Convergence on open-minus-countable topology Consider the topology on $\mathbb{R}$ in which a set is open if $U = V \setminus C$, where $V$ is open in the usual topology and $C$ is a countable set.
Prove that in this space a sequence converges iff it is eventually constant.
 A: We know that a sequence $\{x_n\}$ converges iff there exists a neighborhood base containing eventually the sequence. So, we can try to consider a neighborhood base $\mathcal{N}_x$ formed by $V_\epsilon = B_\epsilon (x) \setminus C$, where $B_\epsilon (x)$ is the ball with the usual topology and $C$ a countable set such that $x \notin C$.
Then, if a sequence is eventually equal to $x$, it eventually belongs to any $V_\epsilon$; conversely, if a sequence isn't eventually equal to $x$, choosing $C = \cup_{n \in \mathbb{N}} \{x_n\} \setminus \{x\}$ we have that $\{x_n\}$ doesn't belong to any $V_\epsilon$.
A: Recall the definition of convergence of a sequence in a general topological space:

$\{x_n\}_{n\in\mathbb N}$ converges to $x$ if and only if for every open set $U$ such that $x\in U$ there is some $k\in\mathbb N$ such that for all $n>k$ we have $x_n\in U$. That is, $\{x_n\}$ is eventually in $U$ for every open neighborhood $U$ of $x$.

One implication is trivial, of course that eventually constant sequences converge. (You should still write that, though.)
Now suppose that $\{x_n\}$ is not eventually constant. Suppose that $x$ was the limit of $\{x_n\}$, let $V$ be an open interval around $x$, then $U=V\setminus\{x_n\neq x\mid n\in\mathbb N\}$ is an open set in our topology. Because the sequence is not eventually constant we can find arbitrarily large $n\in\mathbb N$ such that $x_n\notin U$. However $x\in U$ and $U$ is open so $x_n$ is not eventually in $U$, which is a contradiction to the definition of convergence.
