I've been looking for a proof of one particular direction of this theorem for metric spaces. I've looked online, but everyone seems to use different terminology/notation to state the theorem, so I'd like to ask for an outline of a proof specific to my text's version, which my text "leaves to the reader".
Let $A \subset M$ where $A$ is compact and $M$ is a metric space.
Let $B \subset C_{b}(A,N)$ where $N$ is a metric space and $C_{b}(A,N)$ is the set of all bounded continuous functions from $A$ to $N$.
Then $B$ is compact if and only if $B$ is equicontinuous, closed, and pointwise compact.
I'm trying to prove the forward direction. The reverse direction (showing compactness) is based on the diagonalization argument, which is described well in the textbook, but the text makes no remarks on the forward direction. I already managed to prove pointwise compactness, and closure, which were trivial, but equicontinuity seems difficult. Could someone provide an outline of the proof, or link me to the proof of this particular version of the theorem?
Remarks:
- The metric on $B$ is defined by $d_{B}(f,g) = sup(d_{N}(f(x),g(x)|x \in A) $ where $d_{N}$ is the metric on $N$. So compactness of $B$ is defined relative to this metric $d_{B}$. (Showing $d_{B}$ is a metric is trivial).