Filter of sets containing a subset converges I'm just learning about filters, and I came across the following exercise in Willard's Topology: 

Let $X$ be a topological space and $A \subset X$. The cluster points
  of the filter $\mathcal{F} = \{ U \subset X \mid A \subset U \}$
  include each point of $\overline{A}$. Under what conditions (on $A$ or
  on the topology) will $\mathcal{F}$ converge to some point?

The answer I got was: $\mathcal{F}$ will converge to some point $x \in A$ iff it converges to every point $x \in A$ iff the subspace topology on $A$ is the indiscrete topology. In particular, if $A$ is not a singleton, $X$ cannot be Hausdorff if $\mathcal{F}$ converges to some point.
Does anyone see any mistakes, or have anything to add?
 A: Suppose that $\mathscr{F}$ converges to $x\in X$; then by definition $\mathscr{U}_x\subseteq\mathscr{F}$, where $\mathscr{U}_x$ is the nbhd filter at $x$. Thus, every nbhd of $x$ contains the set $A$, and hence $A\subseteq\bigcap\mathscr{U}_x$. Conversely, if $A\subseteq\bigcap\mathscr{U}_x$, then $\mathscr{U}_x\subseteq\mathscr{F}$, and $\mathscr{F}$ converges to $x$. This can happen even if the relative topology on $A$ is not indiscrete. For example, let $X=\Bbb N$. For $n\in\Bbb N$ let $U_n=\{k\in\Bbb N:k\ge n\}$, and let 
$$\tau=\{\varnothing\}\cup\{U_n:n\in\Bbb N\}$$
be the topology on $X$. Let $A=\Bbb Z^+$. Then $\mathscr{F}=\{U_0,U_1\}\supseteq\{U_0\}=\mathscr{U}_0$, so $\mathscr{F}\to 0$, but the relative topology on $A$ is not indiscrete; indeed, it’s $T_0$.
It is true, however, that if $X$ is Hausdorff, and $\mathscr{F}$ converges to some point $x$, then $A=\{x\}$. In fact, it’s enough for $X$ to be $T_1$. For suppose that $X$ is $T_1$, $\mathscr{F}\to x$, and $y\in X\setminus\{x\}$. Then there is a $U\in\mathscr{U}_x$ such that $y\notin U$, so $y\notin\bigcap\mathscr{U}_x\supseteq A$, and hence $A=\{x\}$.
