TITLE: The topology of countable complements
QUESTION [5 upvotes]: Take the topology on $\Bbb R$, the real line, which is, $\tau=\{A\subseteq\Bbb R\mid\Bbb R\setminus A\text{ is countable}\}\cup \{\varnothing\}$. Can one find a convergence sequence in this topology? Because, Take a sequence $\{a_n\}$, And suppose $a=\lim a_n$. Take now the set $\Bbb R\setminus\{a_n\}$. Then this is an open neighborhood of $a$ that doesn't contain any of $\{a_n\}$ which is a contradiction...
It feels like I am missing something basic in here but I can't put my finger on it.
Thanks, Shir

REPLY [2 votes]: Stefan's answer is great. Here's an exercise relevant to your question which I think is worth thinking about:
Exercise 1: Let $\{a_n\}_{n\geq 1}$ be a sequence in the finite complement topology on $\mathbb{R}$. Under what conditions does $\{a_n\}_{n\geq 1}$ converge and how many limits does it have?
The situation is, in some vague sense, "dual" to the case you're considering of the countable complement topology. I hope this exercise will be fun to think about for you!