Convergence in compact open topology implies convergence pointwise I'm a beginner in topology, so pardon me if this problem is too easy.

Let $X$ and $Y$ be topological spaces. Let $C(X,Y)$ be the set of all continuous functions from $X$ to $Y$. Show that if a net $(f_i)$ converges to $f$ in the compact open topology of $C(X,Y)$ then it converges to $f$ pointwise.

The part I'm confused is the compact open topology, I mean I know it's the topology generated by $S(A,U)=\{f \mid f(A) \subset U\}$ where $A$ compact and $U$ open. But I don't know how to explain $f_i$ converges to $f$ in definition because the open set of this topology seems to be complicated. So can you solve this problem for me? Thanks for helping.

Suppose $f_n \to f$ in the compact-open topology. 
Then for any $x \in X$, take an open set $U \subseteq Y$ that contains $f(x)$.
Then $f \in S\{x\}, U)$, (this is open in $C(X,Y)$ as finite sets are always compact) which means that there exists $N$ such that for all $n \ge N$ we have $f_n \in S(\{x\}, U)$, which means exactly that for $n \ge N$ we have $f_n(x) \in U$. So as $U$ was an arbitrary open neighbourhood of $f(x)$: $f_n(x) \to f(x)$.
