Equicontinuity on a Compact Metric Set implies Uniform Equicontinuity

Equicontinuity: A collection $\mathcal{F}$ of real-valued functions on a metric space $X$ is equicontinuous at the point $x\in X$ provided for every $\epsilon>0,\ \exists$ $\delta>0$ such that $\forall\ f\in\mathcal{F}$ and $y\in X$ $$d(x,y)<\delta \implies |f(x)-f(y)|<\epsilon$$ The collection $\mathcal{F}$ is said to be equicontinuous on $X$ provided it is equicontinuous at every point in $X$

Uniform Equicontinuity: A equicontinuous collection $\mathcal{F}$ of real-valued functions on a metric space $X$ is uniformly equicontinuous if for every $\epsilon>0,\ \exists$ $\delta>0$ such that $\forall\ f\in\mathcal{F}$ and $x,y\in X$ $$d(x,y)<\delta \implies |f(x)-f(y)|<\epsilon$$

Just like we know that a continuous function on a compact set is uniformly continuous here we are showing that for a collection of functions the same $\epsilon-\delta$ pair works.

Now coming to the proof, since $\mathcal{F}$ is equicontinuous for every $\epsilon>0,\ \exists$ $\delta>0$ such that $\forall\ f\in\mathcal{F}$ and $y\in X$ $$d(x,y)<\delta \implies |f(x)-f(y)|<\epsilon$$ Hence if we fix $\epsilon$ for every $x\in X$ we get a $\delta_x$ such that for all $y\in B_{\delta_x}(x)$ $|f(x)-f(y)|<\epsilon$. Now these $\{B_{\frac{\delta_x}{2}}(x)\}$ covers $X$. Since $X$ is compact it has a finite subcover $$B_{\frac{\delta_1}{2}}(x_1),B_{\frac{\delta_2}{2}}(x_2),\dots,B_{\frac{\delta_n}{2}}(x_n)$$ Now take $$\delta=\frac{1}{2}\min\left\{\delta_i\mid i\in\{1,2,\dots,n\}\right\}$$ Now if $d(x,y)<\delta$ suppose $x\in B_{\frac{\delta_i}{2}}(x_i)$ then $y\in B_{\delta_i}(x_i)$ because  $$d(x_i,y)\leq d(x_i,x)+d(x,y)<\frac{\delta_i}{2}+\delta\leq \frac{\delta_i}{2}+\frac{\delta_i}{2}=\delta_i$$ Hence $$d(x,y)<\delta \implies |f(x)-f(y)|<\epsilon$$ Since this is true for any $f\in\mathcal{F}$, for every $\epsilon>0,\ \exists$ $\delta>0$ such that $\forall\ f\in\mathcal{F}$ and $x,y\in X$ $$d(x,y)<\delta \implies |f(x)-f(y)|<\epsilon$$ Therefore $\mathcal{F}$ is uniformly equicontinuous. $\blacksquare$

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