Exercise 6.3.4

Proof. First we show by induction that every $V_n$ is finite (for $n\in 𝑵$). For $n=0$ we have $V_n=V_0=\varnothing$, which is clearly finite. Now suppose that $V_n$ is finite then we have $V_{n+1}=𝒫\left(V_n\right)$, which is finite by Theorem 4.2.8.

Now consider any $x\in V_{\omega }={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n\in \omega }{V}_n$ so that there is an $n\in \omega$ such that $x\in V_n$. We note that $n\ne 0$ since $V_0=\varnothing$ so it cannot be that $x\in V_0=\varnothing$. Hence $V_{n-1}$ is a set and moreover $V_n=𝒫\left(V_{n-1}\right)$. So since $x\in V_n$ it follows that $x\in 𝒫\left(V_{n-1}\right)$ so that $x\subseteq V_{n-1}$. Thus it follows that $|x|\le |V_{n-1}|$ so that clearly $x$ is finite since $V_{n-1}$ is (shown above). □

Proof. Consider any $x\in V_w$. Then by the same argument as in part (a) above it follows that $x\in V_n$ where $n\ne 0$. Hence again $V_{n-1}$ is a set and $V_n=𝒫\left(V_{n-1}\right)$ so that $x\subseteq V_{n-1}$. Then for any $y\in x$ we have that $y\in V_{n-1}$, from which it follows that clearly $y\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k\in \omega }{V}_k=V_{\omega }$. Hence since $y$ was arbitrary $x\subseteq V_{\omega }$, and since $x$ was arbitrary this shows that $V_{\omega }$ is transitive by definition. □

Proof. First we show by induction that each $V_n$ (where $n\in \omega$) is transitive. For $n=0$ we have $V_n=V_0=\varnothing$, which is clearly vacuously transitive. Now suppose that $V_n$ is transitive and consider any $x\in V_{n+1}=𝒫\left(V_n\right)$ so that $x\subseteq V_n$. Now consider any $y\in x$ so that also $y\in V_n$. But since $V_n$ is transitive $y\subseteq V_n$ so that $y\in 𝒫\left(V_n\right)=V_{n+1}$. Hence since $y$ was arbitrary this shows that $x\subseteq V_{n+1}$ and since $x$ was arbitrary this shows by definition that $V_{n+1}$ is transitive, thereby completing the inductive proof.

Now we show that $V_{\omega }$ is inductive. So first note that $V_1=𝒫\left(V_0\right)=𝒫\left(\text{∅}\right)=\left\{\text{∅}\right\}$ so that $0=\varnothing \in V_1$. From this is clearly follows that $0\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n\in \omega }{V}_n=V_{\omega }$.

Now suppose that $n\in V_{\omega }={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k\in \omega }{V}_k$ so that there is an $m\in \omega$ such that $n\in V_m$. Since it was shown above that $V_m$ is transitive we have that $n\subseteq V_m$ as well. So consider any $x\in n+1=n\cup \left\{n\right\}$. If $x\in n$ then also $x\in V_m$ since $n\subseteq V_m$. On the other hand if $x\in \left\{n\right\}$ then $x=n\in V_m$. Since $x$ was arbitrary this shows that $n+1\subseteq V_m$ so that $n+1\in 𝒫\left(V_m\right)=V_{m+1}$. From this it clearly follows that $n+1\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k\in \omega }{V}_k=V_{\omega }$. This shows that $V_{\omega }$ is inductive by definition. □