Equivalent condition of convergence of Filter Let $\mathcal F$ be a filter in a topological space $X$.
Denote $E_\mathcal F$ the set of elementary filters that are refined by $\mathcal F$. Elementary filters are the filters associated to sequences.
Clearly, if some filter $\mathcal E \in E_\mathcal{F}$ converges to a point $x$, then $\mathcal F$ must also converge to $x$, as $\mathcal F$ refines $\mathcal E$.
Is the converse also true? If $\mathcal F$ converges to $x$ does there exist an elementary filter $\mathcal E$ which is refined by $\mathcal F$ and converges to $x$?
I'm thinking it is false if $X$ is not first-countable (or sequential), is an example easy to construct? And in the case where $X$ is first-countable is it true?
 A: Consider $X=\omega_1 + 1 = [0,\omega_1]$, the space of all ordinals less than or equal to the least uncountable ordinal $\omega_1$, with the order topology. Let $\mathcal{F}$ be the neighborhood filter of $\omega_1$. Of course $\mathcal{F}$ converges to $\omega_1$. However, no filter derived from a sequence refines $\mathcal{F}$.
If $X$ is first countable then the statement is true: Suppose $F$ is a filter converging to $x\in X$. Let $(U_n)$ be a countable neighborhood basis for $x$. Then for every $A\in F$ there is $n$ such that $U_n\subseteq A$. Let $V_n = \bigcap_{i\le n} U_i$. For each $n$ choose $s_n\in V_n$; thus $m\ge n$ implies $s_m\in V_n$. Then the sequence $s = (s_n)$ converges to $x$, and the filter $F_s$ of final segments of $s$ refines $F$: for every $A\in F$ there is $n$ such that for all $m\ge n$, $s_m\in A$.
The converse is not, in general: If $F_s$ is the elementary filter induced by a sequence, then $F_s$ contains countable sets (as well as uncountable sets), so in $\Bbb R$, as you note in your comment, no neighborhood filter can refine it, because all members of a neighborhood filter are uncountable, hence none can be contained in any of the countable members of $F_s$.
