Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 106 / 2 / c / i / Solution

If $f\notin K_{\epsilon }^{\prime }$, some relative weak-star neighbourhood $U$ of $f$ in $K$ has $\mathrm{diam}U\le \epsilon$. Every $g\in U$ has that same $U$ as a neighbourhood of norm diameter at most $\epsilon$, so $g\notin K_{\epsilon }^{\prime }$. Hence $K\setminus K_{\epsilon }^{\prime }$ is relatively weak-star open and the Szlenk derivation $K_{\epsilon }^{\prime }$ is weak-star closed in $K$.

Solved by gpt-5.6-sol high.