My vague intuition is that not only it is common for a simple arithmetic proposition $p$ to be independent of ZFC, but it is common for the statement "$p$ is independent of ZFC" to be independent, and so on. If we let $I(p)$ be the statement that $p$ is independent of ZFC, then this "iterative independence" property is $I^\*(p) = \forall n \ge 0, I^n(p)$.


Of course, it would be impossible to prove $I^\*(p)$ for any particular statement $p$, at least under ZFC. So the best one could do is find some set $P$ of propositions s.t. we can prove at least one $p \in P$ has $I^\*(p)$, but where we don't know which one it is.


Is it possible to find such a set $P$ containing at least one iteratively independent statement?



Arithmetic statement which is independent, and whose independence is independent, and so on?
If we fix things to avoid Will Sawin's observation, then the answer is **yes** under any reasonable interpretation I can think of.


For example, consider the following: let $J(p)$ be the sentence "If $\mathsf{ZFC}$ is consistent then $p$ is independent over $\mathsf{ZFC}$." We can indeed prove in $\mathsf{ZFC}$ many instances of $J$ (e.g. that $J$ holds of the Godel-Rosser sentence for $\mathsf{ZFC}$), so this is nontrivial.


However, note that if we assume that $\mathsf{ZFC}$ is sound *(this is overkill but let's do it anyways)*, then we have $\mathsf{ZFC}\vdash J^n(p)$ for some finite $n$ only if $p$ really is independent of $\mathsf{ZFC}$. Since the set of sentences which are independent over $\mathsf{ZFC}$ is not c.e., there must be many $p$s such that $\mathsf{ZFC}$ doesn't prove any of the $J^n(p)$s.


* The non-c.e.-ness claim above is an instance of a more general fact: if $T$ is *any* theory to which Godel's first incompleteness theorem applies, then the set of $T$-independent sentences is not c.e. This is because if it were, then the set of $T$-theorems would be both c.e. (because it obviously is) and co-c.e. (because every non-$T$-theorem is either $T$-disprovable, which is obviously a c.e. condition, or is $T$-independent which is a c.e. condition by assumption). But that would make the set of $T$-theorems computable, which can't be the case (e.g. if it happened we'd be able to whip up a computable complete extension of $T$).


Basically, any reasonable iterated independence property can't fully capture independence.


