Re: induction vs Cantor

From: Ron Sperber (ronsperber_at_optonline.net)
Date: 11/26/04 

Date: Fri, 26 Nov 2004 14:59:29 -0500

Poker Joker wrote:
> "Chairman of the David Hilbert Appreciation Society"
> <mathgeekxxiiii@hotmail.com> wrote in message
> news:8JqdnQxnYLcA6zrcRVn-3w@giganews.com...
>
>>Poker Joker wrote:
>>
>>>Let L_1 be a list of reals that implies a mapping F_1
>>>between the naturals and reals.
>>
>>Ok
>>
>>
>>>Let D_n be a Cantor anti-diagonal number that can be
>>>formed using the mapping F_n
>>
>>Ok
>>
>>
>>>Let L_n+1 be a list of reals by inserting D_n into L_n
>>>at row 2n and shifting down all the previous rows at 2n
>>>and above. This process is clearly an inductive process
>>>that creates a new mapping for each natural number.
>>>(L_n+1 could also be formed by prepending D_n to
>>>L_n.)
>>
>>Right. You outline a process to create infinitely many L_n,
>>none of which contains all of the reals.
>>
>>
>>>All of the D_n can be found in "infinitely many" mappings
>>>between the naturals and the reals.
>>
>>This part is a bit fuzzy. At this stage you don't have a proof
>>that any one L_n contains all of the reals; which is what you
>>would need to declare that Cantor made an error. That you are
>>hand-waving about something included in "infinitely many" mappings
>>is irrelevant since in no obvious way does such a thing relate
>>to a list, or a bijection.
>
>
> This part is not fuzzy. For all j > n, D_n is in L_j.
>
> The union of all the L_n, (a countable set) contains all the D_n.
>
>
>
So? I now can construct a NEW "antidiagonal". We have U(L_n) is
countable, so let N->U(L_n) by my bijection and I can find use the usual
argument to find an element not in this list.

The point to Cantor's proof is this. Given any f:N->R there exists some
r not in the image. The idea that you can keep adding real numbers to a
list doesn't change the proof.

  -Ron

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