Consider the topological space $(\mathbb{R}, \mathcal{T})$, with $\mathcal{T}\mathcal=\lbrace U \subset \mathbb{R} \mid 0 \in U \Rightarrow (\mathbb{R} \setminus U) \subset \mathbb{Q} \rbrace$. Define the stack of $\lbrace \mathbb{R}\setminus \mathbb{Q} \rbrace \in \mathcal{P}(\mathbb{R})$ as follows:
$$\text{stack}\lbrace \mathbb{R}\setminus \mathbb{Q} \rbrace = \lbrace F \subset \mathbb{R} \mid (\mathbb{R}\setminus \mathbb{Q}) \subset F \rbrace.$$
I have to determine all the convergence and adherent points of the filter $\text{stack}\lbrace \mathbb{R}\setminus \mathbb{Q} \rbrace$. I've already found that $\text{stack}\lbrace \mathbb{R}\setminus \mathbb{Q} \rbrace$ converges to $x$ (and consequently $\text{stack}\lbrace \mathbb{R}\setminus \mathbb{Q} \rbrace \dashv x$), for all $x \in \mathbb{R}\setminus \mathbb{Q}$.
I'm stuck on the case where $x \in \mathbb{Q}$. Any help?
To reformulate the definition of the topology : any set $O$ that does not contain $0$ is open (as it voidly fulfills the condition), so all points of $\mathbb{R} \setminus \{0\}$ are isolated. The only open neighbourhoods of $0$ are of the form $O_A(0) = \{0\} \cup (\mathbb{R} \setminus \mathbb{Q}) \cup A$, where $A \subseteq \mathbb{Q}$.
So indeed all neighbourhoods of $0$ are in the stack and so the stack converges (and is adherent to) to $0$. If $x$ is irrational, it has a neighbourhood $\{x\}$ that is not in the stack so the stack does not converge to $x$, but every neighbourhood of $x$ intersect any member of the stack (in $x$ itself) so all irrationals are adherent to the stack.
Any rational $x \neq 0$ has a neighbourhood $\{x\}$ that is not in the stack, and even is disjoint from a member of the stack ($\mathbb{R} \setminus \mathbb{Q}$ (e.g.) so such $x$ are neither limits nor adherent points.
In short, $0$ and all irrationals are adherent, only $0$ is a limit.