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

On Nov 20, 2007, at 11:42 AM, 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?
> -- 
> Torgny
> >

Dr. Barry Brent

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