2
$\begingroup$

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?

$\endgroup$
2
  • 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). $\endgroup$ Nov 3 '19 at 14:29
  • $\begingroup$ @Robert: Since you mentioned Solovay's model, $\ell^\infty/c_0$ has a trivial dual there. $\endgroup$
    – Asaf Karagila
    Nov 3 '19 at 15:40
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.