It is a well-known fact that closed convex sets in Banach spaces are weakly closed. The common proof is based on the Hahn-Banach theorem that uses the axiom of choice. Is there any proof of this fact that avoids the axiom of choice and uses only ZF?
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1$\begingroup$ Asaf's answer is simple and to the point, but I can't help myself from giving another example. In Solovay's model (with all subsets of $\mathbb{R}$ Lebesgue measurable), the dual space of $\ell^\infty$ is $\ell^1$, so the weak topology on $\ell^\infty$ is the same as the weak-* topology (as the dual of $\ell^1$). Therefore the unit ball of $c_0$ is norm-closed in $\ell^\infty$, but not weakly closed, because it is not weak-* closed (proving these facts does not require choice). $\endgroup$– Robert FurberCommented Nov 3, 2019 at 14:29
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$\begingroup$ @Robert: Since you mentioned Solovay's model, $\ell^\infty/c_0$ has a trivial dual there. $\endgroup$– Asaf Karagila ♦Commented Nov 3, 2019 at 15:40
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Of course not.
The Hahn–Banach is equivalent to the assertion that $X^*$ is nontrivial for any nontrivial Banach space (or normed space, if you prefer).
This means that if HB fails, there is a nontrivial Banach space $X$ whose weak topology is indiscrete, and in particular no set (other than $X$ and $\varnothing$) is weakly closed.
answered Nov 3, 2019 at 8:08