Past exam of the analytic topology and axiomatic set theory course of the University of Oxford / 2007 / B / 8

For a set $A$, its definable power set isOnly finitely many parameters occur in each definition. Satisfaction for the set structure is definable, so this specifies a uniformly definable class function assigning a set to every set $A$, with $\mathrm{Def}(A)\subseteq \mathcal{P}(A)$. Every subset definable over the structure with its parameters occurs. The full set $A$ itself occurs by the formula $x=x$. If $A$ is transitive, then $A\subseteq \mathrm{Def}(A)$, since any $p\in A$ is defined as a subset of $A$ by $x\in p$. In that case $\mathrm{Def}(A)$ is transitive as well: a member of one of its elements lies in $A$, hence in $\mathrm{Def}(A)$.

The constructible hierarchy and universe areThe displayed properties prove by induction that levels are transitive and increasing, so $L$ is transitive. It contains every ordinal: inductively $L_{\alpha }\cap \mathrm{Ord}=\alpha$, and the subset of $L_{\alpha }$ consisting of ordinals is definable there, giving $\alpha \in L_{\alpha +1}$. Ordinalhood is absolute here because the levels are transitive. Both the level function and membership in $L$ are definable in $ZF$. This is Gödel's constructible universe, which can contain fewer subsets than the full universe.

For each $x\in a$, let ${\beta }_x$ be the least ordinal with $x\in L_{{\beta }_x}$. It exists because $a\subseteq L$, and it is uniquely definable. Replacement therefore gives the set $\{{\beta }_x:x\in a\}$. Choose an ordinal $\alpha$ at least their supremum. Increasingness of the hierarchy givesIf $a$ is empty, $\alpha =0$ works. No assumption that $a$ itself belongs to $L$ is used; it is enough that its elements do.

Fix $a\in L$. In the ambient universe, Separation forms the setIt is a set of constructible sets, so the previous part gives an ordinal bound on all its members. Increase the bound to a level $L_{\beta }$ also containing $a$. Transitivity ensures $a\subseteq L_{\beta }$. ThenThe inclusion test is absolute for the transitive level $L_{\beta }$, since all members of $x$ and $a$ lie there. The right side is consequently a subset definable over $L_{\beta }$ with parameter $a$, soFor $x\in L$, transitivity of the constructible universe makes internal and ambient subsethood of $a$ agree. Hence $L$ satisfies that $S$ consists exactly of all its subsets of $a$. This is the Power set Axiom in $⟨L,\in ⟩$. It is not the assertion that every ambient subset of $a$ belongs to $L$.

Use the constructible universe theorem that $L$ satisfies $ZFC$ and GCH. In particular, internally it has a bijectionThis bijection belongs to $L$ and is also an actual bijection in the ambient universe, because its domain, range and graph are transitive-model sets with the same membership relation.

We must compare the ordinals rather than assume cardinality is absolute. Every ordinal below ${\omega }_1^L$ is countable in $L$, and its witnessing enumeration belongs to the ambient universe too. Thus every such ordinal is countable externally, proving ${\omega }_1^L\le {\omega }_1$. Consequently the ambient cardinality of the constructible reals is at most ${\aleph }_1$.

If every subset of $\omega$ were constructible, the displayed bijection would enumerate the full ambient power set using at most ${\aleph }_1$ points. This contradicts the assumption $2^{{\aleph }_0}={\aleph }_2$. ThereforeThe argument permits ${\omega }_1^L$ to be strictly smaller than the ambient ${\omega }_1$, exactly the cardinal-absoluteness issue highlighted in the source.