{c}

Let $E=\{x:f(x)\ne f(a)\}$ be countable. For a closed subset $C$ of the real line, if $f(a)\in C$, its inverse image contains $a$ and is closed by definition. If $f(a)\notin C$, its inverse image lies inside $E$ and is countable, hence is again closed. Every closed inverse image is therefore closed, proving \b[$f$ is continuous].


Solution

Past exam of the probability, statistics and options I course of the University of Oxford / 2009 / E / 1 / b / ii / Solution

 Solution (source code)