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Their equality would imply s(m)(m)=0.5, which is impossible. So g and s(m) differ at the mth term. Therefore g!=s(m) for any m, thus s is not surjective. QED [[User:Scineram|Scineram]] ([[User talk:Scineram|talk]]) 09:57, 30 October 2009 (UTC)
== proof or illusion? ==
Statement from 'Cantor's diagonal argument', in section 'An uncountable set':
"Therefore this new sequence s0 is distinct from all the sequences in the list."
Should be:[all the sequences in the '''sample''' list.] You can't reach a conclusion for all, based on a comparison of some. For example, a doctor can't state with certainty that 100 people don't have disease A, based on examination of 10 from that group.
The problem with isolated experiments is the denial of information that would assist in making an accurate conclusion. If the list L is ordered, and divided into two subsets, L0, all sequences beginning with '0', and L1, all sequences beginning with '1', the first comparison and symbol swap would exclude the subset L0, and it is only necessary to compare s0 to a member of the subsets L10 and L11 for the next position. The process becomes more efficient by eliminating redundant comparisons. A simple pattern emerges with the choice for each position always the same, '0' or '1'. If s0 is compared to a member of one subset, its symbol swap will place it in the complementary subset. Continuing with the current 7-symbol sequence, the next comparison would be with one of the following subsets, each containing an unlimited number of sequences not compared, that match s0 to the current position.
L10111010={10111010...} (all sequences beginning with s0 and appended with 0)
L10111011={10111011...} (all sequences beginning with s0 and appended with 1)
conclusion:
It's obvious that no matter which choice, there is an existing sequence matching s0.
The refined process is equivalent to forming any sequence.
The original process is not forming a 'new' sequence, it's just copying an existing one.
The diagonal and partial list are instances of misdirection.
The remainder of the article has now lost its foundation.
Assuming in principle, if one complete sequence can be formed, then all possible sequences can be formed and presented as a complete ordered list L, otherwise there is no list.
In either case the diagonal proof fails.
[[User:Phyti|Phyti]] ([[User talk:Phyti|talk]]) 18:59, 21 March 2013 (UTC)
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