Open sets in continuous function space Let $Y,Z$ Hausdorff spaces, $Y$ compact, let $F\subseteq Z$ closed,  $G \subseteq Y$ open.
Then $\{ f | f^{-1} (F)\subseteq G \} $  is open in $\mathbb{C} (Y,Z) $ with the open-compact topology.
I don't know how to prove this.
 A: $\{ f | f^{-1} (F)\subseteq G \}=\{ f | f(Y\setminus G)\subseteq Z\setminus F\}. $ 
A: The basic idea is Alex' identity:
$$\{ f \mid f^{-1}[F]\subseteq G \}=\{ f \mid f[Y\setminus G]\subseteq Z\setminus F\}$$
Proof: if $f$ is in the left hand set, take $y \in Y \setminus G$ arbitrarily. Then $f(y) \notin F$, or otherwise (by definition) $y \in f^{-1}[F]$ and thus $f(y) \in G$ by $f$ being in the left hand set, contradiction, so $f[Y \setminus G] \subseteq Z \setminus F$, hence $f$ in the right hand set.
If $f$ is in the right hand set, let $y \in f^{-1}[F]$ be arbitrary. If $y \notin G$ we'd have $y \in Y\setminus G$ and so $f(y) \in f[Y\setminus G] \subseteq Z \setminus F$ and so $f(y) \notin F$, contradicting $y \in f^{-1}[F]$. So $y \in G$ and so $f$ is in the left hand set.
As $Y\setminus G$ is closed in $Y$ and $Y$ is compact, so is $Y\setminus G$. Moreover $Z \setminus F$ is open in $Z$, as $F$ is closed. So the right hand set is by definition open in the compact-open topology on $C(Y,Z)$ and hence so is the set we are interested in. 
