[FOM] Proof via equiconsistency of ZFC with ZFC + ~ACR?
rexbutler at hotmail.com
Mon Jun 27 23:06:59 EDT 2005
ZFC is equiconsistent with ZFC plus the negation of the Axiom of Choice for
the real numbers. As such, it is impossible to define in ZFC a function F
from the nonempty subsets of [0,1] to [0,1] such that F(A) is an element of
A for all nonempty A subset [0,1].
However, the other day I wondered if there might be functions F as above
that are 'nearly choice functions' in the sense that the existence of an A
such that F(A) is not in A becomes a nontrivial result. The hope of course
being that one could creatively construct such F in the hope of proving
Alternatively, one could attempt proving the equiconsistency of ZFC with ZFC
+ ~C where C is the assertion of a choice function for a smaller class of
nonempty subsets of [0,1], and then proceed from there.
Are there any prospects for this? Thanks.
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