5 Mar
2009
5 Mar
'09
12:05 p.m.
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