Convergence of graphs vs pointwise convergence

Question

Suppose that $X$ and $Y$ are compact metric spaces. Let $C(X,Y)$ be the set of continuous maps $f:X\to Y$. For $f\in C(X,Y)$, the graph of $f$ is closed in $X\times Y$. Suppose that $\{f^n\}$ is a sequence in $C(X,Y)$ and suppose that the corresponding sequence of graphs converges to a closed set in $X\times Y$ in the Hausdorff metric topology. Does that imply that there exists a map $f:X\to Y$ that is the pointwise limit of some subsequence of the sequence of functions $\{f^n\}$?

Answer 1

[Update] OP required that the graph of the limit funtion need not be the same as the limit of the closed sets.

Let $X=Y=[-\frac{\pi}{2},\frac{\pi}{2}]\subset\mathbb{R}$, and consider the sequence $$ \bigl\{ f_{2n}(x)=\operatorname{arctan}((2n)x) \mid n\geq 1 \bigr\} $$

and $$ \bigl\{ f_{2n-1}(x)=\operatorname{arctan}((2n-1)x-1) \mid n\geq 1 \bigr\} $$

Then the graph of $f_n$ converges to the union of three line segments, $$ \bigl([-\frac{\pi}{2},0]\times\{-\frac{\pi}{2}\}\bigr) \cup \bigl(\{0\}\times[-\frac{\pi}{2},\frac{\pi}{2}]\bigr) \cup \bigl([0,\frac{\pi}{2}]\times\{\frac{\pi}{2}\}\bigr) $$ But the sequence $\{f_n\}$ does not converge at $x=0$.