every element of $V_{\omega}$ is definable My attempt by $\in$-induction. I am trying find formula that will work:
$N=(V_{\omega},\in)\models rank(\varnothing) =0<\omega$
Assume,given $x\in V_\omega$ that $\forall y\in x$ are definable too $N\models rank(y)<\omega$. Then since $x\in V_\omega$, $|x|<\omega\Rightarrow x$ is finite $\Rightarrow rank(x)=rank(y_{1})+...+rank(y_{n})<\omega$. Is the last equality valid i.e. $rank(x)=rank(y_{1})+...+rank(y_{n})$?
thanks

Let $N$ be an arbitrary structure of an arbitrary language. Recall that $n\in N$ is called definable (without parameters) if there exists a formula $\varphi(x)$ such that $N\models\varphi(u)\iff u=n$.
We want to show that in $V_\omega$ in the language including only $\in$, every element is definable. We do this by $\in$-induction:
Suppose that $x$ is such that for all $y\in x$, $y$ is definable. Then for every such $y$ there is some formula defining it, $\varphi_y$. We can therefore define $x$ to be the unique set whose elements are defined by one of these $\varphi_y$'s. That is: $$\varphi_x(u):=\forall v(v\in u\leftrightarrow\bigvee_{y\in x}\varphi_y(v))$$
The disjunction occur in the meta-theory, where we know that $x$ is finite, and what are its elements. Therefore there is no circularity arguments here.
