Let $S=\{x:f(x)\ne f(a)\}$, which is countable. For any <closed set> $C\subseteq\mathbb R$, consider its <preimage>. If $f(a)\in C$, then $a\in f^{-1}(C)$ and that <preimage> is closed by definition. If $f(a)\notin C$, then $f^{-1}(C)\subseteq S$, so it is countable and again closed. Therefore every <closed set> has a closed <preimage>, proving \b[<continuity>].

Solution

Past exam of the options I course of the University of Oxford / 2009 / E / E1 / b / ii / Solution

 Solution (source code)