Let $x\in V_{\omega}$, is $\operatorname{rank}(\operatorname{tc}(x))=\operatorname{rank}(x)$? while $\operatorname{tc}(x)$ is the minimal transitive set ${a}$ such that $x\subseteq a$
i what to show this so i could prove that in the case as in the above $\operatorname{tc}(x)$ is finite. and my claim is that since $x\in V_{\omega}$ so $\operatorname{rank}(x)<\omega$ but in our case it must be that rank is a function from $\operatorname{tc}(x)$, since rank is defined only on set's with well founded relation on them, and $x\in WF$ so from my definition $\in$ is a well founded relation on $\operatorname{tc}(x)$ and not at $x$.
and it is easy to see in our case that $x=\sup\operatorname{tc}(x)$ and $x\in tc(x)$ thus since rank is a function in to $\mathrm{On}$ so it must be that $\operatorname{rank}(x)=\operatorname{rank}(\operatorname{tc}(x))$ or is it that $\operatorname{rank}(\operatorname{tc}(x))=\operatorname{rank}(x)+1$?
$\DeclareMathOperator{\rank}{rank}\DeclareMathOperator{\tc}{tc}$ It is true in general that $\rank(x)=\rank(\tc(x))$.
You can easily show that $\tc(x)=\bigcup\{\tc(y)\mid y\in x\}\cup x$. Then $$\rank(\tc(x))=\max\Bigl\{\sup\{\rank(\tc(y))+1\mid y\in x\},\sup\{\rank(y)+1\mid y\in x\}\Bigr\}$$
Now if the claim holds for all the members of $x$, the the two $\sup$'s are taken on the same set, essentially, which means that they have the same result, but this means that $\rank(\tc(x))=\rank(x)$.
So we proved this using $\in$-induction. But of course that this can be transformed into a usual induction for finite ranks, or transfinite induction for arbitrary ranks.