Past exam of the axiomatic set theory course of the University of Oxford / 2011 / 1

The cumulative hierarchy is the transfinite recursionHere $\mathcal{P}$ is the power set. The class $V={\bigcup }_{\alpha \in \mathrm{Ord}}{V}_{\alpha }$ consists of all sets in ZF, by the Axiom of foundation and Axiom schema of replacement.

Two facts about a transitive set $A$ drive the transfinite induction. First, $A\subseteq \mathcal{P}(A)$: every $x\in A$ is a subset of $A$. Second, $\mathcal{P}(A)$ is a transitive set: if $x\in y\in \mathcal{P}(A)$, then $x\in A$, hence $x\subseteq A$, hence $x\in \mathcal{P}(A)$.

Starting with $V_0=\varnothing$, these facts prove the successor step. At a limit ordinal $\lambda$, the union of the earlier increasing transitive sets is transitive: if $x\in y\in V_{\lambda }$, some $\alpha <\lambda$ contains $y$ in $V_{\alpha }$, and transitivity puts $x$ there too. The union contains every earlier stage. Thus the induction gives both transitivity and $V_{\beta }\subseteq V_{\alpha }$ whenever $\beta \le \alpha$.

Finally $V_{\alpha }\subseteq V_{\alpha }$, so $V_{\alpha }\in \mathcal{P}(V_{\alpha })=V_{\alpha +1}$. The Axiom of foundation gives $V_{\alpha }\notin V_{\alpha }$. This witnesses strictness:This is the transitivity and strict growth of the cumulative hierarchy.

Fix $x\in V_{\delta }$. Since $\delta$ is a limit ordinal, choose $\beta <\delta$ with $x\in V_{\beta }$. Transitivity gives $x\subseteq V_{\beta }$. Every subset $u\subseteq x$ therefore belongs to $V_{\beta +1}=\mathcal{P}(V_{\beta })$, and consequentlyThe last inclusion uses $\beta +2<\delta$, since a finite number of successor steps remains below a limit ordinal. In particular every subset of $x$ is also an element of the transitive set $V_{\delta }$.

Take $y=\mathcal{P}(x)$. For any $u\in V_{\delta }$, transitivity ensures that the internal statement $u\subseteq x$ agrees with the external one. HenceSuch a $y\in V_{\delta }$ exists for every $x\in V_{\delta }$, so $(V_{\delta },\in )$ satisfies the Axiom of power set. This power-set axiom at limit ranks actually holds for every nonzero limit ordinal; the stated restriction $\delta >\omega$ is sufficient.

Every finite ordinal belongs to $V_{\omega }$, so $\omega \subseteq V_{\omega }$ and $\omega \in \mathcal{P}(V_{\omega })=V_{\omega +1}$. However, $\omega \notin V_{\omega }$, whose elements are the hereditarily finite sets.

Suppose $y\in V_{\omega +1}$ were an internal power set of $\omega$. Internally $\omega \subseteq \omega$, and $\omega$ is an available element of the structure. Thus the Axiom of power set would require $\omega \in y$. But $y\in \mathcal{P}(V_{\omega })$ means $y\subseteq V_{\omega }$, which would put $\omega \in V_{\omega }$, a contradiction. ThereforeThis proves power-set failure at the first rank above V omega by an internal obstruction, including the subset $\omega$ itself.