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 "Can a subset's topology detect continuous extensions?" (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 $\mathcal{T}$.

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 "Can a subset's topology detect continuous extensions?" (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 $\mathcal{T}$.

Here is another example, but with a somewhat different flavor than the others posted here. Namely, let Q be the set of rational numbers, and consider the collection of all functions from Q to Q which can be extended to continuous functions from R to R. (Here R is the set of real numbers, endowed with the Euclidean topology.) The article "Can a Subset's Topology Detect Continuous Extensions?" (The College Mathematics Journal, Volume 49, 2018 - Issue 2) contains a proof that there is no topology T for which this collection equals the collection of all continuous self-maps of Q with respect to T.

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 "Can a subset's topology detect continuous extensions?" (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 $\mathcal{T}$.