= Fortissimo space
Choose an uncountable <set> $X$ and a point $a\in X$. Every point other than $a$ is isolated, and a set containing $a$ is <open> precisely when its complement is countable. Equivalently, the <closed sets> are all subsets containing $a$ together with all countable subsets. Arbitrary intersections and finite unions preserve this latter condition, so it defines a <topology>.
Every convergent <sequence> in this space is eventually equal to its limit. Indeed the set of its values different from its limit is countable and <closed>; its complement is an <open neighbourhood> of the limit. Eventual membership in this complement forces eventual equality. Nevertheless $a$ is not isolated, because its singleton has uncountable complement. In particular <convergence of a sequence in a topological space>[sequence convergence] alone does not determine this topology.
A real-valued <function> on this space is <continuous> exactly when $\{x:f(x)\ne f(a)\}$ is countable. Sufficiency follows by considering the <preimages> of closed subsets of the <real line>: each either contains $a$ or is a subset of that countable exceptional set. For necessity, each set $\{x:|f(x)-f(a)|\geq1/n\}$ is closed and excludes $a$, hence is countable; their countable union is the exceptional set.
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