Assuming the set theory we're working with allows self-containment, as well as arbitrary set building of the form $\{x:\Phi(x)\}$, if we define $S=\{x:x\in x\}$, is $S\in S$ knowable?
As we see from assuming self-containment of $S$ or not, it seems to not give any new information:
$$S\in S \implies S\in S$$
$$S\notin S \implies S\notin S$$
However, unlike Russel's Paradox, there does not seem to be a contradiction for this set's existence (at least naively).
Edit: For clarity, I'm asking if the provability (or rather unprovability of) $S\in S$, is itself, provable.
Let me interpret this as asking the question in $\sf ZF-Foundation$ as our set theory. Can it decide this statement?
The answer is no. Since $\sf Foundation$ is consistent with our set theory, it is consistent that $\{x\mid x\in x\}=\varnothing$, and of course, $\varnothing\notin\varnothing$. So any chance for deciding anything must be for the negative.
Alas, we can arrange for a situation where the only set that contains itself is a unique Quine atom, that is $a=\{a\}=\{x\mid x\in x\}$, and in that case the set of all sets that contain themselves contains itself.
But we're not over yet. It is consistent also to have a proper class of Quine atoms, which means that the class $\{x\mid x\in x\}$ is not even a set to begin with. So we can't even prove that this collection is a set.