[Math] Why is an equicontinuous and pointwise bounded sequence of real-valued functions on a compact metric space uniformly bounded

Question

Why is an equicontinuous and pointwise bounded sequence of real-valued functions on a compact metric space uniformly bounded?
I couldn't get my head around this. Can anyone please explain?

Answer 1

Heuristically, if you have a family of equicontinuous functions, then you could essentially treat them as a "single continuous function" because for any $\epsilon>0$ you could always find a single $\delta>0$ that will work for your entire family. Moreover, we know a continuous function on a compact metric space attains its maximum, i.e. bounded. Hence together you can see why the family should be uniformly bounded.

Edit:

Here's the proof.

Let $(X, d)$ denote the compact metric space and $\mathcal{F}$ is our equicontinuous family of functions.

Fix $\epsilon>0$. Then for every fixed $x \in X$, by the equicontinuity property of $\mathcal{F}$, we can find a $\delta_x>0$ (i.e. $\delta$ depending on $x$) such that \begin{align} |f_n(x)-f_n(y)|<\epsilon \ \ \text{ whenever } \ \ |x-y|<\delta_x. \end{align} In particular, it follows \begin{align} |f_n(y)| < \epsilon +|f_n(x)| \leq \epsilon +M_x \ \ \text{ whenever } \ \ |x-y|<\delta_x. \end{align} Note that we have used pointwise boundedness of $\mathcal{F}$. Next, observe $X \subset \bigcup_{x \in X} B(x, \delta_x)$ is an open cover of our compact metric space. Hence by compactness there exists a finite subcover say $\bigcup^N_{i=1} B(x_i, \delta_i)$. Let $M = \sup_{1\leq i \leq N} M_{x_i}<\infty$, then we see that \begin{align} |f_n(y)| <\epsilon+M \end{align} for every $y \in X$ and $n\in \mathbb{N}$.