of the infinitely many natural numbers associated with
those infinitely many endpoints.
But there is no natural number larger than infinitely many different
natural numbers.
This is a contradiction which can only have been caused by our original
assumption that the reals were countable, so it proves they are not
countable.
QED!
NOTE: While I have never seen this particular proof in the literature of
countability,
it is so obvious that I doubt that it is original with me.
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