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It's known to be consistent with ZF+DC that every subset of $\mathbb{R}$ has the Baire property (BP). (E.g. Shelah's model). If so, then every subset of every complete separable metric space has the BP.

Can we drop the word "separable" here?

Is it consistent with ZF+DC that every subset of every complete metric space has the BP?

In other words, working in ZF+DC, can we prove there exists a complete (non-separable) metric space $X$ and a subset $E \subset X$ without the BP?

I'm not sure which way my intuition goes. On the one hand, non-separable metric spaces are big and maybe they can be weird, even without AC. On the other hand, a counterexample would have the property that the intersection of $E$ with every separable $S \subset X$ would (consistently) have the BP in $S$. That seems unlikely but I don't see how to disprove it.

As one possible approach, a nonmeager proper linear subspace of any Banach space $B$ must not have the BP (by the Pettis lemma). Can we create such a beast in ZF+DC, for some non-separable $B$? We can't do it by taking the kernel of a discontinuous linear functional $f$, because it's consistent with ZF+DC that there aren't any (by considering sequences, the restriction of $f$ to some closed separable subspace $S$ would be discontinuous, making its kernel nonmeager in $S$).

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    $\begingroup$ One specific idea I had: let $A$ be an uncountable set, let $B = \ell^\infty(A)$, and let $E \subset B$ be the space of all bounded $f : A \to \mathbb{R}$ having countable range. Is $E$ meager? $\endgroup$ – Nate Eldredge Jul 13 '16 at 23:11

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