[FOM] Proof via equiconsistency of ZFC with ZFC + ~ACR?

Rex Butler 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 
something nontrivial.

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.

Rex Butler

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