Bayesianism in Math: No Dice

Attention conservation notice: Sniping at someone else's constructive attempt to get the philosophy of mathematics to pay more attention to how mathematicians actually discover stuff, because it uses an idea that pushes my buttons. Assumes you know measure-theoretic probability without trying to explain it. Written by someone with absolutely no qualifications in philosophy, and precious few in mathematics for that matter. Largely drafted back in 2013, then laid aside. Posted now in lieu of new content.

Wolfgang points to an interesting post [archived] at "A Mind for Madness" on using Bayesianism in the philosophy of mathematics, specifically to give a posterior probability for conjectures (e.g., the Riemann conjecture) given the "evidence" of known results. Wolfgang uses this as a jumping-off point for looking at whether a Bayesian might slide around the halting problem and Gödel's theorem, or more exactly whether a Bayesian with \( N \) internal states can usefully calculate any posterior probabilities of halting for another Turing machine with \( n < N \) states. (I suspect that would fail for the same reasons my idea of using learning theory to do so fails; it's also related to work by Aryeh "Absolutely Regular" Kontorovich on finite-state estimation, and even older ideas by the late great Thomas Cover and Martin Hellman.)

My own take is different. Knowing how I feel about the idea of using Bayesianism to give probabilities to theories about the world, you can imagine that I look on the idea of giving probabilities to theorems with complete disfavor. And indeed I think it would run into insuperable trouble for purely internal, mathematical reasons.

Start with what mathematical probability is. The basics of a probability space are a carrier space \( \Omega \), a \( \sigma \)-field \( \mathcal{F} \) on \( \Omega \), and a probability measure \( P \) on \( \mathcal{F} \). The mythology is that God, or Nature, picks a point \( \omega \in \Omega \), and then what we can resolve or perceive about it is whether \( \omega \in F \), for each set \( F \in \mathcal{F} \). The probability measure \( P \) tells us, for each observable event \( F \), what fraction of draws of \( \omega \) are in \( F \). Let me emphasize that there is nothing about the Bayes/frequentist dispute involved here; this is just the structure of measure-theoretic probability, as agreed to by (almost) all parties ever since Kolmogorov laid it down in 1933 ("Andrei Nikolaevitch said it, I believe it, and that's that").

To assign probabilities to propositions like the Riemann conjecture, the points in the base space \( \omega \) would seem to have to be something like "mathematical worlds", say mathematical models of some axiomatic theory. That is, selecting an \( \omega \in \Omega \) should determine the truth or falsity of any given proposition like the fundamental theorem of algebra, the Riemann conjecture, Fermat's last theorem, etc. There would then seem to be three cases:

1. The worlds in \( \Omega \) conform to different axioms, and so the global truth or falsity of a proposition like the Riemann conjecture is ambiguous and undetermined.
2. All the worlds \( \Omega \) conform to the same axioms, and the conjecture, or its negation, is a theorem of those axioms. That is, it is true ( or false) in all models, no matter how the axioms are interpreted, and hence it has an unambiguous truth value.
3. The worlds all conform to the same axioms, but the proposition of interest is true in some interpretations of the axioms and false in others. Hence the conjecture has no unambiguous truth value.

Case 0 is boring: we know that different axioms will lead to different results. Let's concentrate on cases 1 and 2. What do they say about the probability of a set like \( R = \left\{\omega: \text{Riemann conjecture is true in}\ \omega \right\} \)?

Case 1: The Conjecture Is a Theorem

Case 1 is that the conjecture (or its negation) is a theorem of the axioms. Then the conjecture must be true (or false) in every \( \omega \), so \( P(R) = 0 \) or \( P(R) = 1 \). Either way, there is nothing for a Bayesian to learn.

The only escape I can see from this has to do with the \( \