[FOM] ZF versus subsystems of Z_2
Timothy Y. Chow
tchow at alum.mit.edu
Thu Sep 4 10:43:40 EDT 2008
Does ZF prove everything that Z_2 does? I would think not, because
various choice principles are provable in Z_2, whereas I believe that ZF
does not even prove Koenig's lemma. On the other hand, I didn't think
Brouwer's fixed-point theorem required AC, so clearly I'm confused about
something.
This question occurred to me when reading Tim Gowers's blog (which I just
discovered), where he asks (roughly speaking) what we would lose if we
dropped even countable choice and insisted that allegedly countable sets
always come equipped with a bijection with N.
http://gowers.wordpress.com/2008/08/10/a-small-countability-question/
Tim
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