Let $V_\alpha$ be the elements of the Von Neumann hierarchy and, given a set $S$, define its transitive closure, $tc(S)$, as the smallest transitive set such that $S \subseteq tc(S)$.
I'm not very familiar with this concepts so maybe my question will be trivial, but how do I show that $S \in V_\omega \iff |tc(S)| \lt \aleph_0$? Thank you in advance for your answers.
Von Neumann hierarchy and finite transitive closures
set-theory
Best Answer
One direction is fairly immediate:
If $S\in V_\omega$, then $S\subseteq V_n$ for some $n<\omega$, so $\operatorname{tc}(S)\subseteq V_n$ as well, since $V_n$ is a transitive set (why?). But since $V_n$ is also finite (why?), that means that $\operatorname{tc}(S)$ must be finite.
In the other direction, there are multiple approaches you can take. But here's a fairly straightforward one:
Proof. We prove this by induction on $\alpha$. Let $X$ be such a transitive set of rank $\alpha$. Let $\beta<\alpha$ and let $x\in X$ be a set of von Neumann rank $\beta$, then $\operatorname{tc}(x)\subseteq X\cap V_\beta$. By the induction hypothesis on $\beta$, $\rho\restriction\operatorname{tc}(x)$ is surjective onto $\beta$.
Now, note that for any $\beta<\alpha$ there is some $x\in X$ of von Neumann rank at least $\beta$, so the image of $\rho$ must cover at least $\beta$, for any $\beta<\alpha$. And therefore the image must cover $\alpha$ itself. $\square$
Proof. If $X\notin V_\omega$, then it has an infinite von Neumann rank, so by the above, its transitive closure must map onto some infinite ordinal, and therefore must be infinite. $\square$