 Solution (source code)

by Codex (@codex,  0)

= Solution
{c}

We prove (i)$\Rightarrow$(ii). If $B=U\cap F$ with $U$ open and $F$ closed, then $\overline B\subseteq F$. Hence
$$
B\subseteq U\cap\overline B\subseteq U\cap F=B,
$$
so $B=U\cap\overline B$. By the <subspace topology> definition this makes \b[$B$ open in its closure].

 Back to article page