[FOM] Question about Freiling's axiom of symmetry
ali.enayat at gmail.com
Wed Aug 3 01:16:58 EDT 2011
This is a reply to Tim Chow's query (Aug 1), who has asked whether ZF
+ AS + LM + CH and ZF + AS + LM + ~CH are both
AS: Freiling's axiom of symmetry.
LM: All sets are Lebesgue measurable.
CH: The continuum hypothesis.
1. The usual formulation of CH as "R (reals) can be put in 1-1
correspondence with aleph_1" readily implies that R is
well-orderdable; and of course the well-orderability of R has been
known to contradict LM ever since Vitali's construction of
nonmeausrable sets, so ZF + LM + CH is already inconsistent since ZF +
LM proves ~CH.
2. ZF + AS + LM + ~CH holds in Solovay's model (in which LM holds); AS
holds there by Theorem 4 of Galen Weitkamp's paper [The Sigma^1_2-
theory of axioms of symmetry. J. Symbolic Logic 54 (1989), no. 3,
727–734] which shows that AS follows from LM, within ZF + DC
(dependent choice, which of course holds in Solovay's model).
3 However, if the continuum hypothesis is formulated in the weaker
form WCH := "every uncountable subset of R can be put into 1-1
correspondence with R", then ZF + AS + LM + WCH holds in Solovay's
model since in Solovay's model every uncountable subset of R has a
4. In ZFC, WCH and CH are equivalent, so even though ZFC + AS proves
~WCH (as shown by Freiling), (3) shows that ZF + DC + AS + WCH is
5. I do not know the status of ZF + LM + ~WCH.
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