Closure of $(a,b)$ in one-point compactifcation of $\mathbb{Q}$ Take $\mathbb{Q}$ to be the rational numbers with the usual topology induced by the absolute value.
Let $\mathbb{Q}^*=\mathbb{Q} \cup \{ \infty \}$ be the one-point compactification of $\mathbb{Q}$.
Let $a$, $b \in \mathbb{R}$ be any real numbers, $a < b$.

What is the closure of the set $(a,b) \cap \mathbb{Q}$ in $\mathbb{Q}^*$?

I suspect that the answer is not $[a,b] \cap \mathbb{Q}$.
My reasoning is that the complement of that set, i.e. $\mathbb{Q}^* \setminus ( [a,b] \cap \mathbb{Q})= \{ \infty \} \cup (\mathbb{Q} \setminus [a,b] )$, is not open.
That is because $[a,b]$ is not a compact set in $\mathbb{Q}$.
Clearly this is not a proof of my suspicion, but I think it could be expanded to a proof.
However this is very unintuitive, as $[a,b] \subset \mathbb{R}$ is in fact the closure of $(a,b) \subset \mathbb{R}$ in $\mathbb{R}^*$.
Is my guess correct?
 A: It's clear that if $x \in \mathbb{Q}$ and $x < a$ or $x > b$ then there is already in $\mathbb{Q}$ a neighbourhood of $x$ that misses $(a,b)$ (and these are also open in $\mathbb{Q}^\ast$, so they're not in the closure we are looking for), and it's also easy to see that $a$ and $b$ will be in the closure, as their neighbourhoods are the standard ones from $\mathbb{Q}$. 
So the first candidate for the closure of $(a,b) \cap \mathbb{Q}$ will be $[a,b] \cap \mathbb{Q}$, and we have to ask ourselves: is this closed in $\mathbb{Q}^\ast$?
Or, is its complement in $\mathbb{Q}^\ast$ open in $\mathbb{Q}^\ast$?
But the complement contains $\infty$, the compactifying point, and for a set of the form $\mathbb{Q}^\ast \setminus C$ (where $C \subseteq \mathbb{Q}$) to be an open neighbourhood of $\infty$ it is by definition only the case if $C$ is closed and compact in the rationals. But it's classical that $[a,b] \cap \mathbb{Q}$ is not compact in the rationals. (E.g. sequences in it can converge to irrationals and don't have convergent subsequences). So the closure cannot be $[a,b] \cap \mathbb{Q}$, as all other rationals are out, the only remaining candidate is $C = ([a,b] \cap \mathbb{Q}) \cup \{\infty\}$ and its complement in $\mathbb{Q}^\ast$ is just $\mathbb{Q}\setminus ([a,b] \cap \mathbb{Q})$ which is just a regular open set of the rationals so also open in $\mathbb{Q}^\ast$. So $$\overline{(a,b) \cap \mathbb{Q}} = ([a,b] \cap \mathbb{Q}) \cup \{\infty\}\text{.}$$ 
