Here is another example, but with a somewhat different flavor than the others posted here.  Namely, let $\mathbb{Q}$ be the set of rational numbers, and consider the collection of all functions from $\mathbb{Q}$ to $\mathbb{Q}$ which can be extended to continuous functions from $\mathbb{R}$ to $\mathbb{R}$.  (Here $\mathbb{R}$ is the set of real numbers, endowed with the Euclidean topology.)  The article <i>"Can a subset's topology detect continuous extensions?"</i> (The College Mathematics Journal, Volume 49, 2018 - Issue 2) contains a proof that there is no topology $\mathcal{T}$ for which this collection equals the collection of all continuous self-maps of $\mathbb{Q}$ with respect to $\mathcal{T}$.