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.
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
>> 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
> 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?
Dr. Barry Brent
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