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Re:Cantor diagonalisation by danharaj on Friday March 10, 2017 @09:04PM (#13843736)
I think this blog post will delight anyone who liked this submission: http://math.andrej.com/2007/04/08/on-a-proof-of-cantors-theo...
Re:Cantor diagonalisation by danbruc on Friday March 10, 2017 @10:17PM (#13844010)
I am already struggling right at the beginning.

If we open a book on set theory, we will find a proof of Cantor’s theorem which shows explicitly that for every map e : A → P(A) there is a subset of A outside its image, namely S = { x ∈ A ∣ x ∉ e(x) }

What about e(x) = { x }? In that case S would be the empty set, wouldn't it? Does map imply some additional properties in this case? Am I misunderstanding the statement and it does not try to imply that S is non-empty for every e?

Re:Cantor diagonalisation by danharaj on Friday March 10, 2017 @10:34PM (#13844063)
Yes, in that case S is {}. {} ∈ P(A) but is not in the image of e.
Re:Cantor diagonalisation by danbruc on Friday March 10, 2017 @10:48PM (#13844103)
Thanks, now it clicked. I misinterpreted the image as the image of x instead of as the image of e. And then did something weired that now makes no longer any sense at all.