Consider the co-countable topology on $\mathbb{R}$, and $A \subset \mathbb{R}$ such that $A = \mathbb{R} \setminus \{p\}$. I was able to show that for any sequence $(a_n) \in \mathbb{R}$ to converge in the co-countable topology, it should eventually be constant. We can easily find a point in $A$ that is the limit point of a sequence in $A$.
But, my question is, can a point outside $A$ be the limit point of a sequence in $A$?
Here is what I've thought of but I don't know if I'm on the right track: The only point that lies outside $A$ is $p$. So, can $p$ be a limit point of a sequence $(a_n) \in A$? For $p$ to be the limit point, if we take an open set $U$, $p \in U$, $U \setminus \{p\} \cap A \neq \emptyset$. But we can never take an open set around $p$ since $\{p\}$ is a closed singleton under this topology. Hence the limit point can never be outside $A$?
I'm new to topology and would appreciate any help!
Best Answer
The answer to the central question is: "no". No matter how you interpret "limit point" it should imply "in the closure of". If $C=\{a_n:n\in\mathbb{N}\}$ is a sequence, a countable set, in $A$ then $C$ is closed in the co-countable topology and if $p\notin A$ then $\mathbb{R}\setminus C$ is a neighbourhood of $p$ that is disjoint from $C$.
So even if $p$ outside $A$ is in the closure of $A$ it is not in the closure of any countable subset of $A$.