Let $X$ be a metric space and $(f_n)$ an equicontinuous sequence of functions from $X$ to $\mathbb{R}$.
We suppose that $A=\{x\in X| \{f_n(x),n=1,2,...\} \text{ is bounded}\}$, is not empty, and that $X$ is compact and connected.
I have to prove that $(f_n)$ has a convergent subsequence.
I have proved in a previous question that the set $A$ is both open and closed, so since $X$ is connected and $A$ is nonempty we'll have $A=X$. But I don't know how to use this to construct a convergent subsequence.
Thank you in advance for your help.
I think I was able to prove that by using Arzelà-Ascoli theorem as suggested by cmk in the comments.
To apply this theorem, we just need to prove that $(f_n)$ is uniformly bounded.
$(f_n)$ is equicontinuous so $\forall x\in X$ there is an open neighborhood $V_x$ of $x$ such that $\forall n\ge 1\ \forall t\in V_x\ |f_n(t)-f_n(x)|<1$
$X=\bigcup V_x$ and $X$ is compact so $\exists x_1,x_2,...,x_p\in X$ such that $X=V_{x_1}\cup...\cup V_{x_p}$.
$(f_n(x_i))$ is bounded so $\exists M_i>0$ such that $|f_n(x_i)|<M_i\ \forall n$.
$\forall t\in V_{x_i}\ \forall n\ |f_n(t)|\le |f_n(x_i)|+|f_n(x_i)-f_n(t)|<M_i+1\le M$
where $M=max(M_1,...,M_p)+1$. So $(f_n)$ is uniformly bounded and by Arzelà-Ascoli's theorem we can extract a convergent subsequence.