I am being asked to prove the following:
Show, by transfinite induction on $\alpha$, that: For all sets $x$, if $x \in V_\alpha$, then $\mathcal{P}(x)\in V_{\alpha+1}$.
So, I am aware to use transfinite induction correctly I must formally show three things. Here is my attempt:
Let $P=\{\alpha: \text{if } \forall x \in V_{\alpha}, \text{then } \mathcal{P}(x)\in V_{\alpha+1} \}$
- Show $0\in P$. Thus we have to show $\text{if }\forall x \in V_{0}, \text{then } \mathcal{P}(x) \in V_{1}.$ Note that $x \in \emptyset$ implies that $x=\emptyset$. So, we have $\mathcal{P}(\emptyset)= V_1$, which makes the statement true.
- Suppose $\beta \in P$ and we want to show that $\beta+1 \in P$. Now, suppose $\forall x \in V_{\beta}\text{, then } \mathcal{P}(x)\in V_{\beta+1}$, and want to show $\forall x \in V_{\beta+1}\text{, then } \mathcal{P}(x)\in V_{\beta+2}$.
I am unsure on how to proceed. Any suggestions are appreciated.
Note: I have a feeling that this is supposed to be the ordinal version of if $X\subseteq Y$, then $\mathcal{P}(x) \subseteq \mathcal{P}(y)$
Note that $V_{\beta +1 } = \mathcal P (V_\beta)$. So $$x \in V_{\beta +1} \iff x \subseteq V_{\beta}$$ Now let $y := \mathcal P(x)$. We have $$ z \in y \iff z \subseteq x $$ Since $x \in V_{\beta+1}$ and $V_{\beta+1}$ is transitive, we have that $x \subseteq V_{\beta+1}$. This, combined with equivalence above shows that $y \in \mathcal P(V_{\beta+1}) = V_{\beta+2}$.
Finally you have to show that $P$ is closed under limits, i.e. if $\lambda$ is a limit ordinal such that $\lambda \subseteq P$, then $\lambda \in P$. Since $V_{\lambda} = \bigcup_{\alpha < \lambda} V_{\alpha}$, this shouldn't cause too much trouble.