Equivalence of continuity with filters In lecture today, my professor posed a related theorem to one in Willard's General Topology. He said he had a proof for his updated statement, but he realized during the presentation that it was either wrong (i.e. he could come up with a counter example) or he didn't write down his proof well enough, so he shoved it aside and said to forget about it. But I'm too curious now!
So, let's allow $X$ and $Y$ be topological spaces.

Theorem 12.8 Let $f:X \to Y$. Then $f$ is continuous at $x_0 \in X$ if and only if whenever $\mathscr{F}$ is a filter on $X$ and $\mathscr{F} \to x_0$ in $X$, then $f(\mathscr{F}) \to f(x_0)$.

My professor's new statement is 

Theorem 12.8' Let $f:X \to Y$. Then $f$ is continuous at $x_0 \in X$ if and only if whenever $\mathscr{F}$ is a filter on $X$, and $\mathscr{F}$ clusters at $x_0$, then $f(\mathscr{F})$ clusters at $f(x_0)$.

I'm not quite sure if Willard's definitions are ubiquitous, but he tells us that given a filter $\mathscr{F}$ on $X$, $\mathscr{F} \to x\in X$ means that the neighborhood filter, $\mathscr{N}_x$, is a subfilter of $\mathscr{F}$ and $\mathscr{F}$ clusters at $x\in X$ if $x \in \overline{F}$ for all $F \in \mathscr{F}$. 
Are there any immediate counter examples to the new theorem?
 A: The new statement is true.  For one direction, suppose $f$ is continuous at $x_0$ and $\mathscr{F}$ clusters at $x_0$.  To show $f(\mathscr{F})$ clusters at $f(x_0)$, we must show every element $A\in f(\mathscr{F})$ intersects every neighborhood $U$ of $f(x_0)$.  Now since $f$ is continuous at $x_0$, $f^{-1}(U)$ is a neighborhood of $x_0$, so it intersects $f^{-1}(A)$ since $f^{-1}(A)\in\mathscr{F}$.  If $x\in f^{-1}(A)\cap f^{-1}(U)$ then $f(x)\in A\cap U$ so the intersection is nonempty, as desired.
For the other direction, suppose $f$ is not continuous at $x_0$, so there is some neighborhood $U$ of $f(x_0)$ such that $f^{-1}(U)$ is not a neighborhood of $x_0$.  This means that $x_0\in\overline{X\setminus f^{-1}(U)}$, so if we let $\mathscr{F}$ be the filter generated by $X\setminus f^{-1}(U)$, then $\mathscr{F}$ clusters at $x_0$.  Also, $Y\setminus U\in f(\mathscr{F})$.  But since $U$ is a neighborhood of $f(x_0)$, $f(x_0)\not\in\overline{Y\setminus U}$, so $f(\mathscr{F})$ does not cluster at $x_0$.
