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

The cumulative hierarchy is defined by transfinite recursion on the ordinals:The recursion theorem of ZF supplies each set stage. The whole universe is $V={\bigcup }_{\alpha \in \mathrm{Ord}}{V}_{\alpha }$, a proper-class union. Equivalently, $x\in V_{\alpha }$ exactly when the rank of $x$ is less than $\alpha$, where $\mathrm{rank}(x)=\sup \{\mathrm{rank}(z)+1:z\in x\}$.

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.

Take a nonzero limit ordinal $\lambda$ and $x\in V_{\lambda }$. Choose $\beta <\lambda$ with $x\in V_{\beta }$. Transitivity gives $x\subseteq V_{\beta }$. Hence every actual subset of $x$ belongs to $\mathcal{P}(V_{\beta })=V_{\beta +1}$, and the actual set $\mathcal{P}(x)$ belongs to $V_{\beta +2}$. Since $\lambda$ is a limit, $\beta +2<\lambda$, so $\mathcal{P}(x)\in V_{\lambda }$. Thus the internal power set contains every actual subset of $x$, and the power-set axiom holds in $V_{\lambda }$.

Here the absoluteness fact used is: every bounded formula is absolute between transitive sets or classes with the actual membership relation. In particular $z\subseteq x$, expressed by $\forall t\in z(t\in x)$, has the same meaning internally and externally. The argument supplies the actual power set as an internal witness. If zero is included in one's convention for limit ordinals, its empty stage has no nonempty first-order domain; the assertion is understood for nonzero limits.

If also $\lambda >\omega$, the preceding ordinal-membership result gives $\omega \in V_{\omega +1}\subseteq V_{\lambda }$. The actual set $\omega$ contains the empty set and is closed under $z\mapsto z\cup \{z\}$. These bounded membership and successor conditions are absolute in the transitive stage, so $\omega$ is an internal inductive set. Therefore Infinity holds as well. In contrast $V_{\omega }$ has only hereditarily finite sets and cannot use $\omega$ as an element.

To prove replacement failure at V omega plus omega, put $M=V_{\omega +\omega }$. The set $\omega$ belongs to $M$, and for every $n\in \omega$ the ordinal $\omega +n$ also belongs to $M$. The relation saying that $y$ is obtained by taking $n$ successive ordinal successors of $\omega$ is definable, with parameter $\omega$, and is single valued in $M$.

For clarity this does not assume Replacement inside $M$. For each finite $n$ its finite computation sequence exists in $M$: its entries have rank below $\omega +n+1$, and the finite coding overhead still has rank below $\omega +\omega$. The bounded formula verifying this finite successor chain is absolute between $M$ and the universe. The ordinary finite induction identifying its unique last entry uses the actual finite ordinals, which are the same in $M$. This justifies the internal definable function $n\mapsto \omega +n$ without invoking the axiom being tested.

If Replacement held there, its range $B=\{\omega +n:n\in \omega \}$ would be an element of $M$. But ordinal rank equals the ordinal, soEvery element of $V_{\omega +\omega }$ has strictly smaller rank. This contradiction proves Replacement fails in $V_{\omega +\omega }$. Each individual output is present; the missing object is the set collecting all outputs.