Torgny Tholerus wrote:
> Bruno Marchal skrev:
>> But then the complementary sequence (with the 0 and 1 permuted) is 
>> also well defined, in Platonia or in the mind of God(s)
>> *0* *1* *1* *0* *1* *1* ...
>> But *this* infinite sequence cannot be in the list, above. The "God" 
>> in question has to ackonwledge that.
>> The complementary sequence is clearly different
>> -from the 0th sequence (*_1_*, 0, 0, 1, 1, 1, 0 ..., because it 
>> differs at the first (better the 0th) entry.
>> -from the 1th sequence (0, *_0_*, 0, 1, 1, 0, 1 ... because it 
>> differs at the second (better the 1th) entry.
>> -from the 2th sequence (0, 0, *_0_*, 1, 1, 0, 1 ... because it 
>> differs at the third (better the 2th) entry.
>> and so one.
>> So, we see that as far as we consider the bijection above well 
>> determined (by God, for example), then we can say to that God that 
>> the bijection misses one of the neighbor sheep, indeed the "sheep" 
>> constituted by the infinite binary sequence complementary to the 
>> diagonal sequence cannot be in the list, and that sequence is also 
>> well determined (given that the whole table is).
>> But this means that this bijection fails. Now the reasoning did not 
>> depend at all on the choice of any particular bijection-candidate. 
>> Any conceivable bijection will lead to a well determined infinite 
>> table of binary numbers. And this will determine the diagonal 
>> sequence and then the complementary diagonal sequence, and this one 
>> cannot be in the list, because it contradicts all sequences in the 
>> list when they cross the diagonal (do the drawing on paper).
>> Conclusion: 2^N, the set of infinite binary sequences, is not 
>> enumerable.
>> All right?
> 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.  So you 
> will not be able to prove that there is no bijection...
> ======
> What is wrong with this conclusion?

You'd have to insert the new sequence in the beginning, as there is no 
"end of the list".

Brent Meeker
> -- 
> Torgny
> >

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