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.

- Recall from The Set of Real-Valued Continuous Functions on a Compact Metric Space X, C(X) that if $(X, d)$ is a compact metric space then the set of real-valued functions $f : X \to \mathbb{R}$ that are continuous is denoted $C(X)$.

- 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:

\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:

\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 $\frac{1}{n}$:

\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:

\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 $B(x, r)$:

\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 $X$.