Let $Y,Z$ be topological spaces with $Z$ a $T_2$-space and $Z^Y$ the space of continuous functions $f:Y\to Z$ with the compact-open topology.
The problem is:
$A:=\{(f,y,z):f(y)=z\}$ is a closed subset of $Z^Y\times Y\times Z$.
I thought maybe this could be easily solved with nets.
If we take a net $(x_{\lambda})$ in $A$ with $x_{\lambda}\to x$, we would want to show that $x\in A$. Say $x_{\lambda}=(f_{\lambda},y_{\lambda},z_{\lambda})$ and $x=(f,y,z)$. Then we have $f_{\lambda}\to f$, $y_{\lambda}\to y$ and $z_\lambda \to z$ where $f_{\lambda}(y_{\lambda})=z_{\lambda}$ for every $\lambda$.
I we choose this way, this is my next question:
Would it be true that $f_{\lambda}(y_{\lambda})\to f(y)$?
If so, as $Z$ is $T_2$ then we would have $f(y)=z$ as desired.
But I can't prove that yet. Take $V\subseteq Z$ open such that $f(y)\in V$. Then $y\in f^{-1}(V)$ and so there exists $\lambda_1$ such that $y_{\lambda}\in f^{-1}(V)$, equivalently, $f(y_{\lambda})\in V$ if $\lambda\ge\lambda_1$. Now, subbasic open subsets in $Z^Y$ are of the form $(K,U)$ where $K\subseteq Y$ is compact and $U\subseteq Z$ is open. If would be very nice if $K=\{y_{\lambda}:\lambda\ge\lambda_1\}\cup\{y\}$ were compact, because then $f\in (K,V)$ and with the fact that $f_{\lambda}\to f$ we would get the desired result, but I think this last is not true.
What do you think?
Thank you.