Every evaluation map $\pi_t$ is continuous in the uniform metric, so $\mathcal F\subseteq\mathcal B$. Conversely, for $g\in C[0,1]$,
$$
\{f:\lVert f-g\rVert_\infty<r\}
=\bigcup_{m\geq1}\bigcap_{q\in\mathbb Q\cap[0,1]}
\{f:|f(q)-g(q)|\leq r-1/m\},
$$
with the harmless restriction to $m$ for which $r-1/m>0$. Thus every open ball belongs to $\mathcal F$. Separability makes every open set a countable union of balls, so $\mathcal B\subseteq\mathcal F$ and equality follows.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 201 / 6 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-201/6/a/solution