Equicontinuity and uniform convergence Let $\{f_n\}$ be a sequence of continuous functions $f_n\colon X \to R$ where $X \subset R$. Prove that if every point $x_0  \in X$ has an interval $
I_{x_0}=B\left({x_0 ,\varepsilon _{x_0}}\right)\cap X$
for some $\varepsilon _{x_0 } >0$, such that $\{f_n\}$ converges uniformly in $I_{x_0 } \cap X$, then $\{f_n\}$ it´s also equicontinuous.
Is this result true for a non-countable set of continuous functions?
Is the reciprocal true?
Is this result true for other kind of topological, or metric spaces?
Please help me with this, to start, it's the only that i could not do.
 A: We have the generalization:

Let $(Y,d)$ a metric space and $X$ a non-empsty subset of $Y$ and $\{f_n\}$ a sequence of continuous functions from $X$ to $\mathbb R$. If for all $x_0\in X$, we can find a ball $B:=B(x_0,\varepsilon_{x_0})$ such that the sequence $\{f_n\}$ converges uniformly to $f$ on $B\cap X$, then the family $\{f_n\}$ is equicontinuous.  

We fix $x_0\in X$ and $\varepsilon>0$. Let $\delta_1$ such that the sequence $\{f_n\}$ converges uniformly to $f$ on $X\cap B(x_0,\delta_1)$. Let $N$ an integer such that $\sup_{x\in B(x_0,\delta_1)}|f(x)-f_n(x)|\leq\frac{\varepsilon}3$. For each $i\in\{1,\ldots,N\}$ we can find a $\eta_i$ such that if $d(x,x_0)\leq \eta_i$ then $|f_i(x)-f_i(x_0)|\leq \frac{\varepsilon}3$. Put $\delta_2:=\min\{\eta_i,1\leq i\leq N\}$. Finally, since $f$ is continuous at $x_0$ as a uniform limit of such functions, let $\delta_3$ such that if $d(x,x_0)\leq \delta_3$ then $|f(x)-f(x_0)|\leq\frac{\varepsilon}3$. Now put $\delta:=\min(\delta_1,\delta_2,\delta_3)$. Let $x\in B(x_0,\delta)$. Let $n\in\mathbb N$. If $n\leq N$ we are done since $\eta_n\leq\delta$, and if $n>N$ we have 
\begin{align}|f_n(x)-f_n(x_0)|&\leq |f_n(x)-f(x)|+|f(x)-f(x_0)|+|f(x_0)-f_n(x_0)|\\&\leq 2\sup_{y\in B(x_0,\delta)}|f_n(y)-f(y)|+|f(x)-f(x_0)|\\
&\leq 3\frac{\varepsilon}3,
\end{align}
hence the family $\{f\}\cup\{f_n,n\geq 1\}$ is equicontinuous at each point of $X$.
