Past exam of the analytic topology and axiomatic set theory course of the University of Oxford / 2009 / B / 4 / ii

A set $T$ is transitive if every element of an element of $T$ is itself in $T$, equivalently each element of $T$ is a subset of $T$. Proceed by transfinite induction. The empty set is transitive. If $T$ is transitive and $u\in v\in \mathcal{P}(T)$, then $v\subseteq T$ gives $u\in T$, and transitivity of $T$ gives $u\subseteq T$. Thus $u\in \mathcal{P}(T)$, proving the successor stage transitive. A union of transitive sets is transitive: if $u\in v\in {\bigcup }_i{T}_i$, choose $i$ with $v\in T_i$ and infer $u\in T_i$. This proves the limit step and hence every $V_{\alpha }$ is transitive.

For $x\in V_{\alpha }$, transitivity of $V_{\alpha }$ gives $x\subseteq V_{\alpha }$. Therefore $x\in \mathcal{P}(V_{\alpha })=V_{\alpha +1}$, provingTogether with the union definition at limits, transfinite induction now gives $V_{\alpha }\subseteq V_{\beta }$ whenever $\alpha \le \beta$. No rank theorem is needed for this monotonicity argument.

Induct on $\alpha$. For every $\beta <\alpha$, the induction hypothesis says $\beta \in V_{\beta +1}$, and monotonicity gives $V_{\beta +1}\subseteq V_{\alpha }$. Thus every element of the ordinal $\alpha$ lies in $V_{\alpha }$, so $\alpha \subseteq V_{\alpha }$. Taking its power set givesFor $\alpha =0$ the inclusion is vacuous and $\varnothing \in \mathcal{P}(\varnothing )$ supplies the base case. This argument uses ordinal membership as strict order and the previously proved hierarchy monotonicity.