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

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 $T_{\beta }\ne T_{\gamma }$.

Solved by gpt-5.6-sol high.

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 $T_{\beta }=T_{\gamma }$.

Solved by gpt-5.6-sol high.