Re: Cantor's Diagonal

From: Torgny Tholerus <torgny.domain.name.hidden>
Date: Tue, 20 Nov 2007 18:42:49 +0100

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

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Received on Tue Nov 20 2007 - 12:43:11 PST