Arzela-Ascoli theorem exercise The question: 

Define a metric space  $C(K)=\left \{ f: K\rightarrow \mathbb{R} 
> \right\} $ , where $f$ is continuous function on $K$. Let  $K\in
 \mathbb{R}$ be compact and let $B\subset C(K)$ be compact. Prove that
  $B$ is equicontinuousas follows:
  
  
*
  
*Prove that the map $F:C(K)\times K\rightarrow \mathbb{R}$ defined
  by $F(f,x)=f(x)$ is continuous.
  
*Use uniform continuity of $F$ restricted to $B\times K$ to deduce
  the result.

My attempt: 


*

*For $F(f,x)=f(x),F(g,x)=g(x)$, $\left | F(f,x)-F(g,x) \right |=\left | f(x)-g(x) \right |\leq \sup\left | f(x)-g(x) \right |=d(f,g)$ Hence, $F$ is continuous on $C(K)\times K
$.

*If I show $B$ is closed and pointwise bounded ,then $B$ is equicontinuous by Arzela-Ascoli theorem. Since $B$ and $K$are compact, cartesian product $B\times K$ is also a compact set. So, $F$ is uniformly continuous on $B\times K$ . $B$ is compact. By Heine-Borel theorem, $B$ is closed and bounded. It suffices to show that $B$ is pointwise bounded. But, I don't know this part using the uniform continuity of F on $B\times K$.
 A: A direct proof. Let $\epsilon>0$. Since $B$ is compact in the metric space $C(K)$ (with the uniform distance), it follows that the open cover $\bigcup_{f\in B}B_f(\epsilon/3) $ has a finite subcover:  there exist $f_1, f_2,\dots, f_N\in B$ such that 
$$B\subset \bigcup_{i=1}^NB_{f_i}(\epsilon/3).$$
Now let $\delta>0$ be such that $f_1, f_2,\dots, f_N$ are $(\epsilon/3,\delta)$-uniformly continuous in the compact set $K$: if $x,y\in K$ with $|x-y|<\delta$ then $|f_i(x)-f_i(y)|<\epsilon/3$ for $i=1,2,\dots,N$.
Hence, if $f\in B$ and $x,y\in K$ with $|x-y|<\delta$ then for some $1\leq i\leq N$, $f\in B_{f_i}(\epsilon/3)$ and
$$|f(x)-f(y)|\leq \underbrace{|f(x)-f_i(x)|}_{\text{$<\epsilon/3$ because $f\in B_{f_i}(\epsilon)$} }+\underbrace{|f_i(x)-f_i(y)|}_{\text{$<\epsilon/3$ because $f_i$ is $(\epsilon/3,\delta)$-u.c. in $K$}} +\underbrace{|f_i(y)-f(y)|}_{\text{$<\epsilon/3$ because $f\in B_{f_i}(\epsilon/3)$} } < \epsilon$$
which means that  $B$ is equicontinuous.
A: You have not specified the metric on $B \times K$; I will take it as $ D((f,x),(g,y))=\max \{||f-g||,d(x,y)\}$ where $||\,.\, ||$ is the sup norm.  Uniform continuity of $(f,x) \to f(x)$ means given $\epsilon >0$ there exists $\delta >0$ such that $\max \{||f-g||,d(x,y)\}<\delta$ implies $|f(x)-g(y)|<\epsilon$. Fix an element $g \in B$ and a point $y \in K$. Then $|f(x)| <\epsilon +|g(y)|$ whenever $f \in B(g,\delta)$ and $x \in B(x,\delta)$ The open sets $B(g,\delta ) \times B(y,\delta )$ cover the compact set $B \times K$ as $g$ and $y$ vary. Extract a  finite subcover $\{B(g_i,\delta ) \times B(y_i,\delta ):1\leq i \leq N\}$. Now can you conclude that $f(x) <\epsilon + \sum_{i=1}^{N} |g_i(y_i)|$ for all $(f,x) \in B\times K$. This proves pointwise boundedness.
