A probability is a finitely additive measure on a boolean algebra with total measure $1$.

A function $P:\scr B \times (\scr B - \{ 0 \})$ is a full conditional probability on $\scr B$ (for a boolean algebra $\scr B$) iff $P(-|A)$ is a probability for all $A$ and $P(A|B)P(B|C)=P(A|C)$ whenever $A\le B\le C$.

Identify an ultrafilter with a $\{0,1\}$-valued probability.

The consider these two claims:

FCP: For any nonempty $\Omega$ there is a full conditional probability $P$ on $2^\Omega$.

FCU: for any nonempty $\Omega$, there is a full conditional probability $P$ on $2^\Omega$ with every $P(-|A)$ being an ultrafilter.

Then I think I can show that BPI$\Rightarrow$FCU$\Rightarrow$FCP.


  1. Does FCU imply BPI?
  2. Does Hahn-Banach imply FCP? (In the first version of this post I said that I thought I had a proof. But there was a mistake in it.)
  3. Does FCP imply Hahn-Banach?

If one replaces $2^\Omega$ in FCU with an arbitrary boolean algebra, then FCU becomes equivalent to BPI, and FCP comes to imply HB.

Some miscellaneous notes:

A. FCP immediately implies the following Probabilistic Choice principle:

(PC) For any set-valued function $f$ on a set $X$ such that $f(x)$ is always non-empty, there is a function $g$ on $X$ such that $g(x)$ is a probability on $f(x)$.

PC is strong enough to imply both the existence of Lebesgue nonmeasurable sets and the Banach-Tarski paradox (the Foreman-Wehrung and Pawlikowski proofs of these things from Hahn-Banach basically only use PC).

I don't know if PC is weaker than FCP nor if it's weaker than HB.

B. Analogously, FCU implies the following Ultrafilter Choice principle:

(UC) For any set-valued function $f$ on a set $X$ such that $f(x)$ is always non-empty, there is a function $g$ on $X$ such that $g(x)$ is an ultrafilter on $f(x)$.

UC does not imply FCU, because Form 344 does not imply Form 30 (linear orderability), while FCU implies linear orderability (see point D).

UC, and hence FCU, immediately implies Choice for families of finite sets. According to Howard and Rubin, it is unknown whether UC implies HB.

C. FCU is equivalent to there being a function $u$ that assigns to each non-empty subset $A$ of $\Omega$ an ultrafilter $u(A)$ with $A\in u(A)$ and the coherence property that if $A\subseteq B$ and $A\in u(B)$, then $u(A)=u(B)$.

D. FCU implies that the order extension principle: every partial order extends to a linear order. To see this, note that replacing each element of the poset by the set of all elements less than or equal to it, we can suppose the partial order is setwise inclusion. Let $\Omega$ be any set and let $P$ be full conditional ultrafilter probabilities on the subsets of $\Omega$. Then define $A \preceq B$ for $A,B\subseteq\Omega$ iff $A=B$ or $P(A-B|A\Delta B) \le P(B-A|B\Delta A)$. This is a total order extending set inclusion.

E. Any well-ordered set $\Omega$ is such that there a principal ultrafilter full conditional probability on $2^\Omega$: just let $P(B|A)=1$ iff $\min A \in B$. Conversely, if all the conditional probabilities are principal, then the order given by $a\le b$ iff $P(\{a\} | \{a,b\})=1$ is a well-order.

Therefore, FCU implies that every non-well-orderable set has a nonprincipal ultrafilter, which implies that some set has a nonprincipal ultrafilter.

F. Amusingly, every amorphous set has a full conditional probability. Just let $P(A|B) = 1$ if $A\cap B$ is infinite and $P(A|B)=|A\cap B|/|B|$ otherwise.

  • $\begingroup$ For those of us that don't have the knowledge at the tips of our frontal lobes, a statement of BPI (Boolean Prime Ideals?) would be appreciated. $\endgroup$ Dec 22, 2013 at 18:12
  • $\begingroup$ Equivalent versions: (1) every filter (collection $F$ of subsets of $X$ such that $A,B\in F$ implies $A\cap B\in F$ and $A\in F$ and $A\subseteq B\subseteq X$ implies $B\in F$) extends to an ultrafilter (maximal filter); (2) every ideal $I$ in a boolean algebra $B$ ($a\in I, b\in B$ implies $a\wedge b\in I$, and $a,b\in I$ implies $a\vee b\in I$) is contained in a prime ideal ($a\wedge b\in I$ implies $a\in I$ or $b\in I$). Form 14 here: consequences.emich.edu/CONSEQ.HTM $\endgroup$ Dec 22, 2013 at 20:13
  • $\begingroup$ I don't have enough rep to comment (and would be happy if someone converted this answer to a comment), but why doesn't FCU imply BPI just by Stone's representation theorem? $\endgroup$
    – user39080
    Dec 22, 2013 at 23:55
  • $\begingroup$ Stone representation is equivalent to BPI, so that doesn't help. (There is a version of Stone that holds in ZF--use a filter instead of an ultrafilter in the proof and get a weaker conclusion--but I don't know how to use that for the purposes needed here.) $\endgroup$ Dec 23, 2013 at 14:41


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.