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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