Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 128 / 1

The definable power set of a set $X$ isThus definability is evaluated internally in the structure $(X,\in )$ and parameters from $X$ are allowed.

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The constructible hierarchy is defined by transfinite recursion:when $\lambda$ is a limit ordinal. Its union over all ordinals is the constructible universe $L$.

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A condensation sentence for the constructible hierarchy may be obtained by taking a single conjunction $\sigma$ that expresses a sufficiently strong finite fragment of set theory, the assertion $V=L$, and that the ordinals have no largest member. The finite fragment is chosen strong enough to define the satisfaction relation needed for the $L$-construction and to prove its absoluteness for transitive sets.

If a transitive set $X$ satisfies $\sigma$, let $\lambda =X\cap \mathrm{Ord}$. The absence of a largest ordinal makes $\lambda$ a limit ordinal. Internal $V=L$ says every $x\in X$ belongs to some internally constructed $L_{\beta }$, while transitivity and absoluteness identify that level with the actual $L_{\beta }$. Conversely the finite closure axioms ensure that every $L_{\beta }$, $\beta <\lambda$, belongs to $X$. Hence $X=L_{\lambda }$.

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Choose a successor ordinal $\beta =\delta +1>\alpha$, and then choose a limit ordinal $\gamma >\beta$. The level $L_{\gamma }$ satisfies the condensation sentence for the constructible hierarchy. The level $L_{\beta }$ cannot satisfy it: otherwise condensation would give $L_{\beta }=L_{\lambda }$ for a limit $\lambda$, butwould imply the impossible equality $\beta =\lambda$. Thus the condensation sentence belongs to $T_{\gamma }$ but not to $T_{\beta }$, and $T_{\beta }\ne T_{\gamma }$.

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The language of set theory has only countably many sentences, so there are at most $2^{{\aleph }_0}$ possible complete sets of sentences $T_{\xi }$. Choose $(2^{{\aleph }_0}{)}^{+}+1$ ordinals above $\alpha$. By cardinal pigeonhole, two of their theories agree. Hence some $\gamma >\beta >\alpha$ satisfy $T_{\beta }=T_{\gamma }$.

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