The reason it isn't a bijection (of a denumerable set with the set of
binary sequences): the pre-image (the left side of your map) isn't
a set--you've imposed an ordering. Sets, qua sets, don't have
orderings. Orderings are extra. (I'm not a specialist on this stuff
but I think Bruno, for example, will back me up.) It must be the
case that you won't let us identify the left side, for example, with
{omega, 0, 1, 2, ... }, will you? For if you did, it would fall under
Cantor's argument.
Barry
On Nov 21, 2007, at 10:33 AM, Torgny Tholerus wrote:
> 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
>
> >
Dr. Barry Brent
[EMAIL PROTECTED]
http://home.earthlink.net/~barryb0/
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