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Wikipedia states that, in $ZF$, the Axiom of Choice ($AC$) implies the Hahn-Banach theorem, but that the Hahn-Banach theorem does not imply $AC$. It also states that in $ZF$, the Hahn-Banach theorem implies the paradoxical decomposition of the sphere. My question is simply this:

In $ZF$, does the Hahn-Banach theorem imply some weak choice principle, and if so, what is it? Also in $ZF$, is this weak choice principle (if it exists) equivalent to Hahn-Banach?

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    $\begingroup$ Have you seen this thread mathoverflow.net/questions/45844/hahn-banach-without-choice ? Does it answer your question (maybe this thread should be marked as a duplicate in this case)? $\endgroup$
    – Johannes Hahn
    Commented Jan 9, 2015 at 16:34
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    $\begingroup$ What do you mean when you write a choice principle? $\endgroup$
    – Asaf Karagila ♦
    Commented Jan 9, 2015 at 16:34
  • $\begingroup$ You should obtain a copy of Herrlich "The Axiom of Choice", I think it might answer a lot of your questions before you even ask them here. $\endgroup$
    – Asaf Karagila ♦
    Commented Jan 9, 2015 at 16:35
  • $\begingroup$ @JohannesHahn: It gets me closer, but not quite there.... $\endgroup$
    – Thomas Benjamin
    Commented Jan 9, 2015 at 17:15

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Over $ZF$, the Hahn-Banach theorem is equivalent to the statement that every Boolean algebra $B$ admits a real-valued measure, i.e., a non-negative real-valued function $\mu$ on $B$ such that $\mu(0)=0$, $\mu(1)=1$ and $\mu(a \vee b)=\mu(a)+\mu(b)$ whenever $a \wedge b = 0$. This statement can be seen as a choice principle via, e.g., Zorn's lemma, which can be used to build such a measure. But it is also possible to construct this measure using just the Boolean Prime Ideal theorem ($BPI$), which is known to be strictly weaker than $AC$. More about this can be found in the usual literature, Howard & Rubin "Consequences of the Axiom of Choice".

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  • $\begingroup$ Does this real-valued measure extend to a probability measure on $B$? Also, what about the equivalence of Hahn-Banach to $WKL_0$ in reverse mathematics? $\endgroup$
    – Thomas Benjamin
    Commented Jan 12, 2015 at 12:24
  • $\begingroup$ It's clear that such a measure must take values within $0$ and $1$, but in order to even state countable additivity you would need $B$ to have countable joins and meets, which is not necessarily going to happen. Regarding $WKL_0$, you need to specify which base theory and which statement of Hahn-Banach theorem you are considering. $ZF$ is too strong since it proves Weak König's lemma, and the general formulation of Hahn-Banach theorem as in the above equivalence involves arbitrary vector spaces not necessarily definable in second order arithmetic. $\endgroup$
    – godelian
    Commented Jan 12, 2015 at 13:06

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