Equicontinuity, compactness and uniform convergence. Let $X$ be a compact topological space and let $K\subset \mathbb{R}^n$ be compact. Let $F = \{f_n\}_{n \ge 1}$ be a sequence of continuous functions where $f_n : X \to K$ that converges pointwise to $f$.
Show that if $F$ is equicontinuous, then $f$ is continuous and $f_n$ converges uniformly to $f$.
Attempt
a) It is enough to show that the convergence is uniform, since the Uniform Limit Theorem then implies that $f$ is continuous, correct?
b) Since the convergence is pointwise, $F$ must be pointwise bounded, correct? (If it's not, then I have a big misunderstanding...)
c) By Arzela's theorem, since $X$ is compact,$F$ pointwise bounded and equicontinuous,  $F$ has a uniformly convergent subsequence, correct?
d) Since $F$ has a uniformly convergent subsequence, can I conclude that it converges uniformly to $f$? Why? Why not?
Any help is highly appreciated.
 A: Hi you are right with point b) and c).
let's show that f is continuous
if $F$ is equicontinuous let $\epsilon >0$ let $x>0$
there exist $\delta >0$ such that  $\forall n \in N$ $|x-y|< \delta =>|f_n(x)-f_n(y)|< \frac{\epsilon}{3}$
So let $y$ such that $|x-y|< \delta$
by pointwise convergence let $n \in N $ such that $|f_n(y)-f(y)|<\frac{\epsilon}{3}$
and $|f_n(x)-f(x)|<\frac{\epsilon}{3}$
then $|f(x)-f(y)|< |f(x)-f_n(x)| + |f_n(x)-f_n(y)| + |f_n(y)-f(y)| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$
then f is continuous.
As mentionned in the comment to conclude you can use this sub-subsequence argument :
let's show that $f_n$ uniformly converge to $f$
We assume that $f_n$ don't uniformly converge to $f$
So there exist $\epsilon$ and there exist $\phi$ such that
$\forall n \in N ~~ |f_{\phi(n)}-f|_{\infty}> \epsilon$
but you can apply arzela's theorem to the subsequence $(f_{\phi(n)})$ because the subsequence verify the theorem's conditions.
So there exist $\psi$ such that $(f_{\phi(\psi(n))})$ uniformly converge to $f$
but as a subsequence of $(f_{\phi(n)})$
$\forall n \in N$ $|f_{\phi(\psi(n))}-f|_{\infty }> \epsilon $
This is a contradiction hence $f_n$ uniformly converge to $f$
