Past exam of the analytic topology and axiomatic set theory course of the University of Oxford / 2008 / B / 5

In first-order set theory a class in set theory is a definable collection, written schematically $\{x:\theta (x,\vec{p})\}$ using a logical formula and set parameters. A class term is an abbreviation for such a defining condition; it does not add a new kind of object to the quantifiers of ZF. A class in set theory is proper if it is not the collection of elements of any set.

Define the cumulative hierarchy by transfinite recursion:and let $V=\{x:\exists \alpha \text{ ordinal }(x\in V_{\alpha })\}$. Without foundation this denotes the well-founded part produced by the hierarchy; it must not simply be assumed to contain every ambient set. Transfinite recursion on ordinals and the relevant set constructions remain available in ZF without foundation.

Prove transitivity by transfinite induction. The empty stage is transitive. Suppose $T=V_{\alpha }$ is transitive. Then each member of $T$ is a subset of $T$, so $T\subseteq \mathcal{P}(T)$. Also $\mathcal{P}(T)$ is transitive: if $u\in v\in \mathcal{P}(T)$, then $u\in T$ and, by transitivity of $T$, $u\subseteq T$, so $u\in \mathcal{P}(T)$. At a limit stage, a union of transitive sets is transitive. This proves transitivity at every stage.

Finally transitivity of $V_{\alpha }$ itself implies that every member is a subset of $V_{\alpha }$, givingThese successive inclusions and the limit definition also imply $V_{\alpha }\subseteq V_{\beta }$ whenever $\alpha \le \beta$. The argument has not used foundation. This leaf includes the unnumbered definitions preceding the source's three claims.

For every $x\in a$, the assumption $a\subseteq V$ supplies an ordinal stage containing $x$. Let ${\alpha }_x$ be the least such ordinal. Leastness makes this a uniquely definable assignment, so ambient replacement gives the setPut $\beta =\sup S$, taking $\beta =0$ if $a$ is empty. Monotonicity of the hierarchy gives $x\in V_{{\alpha }_x}\subseteq V_{\beta }$ for every $x\in a$. Hence $a\subseteq V_{\beta }$, and thereforeThis is a rank-bounding argument: a set-sized collection of elements from the hierarchy has its stages bounded by one ordinal. No simultaneous arbitrary choices are required, since each least stage is uniquely specified.

Fix a logical formula $\theta (x,y,\vec{p})$ with parameters in $V$, and a set $a\in V$. Suppose $⟨V,\in ⟩$ satisfies that each $x\in a$ has a unique $y$ with $\theta (x,y,\vec{p})$. Relativize every unbounded quantifier in $\theta$ to the definable class in set theory $V$, obtaining an ambient first-order formula ${\theta }^V$.

Transitivity of $V$ ensures that every member of $a$ belongs to $V$. The ambient relationtherefore defines a unique ambient set $y$ for every $x\in a$. Apply ambient replacement to obtain its range set $b$. By construction every element of $b$ belongs to $V$, so part (ii) gives $b\in V$.

Membership in this set $b$ is exactly the range relation required by replacement inside the structure $⟨V,\in ⟩$. Since the logical formula and parameters were arbitrary, every instance of the replacement scheme holds in the hierarchy class in set theory. The proof uses formula relativization to a class and the closure property from part (ii), not the unsupported assertion that $V$ is the whole ambient universe.