1
$\begingroup$

Is it consistent that there is a measurable cardinal $κ$, a $κ$-complete normal nonprincipal ultrafilter $U$ on $κ$, and $S∈U$ such that for every $T⊂S$ with $T∈U$ and $T$ ordinal definable from $S$ and a countable sequence of ordinals, $S$ and $T$ satisfy the same $Σ_2^V$ properties with parameters in $V_{\min(S)}$? If so, what is the consistency strength?

Notes:
* Using $Σ_n^V$ ($n>2$) in place of $Σ_2^V$ properties should have a similar (but likely growing with $n$) consistency strength.
* A possible natural additional requirement is that for every $S'∈U$, there is $T∈U$ with $T⊂S∩S'$ and $S$ and $T$ satisfying the same $Σ_2^V$ properties with parameters in $V_{\min(S)}$.

The motivation for the question is (1) it is a very strong form of indiscernibility (given the axiom of choice), and (2) its role in extensions of set theory.

The axiom of choice precludes unconditional infinitary indiscernibles: $κ \not\rightarrow (ω)^ω$. However, it does not usually prevent definable versions of regularity properties, so the above restricts it to (essentially) definable predicates and definable subsets of $S$. For long enough $S$, both restrictions are necessary. However, our use of parameters in $V_{\min(S)}$ (for distinguishing but not defining $T$) is not a problem, or rather $S∈U$ is but a point in a hierarchy of notions, and stronger notions without $V_{\min(S)}$ give weaker notions with $V_{\min(S)}$.

The relevance to extensions of set theory is the following. To increase expressiveness beyond the language of set theory, we can 'choose' a cardinal $S_0$ with sufficiently strong reflection properties, which allows us to express statements in higher order set theory. For further expressiveness, we can choose a larger cardinal $S_1$ in the same manner, and iterate this indefinitely, getting (essentially) a reflective sequence $S$. (Reflective cardinals and reflective sequences are introduced and analyzed in my paper Reflective Cardinals (and its precursor Extending the Language of Set Theory).) We continue through $S_ω$ (which is above $\lim_{n→ω} S_n$), $S_{ω_1}$, $S_{\min(S)}$, eventually reaching a limit point of $S$ that is in $S$, and going further, $S∈U$ is a strong version of claiming that $S$ has many elements.
    To axiomatize such $S$, we assert a strong form of symmetry and indiscernibility, and in particular that if we were stricter about putting elements into $S$ in a sufficiently definable manner (with symmetry allowing the use of definability from $S$) and still get enough elements (formalized through $T∈U$), then we get the same theory (with parameters that are small relative to $\min(S)$). We also want strong reflection properties for $κ$ and $U$, but the form in the question suffices for many basic properties of $S$, assuming there is no falsehood or inconsistency.

Starting with an inner model with a measurable cardinal of order 2 below $ω_1^V$, we can iterate it and the measurables below it until they match the elements of $S$ (but I have not proved that), with $\sup(S)$ being the resulting least measurable of order 2. Thus (assuming the above), $(L[S],∈,S)$ has a well-behaved theory that does not depend on $S$. However, the least measurable in $K^{L[S]}$ appears to be $\lim_{n→ω} S_n$, so in $L[S]$, $S \setminus S_ω$ is distinguishable from $S$. We would need more sets to get the symmetry in the question.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.