MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The standard proof of the Hahn-Banach theorem makes use of Zorn's lemma. I hear that, however, Hahn-Banach is strictly weaker than Choice. A quick search leads to many sources stating that Hahn-Banach can be proven using the ultrafilter theorem, but I cannot seem to find an actual proof. So...

  1. What is the ultrafilter theorem; and
  2. How does the said theorem imply the Hahn-Banach theorem?

Any easily-accessible reference would be quite enough; thanks in advance!

share|cite|improve this question
I can't understand the vote to close. The question is a good one and the questioner asks for a reference of a result he can't find after looking around. So what's the problem? – Todd Trimble Nov 12 '10 at 17:34
I second Todd's remark (though to be fair the software doesn't allow one to rescind votes-to-close which were cast in haste) – Yemon Choi Nov 12 '10 at 18:01
up vote 14 down vote accepted

The ultrafilter theorem is the statement that any filter on a set can be extended to an ultrafilter. It is perhaps more common to see it sated as the (Boolean) Prime ideal theorem: Every Boolean algebra admits a prime ideal.

The Hahn-Banach theorem is actually equivalent to the statement that every Boolean algebra admits a real-valued measure, but this is not entirely straightforward (see Luxemburg, "Reduced powers of the real number system and equivalents of the Hahn-Banach extension theorem", Intern. Symp. on the applications of model theory, (1969) 123-127).

For a discussion of Hahn-Banach vs. Choice and some additional remarks and references, see Jech "The axiom of choice", North-Holland, 1973.

You may also be interested in the references I include in this answer.

share|cite|improve this answer
A curious remark. Hahn-Banach (HB) is strictly weaker than AC, in the sense that ZF+HB does not prove AC. There is another proposition of functional analysis in a similar state: the Krein-Milman theorem (KM), which states a nonempty compact convex set in a Banach space has an extreme point. It is known that ZF+KM does not prove AC. But (the curious bit) ZF+KM+HB does prove AC. – Gerald Edgar Nov 12 '10 at 18:44
Hi Gerald. That's interesting! I don't recall having run into that statement before. Do you know of a reference by any chance? – Andrés E. Caicedo Nov 12 '10 at 19:13
All I could find quickly was MR0282186 in MathSciNet. That seems to say that it is PI+KM that proves AC, while HB is weaker than PI. – Gerald Edgar Nov 12 '10 at 20:03
To complement Gerald's comment: The paper Gerald mentions is Bell-Jellett. "On the relationship between the Boolean prime ideal theorem and two principles in functional analysis", Bull. Acad. Polon. Sci. Ser. Sci. Math. Astronom. Phys. 19 1971 191–-194. From the review by Luxemburg: "The authors’ main purpose is to show that the Hahn-Banach extension principle (HB) and an extended version of the Kreın-Milman theorem (VKM), stating the existence of extreme points for quasicompact convex subsets of a locally convex Hausdorff space, imply the prime ideal theorem for Boolean algebras (PI)." – Andrés E. Caicedo Nov 12 '10 at 22:55
Also, "From the folk theorem stating that the Kreın-Milman theorem (KM) and (PI) effectively imply the axiom of choice (AC), it follows immediately also that (HB) and (VKM) imply (AC)." Since (PI) does not imply (AC), we have that (KM) and (HB) do not suffice to prove (AC), in particular, (VKM) is strictly stronger than (KM). – Andrés E. Caicedo Nov 12 '10 at 22:55

An amusing related fact is that the non linear Hahn-Banach theorem (that every Lipschitz one function into the reals from a subset of a metric space can be extended to a Lipschitz one function on the entire space) is provable in ZF. I believe it is proved this way in the book of Araujo and Gine.

share|cite|improve this answer
Well, if I understand what you mean, one way to extend a Lipschitz-1 $f: A \to \mathbb{R}$ along a subspace embedding $i: A \to X$ is via the function $g: X \to \mathbb{R}$ defined by $g(x) = \inf_{a \in A} f(a) + d(x, i(a))$. Believe it or not, this construction is a special case of an enriched Kan extension in category theory: (page 32 of 37 mentions this special case). – Todd Trimble Feb 1 at 3:17

This might be irrelevant, but I would like to point out that if one restates Hahn-Banach theorem in the language of locales (replacing topological spaces), one can get rid of the axiom of choice and the ultrafilter theorem altogether. See the paper “A Direct Proof of the Localic Hahn-Banach Theorem” by Thierry Coquand and references therein.

share|cite|improve this answer
And of course this is true also for several other choice-y results, most notably Tychonoff… On counterfactual days, I do wonder whether in some alternative histories, the mainstream theory of spaces and continuity might have been founded from the start on locales rather than on topological spaces, and if we would view AC differently in that case. (Or, of course, it might have been based on some theory we still haven't invented yet in this reality!) – Peter LeFanu Lumsdaine Nov 13 '10 at 0:06
Not to mention that recent results of Alex Simpson show that localic measure theory is a much saner thing than ordinary measure theory. There is a locale of random sequences, for example, but there is no space of random sequences. – Andrej Bauer Dec 4 '10 at 7:08

Since I can't (yet) leave a comment to Andres Calceido's answer.

A link to another paper by Luxemburg with a proof of the Hahn-Banach Theorem using ultrapowers

share|cite|improve this answer

As mentioned earlier, the axiom of choice implies the Hahn–Banach theorem. The converse is not true. One way to see that is by noting that the ultrafilter lemma, which is strictly weaker than the axiom of choice, can be used to show the Hahn–Banach theorem, although the converse is not the case. The Hahn–Banach theorem can in fact be proved using even weaker hypotheses than the ultrafilter lemma.[4] For separable Banach spaces, Brown and Simpson proved that the Hahn–Banach theorem follows from WKL0, a weak subsystem of second-order arithmetic.[5]

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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