It is well known that the set of hereditarily countable sets $H(\omega_1)$ —or, if you prefer, $H_{\omega_1}$— has cardinality $2^{\aleph_0}$, and I understand that every countable transitive model (ctm) of ZFC lives in this set. So, if there were a way to give $H(\omega_1)$ a sensible structure (separable metric, for instance) one might try to calculate the probability that a ctm satisfy CH (to name just one example).

Is there a way of defining a sensible probability measure on the set of ctms, provided that this set is nonempty?

I imagine that this idea has been explored and rejected quickly, but I couldn't find anything related to it.

A prior requirement, as I mentioned, would be to have some “definable” topology in the descriptive-set-theoretic sense (Polish, analytic, coanalytic, or the like).

Is there a sensible definable topology on $H(\omega_1)$?

About this, I found in the book Classification and Orbit Equivalence Relations by Hjorth that one can map $H(\omega_1)$ into the set of isomorphism classes of countable binary structures. Now, the countable binary structures form indeed a Polish space, so the odds are that this identification would give as something at least as complicated as $\boldsymbol{\Pi}_1^1(2^{\mathbb{N}\times\mathbb{N}})$ (since one has to say that the binary relation is well founded).

I asked this on Math.SE, and there I just obtained a pair of comments. It was argued that one can indeed put a probability measure on the set of all countable structures in a natural way, but the models of any sufficiently interesting first-order theory will always be null (and meager). In any case, I do not see why this precludes the possibility of having a probability measure on the set of all ctms. (As a silly response, consider the Cantor ternary set in $\mathbb{R}$: It's null and meager; but nevertheless it admits a natural probability measure, when seen as $2^\mathbb{N}$.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.