TITLE: Is a countable subspace of a metric space closed?
QUESTION [6 upvotes]: Let $(X,d)$ be a metric space and $a_1,a_2,\ldots\in X$. Define $A=\{a_n:n\in\mathbf N\}$. Is $A$ closed in $(X,d)$?
And is $A$ closed when X is a topological space?

REPLY [2 votes]: Let $X$ be an infinite topological space.  Then:
1) If $X$ is discrete, then (all subsets are closed, so) every countably infinite subset is closed.
2) If $X$ is non-discrete, separated and first-countable -- in particular, if $X$ is non-discrete and metrizable -- then there is a non-isolated point $x$ and a sequence $\{x_n\}_{n=1}^{\infty}$ of distinct elements of $X \setminus \{x\}$ converging to $x$.  Thus $A_1 = \{ x_n: n \in \mathbb{Z}^+ \}$ is a countably infinite subset which is not closed, and $A_2 = A_1 \cup \{x\}$ is a countably infinite subset which is closed.  
3) Above the hypothesis of first countability can be somewhat weakened: it is enough to assume instead that $X$ is sequential.  But it cannot be dropped entirely: let $X$ be an uncountable set endowed with the cofinite topology, in which a proper subset is closed if and only if it is finite.  Then $X$ is a separated space which has no countably infinite closed subsets.  
At the moment I am not seeing a clean necessary and sufficient condition on a general topological space for it to admit a countably infinite closed subset.  Maybe someone else can do better?