Re: Cantor's Diagonal

From: Torgny Tholerus <torgny.domain.name.hidden>
Date: Wed, 21 Nov 2007 17:33:25 +0100
Bruno Marchal skrev:
Le 20-nov.-07, à 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

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Received on Wed Nov 21 2007 - 11:33:47 PST

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