## Tuesday, November 19, 2013

### An interesting equivalent to the Hahn-Banach theorem

The Hahn-Banach Theorem (HB) cannot be proved without some version of the Axiom of Choice. (Technically, it's stronger than ZF but weaker than BPI.) A cool fact about HB is that it is sufficient for proving the existence of nonmeasurable sets and even for proving the Banach-Tarski paradox.

In 1969, Luxemburg proved that the Hahn-Banach theorem is equivalent to the claim that every boolean algebra has a (finitely additive) probability measure.

He also proved that the Hahn-Banach theorem is also equivalent to the following interesting claim:

1. For any set Ω and proper ideal N of subsets of Ω (i.e., N is closed under finite unions, any subset of a member of N is a member of N, but not every subset of Ω is in N), there is a (finitely additive) probability measure on all subsets of Ω that is zero on every member of N.
One direction can be proved by using the existence of a probability measure on the quotient boolean algebra 2Ω/N via Hahn-Banach. The other direction follows (I don't know if Luxemburg did it this way) from this fact which can be proved without the axiom of choice:
1. Any boolean algebra is isomorphic to the quotient of a boolean algebra of sets.
Given the Axiom of Choice, this is a trivial consequence of the Stone Representation Theorem. Without the Axiom of Choice, (2) is a quick consequence of a theorem of Buskes, de Pagter and van Rooij. It can also be proved directly.[note 1]

(This is part of a general observation that some of the things that can be done with ultrafilters can also be done with filters, though the results may be weaker.)