Let $K \subset R$ be compact and let $B \subset C(K)$ be compact. Prove that $B$ is equicontinuous as follows Let $K \subset R$ be compact and let $B \subset C(K)$ be compact. Prove that $B$ is equicontinuous as follows: ($C(K)$ is the set of all complex valued, continuous, bounded functions with domain K)
(1) Prove that the map $F : C(K) \times K \to R$ defined by $F(f,x)=f(x)$ is continuous.
(2) Use uniform continuity of $F$ restricted to $B \times K$ to deduce the result.
I'm trying to solve this problem but I'm totally lost on how to prove the continuity of this function when there is no metric defined on the domain. I've been trying to prove it by showing that the inverse image of an open set in $R$ is open in $C(K) \times K$ but cannot find a way to show this. How may I solve this problem?
 A: You don't really need a metric, $C(K) \times K$ just has the product topology, which is generated by all sets of the form $U \times V$, where $U \subset C(K)$ is open and $V \subset K$ is also open. Sets of this form form a base for the topology, which means that $O \subset C(K) \times K$ is open iff for all $(f,x) \in O$, there exist $U \subset C(K)$ open and $V \subset K$ open such that $f \in U, x \in V$ and $U \times V \subset O$.
But indeed, as Joel pointed out, we can use a metric (the standard product metric) because both of our spaces have a metric: $D(f,g) = \sup_{x \in K} d(f(x), g(x))$, where $d$ is the metric on $\mathbb{R}$. Then $C(K) \times K$ has a metric too: $d'((f,x),(g,y)) = D(f,g) + d(x,y)$, and the topology induced by $d'$ is exactly the same as the product topology. I will use this metric in the proof, as this is the most concrete.
We can show continuity of $F$ at the point $(f_0,x_0)$ as follows: let $\epsilon > 0$. We need to find $\delta > 0$ such that for all $(f,x) \in C(K) \times K$: $d'((f_0,x_0), (f,x)) < \delta$ then $d(F(f_0,x_0), F(f,x)) = d(f_0(x_0), f(x)) < \epsilon$.
First use that $f_0$ is a continuous function at $x_0$, and find $\delta_1$ such that if $x \in K$ satisfies $d(x_0, x) < \delta_1$, then $d(f(x_0), f(x)) < \frac{\epsilon}{2}$.
Then define $\delta = \min(\frac{\epsilon}{2}, \delta_1)$.
Now if $d'((f_0, x_0),(f,x)) < \delta$, we know in particular that $D(f_0,f) < \frac{\epsilon}{2}$, so that for all $x \in K$, $d(f_0(x), f(x)) \le D(f_0, f) < \frac{\epsilon}{2}$, as $D$ is the $\sup$. Also, $d(x_0, x) < \delta_1$ so we know that $d(f_0(x_0), f_0(x)) < \frac{\epsilon}{2}$. So now we apply the triangle inequality, for when $d'((f_0, x_0),(f,x)) < \delta$:
$$d(f_0(x_0), f(x)) \le d(f_0(x_0), f_0(x)) + d(f_0(x), f(x)) < {\epsilon \over 2} + {\epsilon \over 2} = \epsilon$$
which is as required.
