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.)


Sell said...

This sounds awesome. Sadly, it's incomprehensible to me. Where would you recommend starting to get up-to-speed on contemporary math relevant to contemporary philosophy? (For a graduate student.)

Alexander R Pruss said...

Sider has a good logic book...