On 5 Mar 2009, at 17:06, Daniel Fischer wrote:
Cantor-Bernstein doesn't require choice (may be different for
intuitionists).
http://en.wikipedia.org/wiki/Cantor-Bernstein_theorem
Yes, that is right, Mendelson says that. - I find it hard to figure
out when it is used, as it is so intuitive.
Mendelson says AC is in fact equivalent proving
all x, y: card x <= card y or card y <= card x
Hans Aberg
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