Equicontinuity and modulus of continuity A modulus of continuity for a function $f$ is a continuous increasing function $\alpha$ such that $\alpha(0) = 0$ and $|f(x) - f(y)| < \alpha(|x-y|)$ for all $x$ and $y$. I am trying to prove that an equicontinuous family $\mathcal F$ of functions has a common modulus of continuity. This seems intuitively obvious, but I am having difficulty proving continuity. So far, I have defined 
$\alpha(\delta) = \sup\{|f(x) - f(y)| :  d(x,y) \leq \delta, f\in \mathcal F\} $. 
I want to show that this function is continuous in $\delta$. Any suggestions?
 A: Building on Beni's answer, suppose that this is not right-continuous, i.e. we have some $\delta$ such that $$\sup\{|f(x)-f(y)| : d(x,y)\leq \delta+\epsilon, f\in \mathcal{F}\}-\sup\{|f(x)-f(y)| : d(x,y)\leq \delta, f\in \mathcal{F}\}>z$$
for some fixed $z>0$ and for arbitrarily small $\epsilon>0$. Then for any fixed $\epsilon$ we have some $f\in \mathcal{F},x',y'\in \mathbb{R}$ such that $d(x',y')\leq \delta+\epsilon$ and
$$|f_\epsilon(x')-f_\epsilon(y')|-\sup\{|f_\epsilon(x)-f_\epsilon(y)| : d(x,y)\leq \delta\}>z$$
meaning that if we let $x'',y''\in\mathbb{R}$ be such that $d(x'',y'')\leq \delta$, $d(x',x'')\leq \epsilon$,$d(y',y'')\leq \epsilon$ (which can always be done) we get $$|f_\epsilon(x'')-f_\epsilon(x')| + |f_\epsilon(y'')-f_\epsilon(y')| \geq |f_\epsilon(x')-f_\epsilon(y')| - |f_\epsilon(x'')-f_\epsilon(y'')|>z$$
so one of the summands on the left must be at least $z/2$, meaning that for sufficiently small $\epsilon>0$ (as in smaller than $z/2$) and any $\delta>0$ we have some function $f_\delta\in\mathcal{F}$ for which
$$d(x,y)<\delta\not\Rightarrow d(f_\delta(x),f_\delta(y))<\epsilon$$
and so $\mathcal{F}$ is not equicontinuous.
A: Suppose $\delta_n$ is increasing and $\delta_n \to \delta$. Denote $\alpha(\delta)=\sup \{ |f(x)-f(y)| : d(x,y)<\delta, f \in \mathcal F\}$ (with strict inequality). Then pick $x,y$ with $d(x,y)<\delta$. This means that there exists $n$ such that $d(x,y)<\delta_n$ and therefore $|f(x)-f(y)| \leq \alpha(\delta_n)\leq \lim_{n \to \infty}\alpha(\delta_n)\leq \alpha(\delta)$. Take supremum in the left side and we get $\alpha(\delta)=\lim_{n \to \infty}\alpha(\delta_n)$. This proves that $\alpha$ is left continuous. 
I'm not sure about the right-continuity right now. Maybe it works on the same way, but I didn't manage to get it now.
