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Martin Väth
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Axiom of Countable Choice and meager sets
Asaf, yes that's what I meant exactly: It is consistent (with ZF) that UMM is unprovable. That's why I wrote "quite the opposite" (to the assertion that UMM is provable in ZF).
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Axiom of Countable Choice and meager sets
@D.S. Lipham. Quite the opposite. From GabeGoldberg's (correct) observation, it follows even in ZF that the real line is not meager. Hence the argument in the original question shows indeed that UMM is unprovable in ZF.
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