The Arzelà–Ascoli Theorem Part 1

# The Arzelà–Ascoli Theorem Part 1

We are now going to look at a very famous theorem known as The Arzelà–Ascoli Theorem. We will prove the theorem in three parts on The Arzelà–Ascoli Theorem Part 1, The Arzelà–Ascoli Theorem Part 2, and The Arzelà–Ascoli Theorem Part 3 pages.

First we review some important definitions.

• On the Boundedness of a Subset of C(X) page we saw that a collection of functions $\Gamma \subseteq C(X)$ is said to be bounded if there exists an $M \in \mathbb{R}$, $M > 0$ such that for all $f \in \Gamma$ and for all $x \in X$ we have that:
(1)
\begin{align} \quad \mid f(x) \mid \leq M \end{align}
• On the Equicontinuity of a Subset of C(X) page we saw that a collection of functions $\Gamma \subseteq C(X)$ is said to be equicontinuous on $X$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that for all $f \in \Gamma$ and for all $x, y \in X$, if $d(x, y) < \delta$ then:
(2)
\begin{align} \quad \mid f(x) - f(y) \mid < \epsilon \end{align}
 Theorem 1 (The Arzelà–Ascoli Theorem): Let $(X, d)$ be a compact metric space, and for each $n \in \mathbb{N}$ let $f_n : X \to \mathbb{R}$ be a continuous function. Then if the sequence of functions $(f_n(x))_{n=1}^{\infty}$ in $C(X)$ is bounded and equicontinuous then there exists a uniformly convergent subsequence.
• Proof: We begin the proof by showing that there exists a subset $S \subseteq X$ that is countable and dense.
• Let $n \in \mathbb{N}$ and for each $x \in X$, consider the open ball centered at $x$ with radius $\frac{1}{n}$:
(3)
\begin{align} \quad B \left ( x, \frac{1}{n} \right ) = \left \{ y \in X : d(x, y) < \frac{1}{n} \right \} \end{align}
• Clearly we have that the union of these balls is an open covering of $X$ since each $x \in X$ is the center of one of these open balls, so, $\displaystyle{X \subseteq \bigcup_{x \in X} B \left ( x, \frac{1}{n} \right )}$.
• Since $(X, d)$ is a compact metric space there exists a finite subcollection of $\left \{ B \left (x, \frac{1}{n} \right ) : x \in X \right \}$ that also covers $X$. In other words, there exists $\left \{ B\left ( x_1, \frac{1}{n} \right ), B \left ( x_2, \frac{1}{n} \right ), ..., B\left (x_k , \frac{1}{n} \right ) : x_1, x_2, ..., x_k \in X \right \} \subseteq \left \{ B \left (x, \frac{1}{n} \right ) : x \in X \right \}$ such that:
(4)
\begin{align} \quad X \subseteq \bigcup_{i=1}^{k} B \left ( x_k, \frac{1}{n} \right ) \end{align}
• Let $S_n = \{ x_1, x_2, ..., x_k \}$. Then for each $n \in \mathbb{N}$, $S_n$ is countable, and furthermore, if we let $S = \bigcup_{n=1}^{\infty} S_n$ then $S$ is also countable (as it is a countable union of (finitely) countable sets). We claim that $S$ is also dense.
• Let $x \in X$ and $r > 0$ and consider the intersection of $S$ with the open ball $B(x, r)$:
(5)
\begin{align} \quad S \cap B(x, r) = \left ( \bigcup_{n=1}^{\infty} S_n \right ) \cap \left \{ y \in X : d(x, y) < \frac{1}{r} \right \} \end{align}
• For each $r > 0$, there exists an $n_r \in \mathbb{N}$ such that $\frac{1}{n_r} < \frac{1}{r}$. For each point $y \in B(x, r)$ we have that $d(x, y) < \frac{1}{r}$, and since every point $y \in X$ is of a distance of less than $\frac{1}{n_r}$ of a point in $S_{n_r} \subset S$. So $S \cap B(x, r) \neq \emptyset$, so $S$ is a countable dense subset of $X$.