Most functional analysis books prove that Zorn's Lemma$\implies$Hahn-Banach$\implies$"The weak topology on every Banach space is Hausdorff", however it is known that the Hahn-Banach theorem can be proved from the Boolean Prime Ideal theorem, which is strictly weaker than $\sf ZL$, and that $\sf ZF+HB$ is strictly weaker than $\sf ZF+BPI$.

Let $\sf WH$ be the assertion "The weak topology on every Banach space is Hausdorff", what's the relation between $\sf ZF$,$\sf ZF+WH$ and $\sf ZF+HB$?


1 Answer 1


Well, you just get the Hahn–Banach theorem.

Luxemburg, W. A. J. and Väth, M. The existence of non-trivial bounded functionals implies the Hahn-Banach extension theorem. Z. Anal. Anwendungen 20 (2001), no. 2, 267–279. MR1846601

In the above paper the authors show that the Hahn–Banach theorem is equivalent to the statement "Every Banach space has a nontrivial continuous functional". This means that if the Hahn–Banach theorem fails, there is a Banach space $X$ with a trivial [continuous] dual, and therefore the weak topology on $X$ is the indiscrete one.

So actually what we get is that just "The weak topology on a nontrivial Banach space is not indiscrete" is itself enough to prove the Hahn–Banach theorem, let alone requiring it is Hausdorff.

  • 1
    $\begingroup$ That's a very surprising equivalent of the Hahn-Banach theorem! $\endgroup$ Jan 20, 2018 at 14:22
  • $\begingroup$ Indeed it is! :) $\endgroup$
    – Asaf Karagila
    Jan 20, 2018 at 14:24

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