Exercise 6.3.5

Proof. First we show that if $x\in V_n$ for some $n\in \omega$ then $x\in V_m$ for all $m\ge n$. We show this by induction on $m$. So for $m=n$ clearly $x\in V_n=V_m$. Now suppose that $x\in V_m$. Then it was shown in Exercise 6.3.4 part (c) that $V_m$ is transitive so that $x\subseteq V_m$. Hence $x\in 𝒫\left(V_m\right)=V_{m+1}$, thereby completing the inductive proof.

Now suppose that $x,y\in V_{\omega }$. Then there are $n,m\in \omega$ such that $x\in V_n$ and $y\in V_m$. Without loss of generality we can assume that $n\le m$ (since if this is not the case then we simply reverse the roles of $x$ and $y$). So since $m\ge n$ it follows from what was shown above that $x\in V_m$ as well. Hence we have that clearly $\left\{x,y\right\}\subseteq V_m$ since both $x\in V_m$ and $y\in V_m$. Then $\left\{x,y\right\}\in 𝒫\left(V_m\right)=V_{m+1}$ from which it clearly follows that $\left\{x,y\right\}\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k\in \omega }{V}_k=V_{\omega }$. □

Proof. Suppose that $X\in V_{\omega }={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k\in \omega }{V}_k$. Then there is an $n\in \omega$ such that $X\in V_n$. It was shown in Exercise 6.3.4 part (c) that $V_n$ is transitive so that $X\subseteq V_n$.

First we show that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->X\in V_{\omega }$. So consider any $x\in \bigcup <!--\; FUNCTION\; APPLICATION\; -->X$ so that there is a $Y\in X$ such that $x\in Y$. Then since $X\subseteq V_n$ we have that $Y\in V_n$. Since again $V_n$ is transitive we have that $Y\subseteq V_n$ so that $x\in V_n$ since $x\in Y$. Since $x$ was arbitrary it follows that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->X\subseteq V_n$ so that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->X\in 𝒫\left(V_n\right)=V_{n+1}$. From this it clearly follows that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->X\in {\bigcup }_{k\in \omega }{V}_k=V_{\omega }$.

Next we show that $𝒫\left(X\right)\in V_{\omega }$. So consider any $Y\in 𝒫\left(X\right)$ so that $Y\subseteq X$. Now consider any $y\in Y$ so that also $y\in X$. Since $X\subseteq V_n$ we have that $y\in V_n$. But since $y\in Y$ was arbitrary it follows that $Y\subseteq V_n$ so that $Y\in 𝒫\left(V_n\right)=V_{n+1}$. Then since $Y\in 𝒫\left(X\right)$ was arbitrary it follows that $𝒫\left(X\right)\subseteq V_{n+1}$ so that $𝒫\left(X\right)\in 𝒫\left(V_{n+1}\right)=V_{n+2}$. From this it clearly follows that $𝒫\left(X\right)\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k\in \omega }{V}_k=V_{\omega }$. □

Proof. Note the issue with this part in the errata list. Since $A\in V_{\omega }$ we have by Exercise 6.3.4 part (a) that $A$ is finite. Then by Theorem 2.2.5 it follows that $f[A]$ is finite. Also clearly $f[A]$ is a subset of $V_{\omega }$ and hence is a finite subset. Therefore by part (d) below $f[A]\in V_{\omega }$. □

Proof. Consider any finite $X\subseteq V_{\omega }$. Suppose then that $|X|=n$ for some $n\in 𝑵$. Then for each $x_k\in X$, where $k\in n$, we have that $x_k\in V_{\omega }={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{m\in \omega }{V}_m$ so that there is an $m_k\in \omega$ where $x_k\in V_{m_k}$. Now let $m=\underset{k\in n}{\max <!--\; FUNCTION\; APPLICATION\; -->}{m}_k$, which exists since $n$ is finite. Then, for any $k\in n$, by what was shown in Exercise 6.3.5 part (a) we have $x_k\in V_m$ since $x_k\in V_{m_k}$ and $m\ge m_k$. Hence it follows that $X\subseteq V_m$ so that $X\in 𝒫\left(V_m\right)=V_{m+1}$. Clearly then $X\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k\in \omega }{V}_k=V_{\omega }$. □