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

The empty set is countable, and the whole space contains $a$, so both belong to the proposed closed family. A finite union contains $a$ if any member does; otherwise it is a finite union of countable sets and remains countable. For an arbitrary intersection, if every member contains $a$, so does the intersection. Otherwise some member excludes $a$ and is countable, and the intersection is a subset of it, hence countable. Thus all closed-set topology axioms hold.

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 $f$ is continuous.

The set $S=\{x_n:x_n\ne \ell \}$ is countable, hence closed, and does not contain $\ell$. Its complement is an open neighborhood of $\ell$. Convergence therefore places every sufficiently late term outside $S$, which is possible only if those terms equal $\ell$. Thus the sequence is eventually constant at its limit, and the set of exceptional indices is finite. This conclusion concerns repeated indices as well as distinct values.