[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 

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.



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