every element of $V_{\omega}$ is definable

Question

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

Answer 1

Assuming you want to show that every element of $V_\omega$ is definable in $V_\omega$ without parameters -- that is, for every $x\in V_\omega$ there is a $\phi(y)$, with at most $y$ free, such that $x = \{y: V_\omega\vDash \phi(y)\}$ -- we can argue as follows.

Since $V_0$ contains no elements, it's trivial that every element of $V_0$ is definable. So, suppose this is true of $V_n$. Now consider $x\in V_{n+1}$. By definition of $V_{n+1}$, $x\subseteq V_n$, and since $x\in V_\omega$, $x$ is finite -- for definiteness, we can let $x = \{y_0,...,y_m\}$. Clearly, $y_0,...,y_m$ are in $V_n$, and so by our induction hypothesis they are all definable -- for definiteness, we can let $y_0 = \{y: V_\omega\vDash \phi_0(y)\}$, ..., $y_m = \{y: V_\omega\vDash \phi_m(y)\}$. Now, consider the following formula:

(*) $\Psi(z) = ``z = \{y: \phi_0(y)\}\vee,...,\vee z =\{y: \phi_m(y)\}"$

Since $V_\omega$ is transitive, it is straightforward to show that $x = \{z: V_\omega\vDash \Psi(z)\}$, as required.

Answer 2

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.