Le 20-nov.-07, à 23:39, Barry Brent wrote :

##
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>
> 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).
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.
OK?
Bruno
http://iridia.ulb.ac.be/~marchal/
>> ======
>> What is wrong with this conclusion?
>>
>> --
>> Torgny
>>
>>>
>
> Dr. Barry Brent
> [EMAIL PROTECTED]
> http://home.earthlink.net/~barryb0/
>
>
>
>
> >
>
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