Can the closure of a countable set be characterized sequentially? Suppose that I have a countable subset $S \subset X$, where $(X, \tau)$ is a topological space that is NOT first countable (so that convergence is characterized by nets and not sequences). I'm interested in the closure $cl(S)$, which is defined as the collection of all limit points of $S$, where limit points of $S$ are defined as all elements $x \in X$ such that for every open set of $X$ containing $x$, there exists some element of $A$ within that open set (but does not equal $x$ itself). 
My question is then the following: Because the set $S$ is itself countable, can we say that $cl(S)$ is equal to the set of limit points of all sequences of elements in $S$ ? I understand when considering a potential limit point of $A$, there could be an uncountable many open sets that elements in $A$ must fall within, but on the other hand, there are only a countable many elements of $A$ in the first place. On which side of things does this fall?
Thanks!
 A: No, we cannot say this. A somewhat practical example of this coming up is if we take the space $(L_{\infty}(\mathbb R))_1^{*}$ to be the space of linear functions $F:L_{\infty}\rightarrow\mathbb R$ such that $F(f)\leq \|f\|_{\infty}$ under the weak-* topology. This space is compact by the Banach-Alaoglu theorem.
Define a sequence of functions $F_n(f)=\int_{n}^{n+1} f$. Note that this sequence has no limit points; in particular, if $L_{s_n}$ is a subsequence, define $$f_s(x)=\begin{cases}1 & \text{if }s_{2n}\leq x <s_{2n}+1 \text{ for some n}\\
0 & \text{otherwise}.\end{cases}$$
Then, $F_{s_n}(f_s)$ oscillates between $1$ and $0$, hence fails to converge in the weak-* topology.
From this, we see that the sequential closure of $\{F_n\}$ is itself. However, the set of $\{F_n\}$ is not closed, because if it were, it would be compact. This is clearly not the case, as it is a countable discrete space.
A: No. A classical counterexample is the space described in this question. (Arens space). The closure of all points $\mathbb{N} \times \mathbb{N}\setminus \{(0,0\}$ is all of the space, but no sequence from it converges to $(0,0)$.
See also this blog post
