TITLE: Equicontinuity of the closure of an equicontinuous set
QUESTION [5 upvotes]: If E Is equicontinuous in C(X,R), I need to show that $\bar{E}$ is equicontinuous as well.
Now $\forall f \in \bar{E},  \exists f_n\in E$ s.t. $f_n\rightarrow f$, thus $\forall \epsilon > 0, \exists n_o \text{  s.t. if  } n\geq n_0, |f_n(x)-f(x)|< \epsilon $.
$$ |f(x)-f(y)|\leq|f(x)-f_{n0}(x)|+|f_{n0}(x)-f_{n0}(y)|+|f_{n0}(y)-f(y)|$$.. whats next?

REPLY [2 votes]: Let $x \in \mathbb{R}$ be given. Fix $\epsilon > 0$. Choose $\delta > 0$ such that $|g(x)-g(y)|<\epsilon$ for all $g \in E$ and $y \in \mathbb{R}$ with $|y-x|<\delta$. Let $f \in \overline{E}$ be given. Choose a sequence $(f_n) \to f$ in $E$. Then for any $y \in \mathbb{R}$ with $|y-x|<\delta$ we have
$$
|f(x)-f(y)| = \lim_{n \to \infty} |f_n(x)-f_n(y)|  = \limsup_{n \to \infty} |f_n(x)-f_n(y)| \leq \limsup_{n \to \infty} \epsilon= \epsilon.
$$
It follows that $\overline{E}$ is equicontinuous.