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 
But look what you write:

> On Nov 20, 2007, at 11:42 AM, Torgny Tholerus wrote:

>> An ultrafinitist comment to this:
>> ======
>> You can add this complementary sequence to the end of the list.
>> That will make you have a list with this complementary sequence
>> included.
>> But then you can make a new complementary sequence, that is not
>> inluded.  But you can then add this new sequence to the end of the
>> extended list, and then you have a bijection with this new sequence
>> also.  And if you try to make another new sequence, I will add that
>> sequence too, and this I will do an infinite number of times.

How could an ultrafinitist refute an argument by saying "... and this I 
will do an infinite number of times. "?

>> So
>> you will not be able to prove that there is no bijection...

Actually no. If you do what you described omega times, you will just 
end up with a set which can still be put in 1-1 correspondence with N 
(as shown in preceding posts on bijections)
To refute Cantor, here, you should do what you described a very big 
infinity of times, indeed an non enumerable infinity of times. But then 
you have to assume the existence of a non enumerable set at the start. 


>> ======
>> What is wrong with this conclusion?
>> -- 
>> Torgny
> Dr. Barry Brent
> http://home.earthlink.net/~barryb0/
> >

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