Suppose $\{f_n, n = 1, 2, \dots\}$ is an equicontinuous family of real-valued functions on a compact metric space $(X, d)$. What are some appropriate conditions ensuring that $\cup_{n = 1}^\infty f_n(X)$ is compact? In particular, is it enough that $\sup_n \sup_x \vert f_n(x)\vert \leq M$ for some constant $M< \infty$?
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2 Answers
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Hint: Let $X=[0,1]$ and let $f_n(x) = 1/n + (1-1/n)x.$ What is $\cup\, f_n(X)?$
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No (as zhw.'s hint should allude), to have that the set $\underset{n \in \mathbb{N}}{\cup}\, f_n(X) \subseteq \mathbb{R}$ is compact it is insufficient to merely assume that $\mathscr{F}=\{\, f_n:X \to \mathbb{R} : n \in \mathbb{N}\}$ is a uniformly bounded equicontinuous family of functions on a compact metric space $(X,d)$.
If $(X,d)$ is a compact metric space and $\mathscr{F}=\{\, f_n : n \in \mathbb{N}\}$ is an equicontinuous family of real-valued functions on $X$, then $\underset{n \in \mathbb{N}}{\bigcup} f_n(X)$ is a compact subset of $\mathbb{R}$ if and only if $\mathscr{F}$ is closed and bounded in $C(X)$ (the set of continuous real-valued functions on $X$) with the supremum metric $\rho(\, f, g)= \underset{x \in X}{\sup} |\, f(x)-g(x)|$.
Sticking with $X=[0,1]$, another example to consider is the sequence of real-valued functions on $X$ $\{g_n\}_{n=1}^\infty$ such that $g_n(x)=x+\frac{1}{n}$ ($n=1,2,\ldots$).
