Let $X$ be a topological space and $Y$ a metric space. We give $C(X,Y)$ the compact open topology. If $f_n\to f$ in $C(X,Y)$ then $f_n$ converges uniformly to $f$ on every $K\subseteq X$ compact.
Let $K\subseteq X$ be a compact subset and $\epsilon >0$. We need to prove there exists $N\in\mathbb{N}$ such that if $n\ge N$ then $d(f_n(x),f(x))<\epsilon$ for all $x\in K$.
Open subbasic subsets of $C(X,Y)$ are of the form $(K_0,U):=\{g\in C(X,Y):g(K_0)\subseteq U\}$, where $K_0$ is compact and $U$ is open. We would have to find some of these opens such that $f\in (K_0,U)$ and apply the convergence. I think we choose $K_0=K$, but what open set $U$ would we choose?
Thank you.
Best Answer
As you said, you need to prove there is an $N\in\mathbb{N}$ such that if $n\ge N$ then $d(f_n(x),f(x))<\epsilon$ for all $x\in K$. In other words, you want $f_n$ to be in $B^K_\varepsilon(f) = \{g:X\to Y \mid d(g(x),f(x))<\varepsilon \text{ for all } x\in K \}$. So if you find pairs $(K_1,V_1),\dots,(K_l,V_l)$ such that $f\in W=\bigcap_{i=1}^l(K_i,V_i)$ and $W\subseteq B^K_\varepsilon(f)$, then you can apply the convergence.
Your initial guess was to take $l=1$ and $K_1=K$, but this does not quite work. Consider for example $X=Y=K=[0,1]$ and $f=id$ and $\epsilon=0.5$. Then no open set $V$ exists such that $(K,V)$ is in $B_{0.5}(id_X)$.
Instead, note that for every $x\in K$ there is an open neighborhood $U_x$ of $x$ such that $f(U_x) \subseteq B_{\epsilon/3}(f(x)).$ Then $$ f\left(\overline{U_x}\right) \subseteq \overline{f(U_x)} ⊆ \overline{B_{\epsilon/3}(f(x))} ⊆ B_{\epsilon/2}(f(x)) $$ Since $K$ is compact and the $U_x$ cover $K$, there are $x_1,\dots,x_l$ such that $U_i=U_{x_i},\ i=1,\dots,l$ cover $K$. Can you derive from these the pairs $(K_1,V_1),\dots,(K_l,V_l)$ such that each $g\in\bigcap_1^l(K_i,V_i)$ is an element of $B^K_\epsilon(f)$?