Re: Cantor's Diagonal

From: Barry Brent <barryb0.domain.name.hidden>
Date: Wed, 21 Nov 2007 16:14:21 -0600

That isn't a bijection.

Barry

On Nov 21, 2007, at 10:33 AM, Torgny Tholerus wrote:

> Bruno Marchal skrev:
>> Le 20-nov.-07, 23:39, Barry Brent wrote :
>>> You're saying that, just because you can *write down* the missing
>>> sequence (at the beginning, middle or anywhere else in the list),
>>> it follows that there *is* no missing sequence. Looks pretty
>>> wrong to me. Cantor's proof disqualifies any candidate
>>> enumeration. You respond by saying, "well, here's another
>>> candidate!" But Cantor's procedure disqualified *any*, repeat
>>> *any* candidate enumeration. Barry Brent
>> Torgny, I do agree with Barry. Any bijection leads to a
>> contradiction, even in some effective way, and that is enough (for
>> a classical logician).
>
> What do you think of this "proof"?:
>
> Let us have the bijection:
>
> 0 -------- {0,0,0,0,0,0,0,...}
> 1 -------- {1,0,0,0,0,0,0,...}
> 2 -------- {0,1,0,0,0,0,0,...}
> 3 -------- {1,1,0,0,0,0,0,...}
> 4 -------- {0,0,1,0,0,0,0,...}
> 5 -------- {1,0,1,0,0,0,0,...}
> 6 -------- {0,1,1,0,0,0,0,...}
> 7 -------- {1,1,1,0,0,0,0,...}
> 8 -------- {0,0,0,1,0,0,0,...}
> ...
> omega --- {1,1,1,1,1,1,1,...}
>
> What do we get if we apply Cantor's Diagonal to this?
>
> --
> Torgny
>
> >

Dr. Barry Brent
barrybrent.domain.name.hidden
http://home.earthlink.net/~barryb0/




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Received on Wed Nov 21 2007 - 17:15:25 PST

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