Ascoli-Arzela Theorem

The content of this blog is almost a copy of the Wikipedia page on this topic. The proof given in Wikipedia is for real intervals. Here, a general compact set is considered. The sketch of the proof for general compact sets is already given in the Wikipedia page. This is merely a completion for the sake of reference.

Let $X$ be a compact space and $C(X;\mathbb{R}^n)$ be the set of continuous functions from $X$ to $\mathbb{R}^n$. Let $F \subset C(X;\mathbb{R}^n)$ be such that it satisfies the following two properties

• Equicontinuity : For each $\epsilon > 0$, for all $x\in X$, there exists a neighbourhood $U_x$ such that $\|f(y) - f(x)\| < \epsilon$ for all $y \in U_x$ and all $f \in F$
• Pointwise Boundedness : For each $x\in X$, $\sup\{\|f(x)\| : f \in F\} < \infty$

Then $\bar{F}$ is compact. (Here $C(X;\mathbb{R}^n)$ is given the metric $\|f\| = \sup\{\|f(x)\| : x \in X\}$

Let $\epsilon_n = \frac{1}{2^n}$. For each $x$, choose $U_x$ (from the equicontinuity of $\mathcal{F}$) such that the oscillation of any function in $\mathcal{F}$ is less than $\epsilon_1$. As $U_x$ form an open cover of $X$ and since $X$ is compact there exists a finite subcover which covers $X$. Denote this cover by $U_{x_{11}},U_{x_{12}},\cdots,U_{x_{1N_1}}$. With this process, we obtain for every $n$ a finite open cover $U_{x_{n1}},U_{x_{n2}},\cdots,U_{x_{nN_n}}$.

Rename the points $x_{11},x_{12},\cdots,x_{1N_1},x_{21}\cdots,x_{2N_2}\cdots$ as $x_1,x_2,\cdots$

Let $\{f_n\}$ be a sequence in $\mathcal{F}$. We want to show that there exists a sub-sequence which converges uniformly. As the space $\bar{\mathcal{F}}$ is a metric space (with the metric mentioned above) we get that it is compact.

Step 1 : As $\{\|f_n(x_1)\|\}$ is bounded, there exists a sub-sequence $\{f_{n_1}\}$ such that $f_{n_1}(x_1)$ converges. Now, we can choose a sub-sequence $f_{n_2}$ of $f_{n_1}$ such that $f_{n_2}(x_2)$ converges. This process is repeated ad-infinitum. Now, the "diagonal" sequence whose $m$th term is $m$th term in the $m$th subsequence $f_{n_m}$ is chosen and denoted by $f_m$. By construction, $f_m(x_i)$ converges for all $i$. This seems to be the central idea. For the next steps fix $l$.

Step 2: From the above, we know that for each $x_k$, there exists an integer $N(\epsilon,x_k)$ such that $\|f_n(x_k) - f_m(x_k)| < \epsilon_l$ for all $n,m > N(\epsilon_l,x_k)$.

Step 3 : Clearly, for $K=\sum_{i=1}^l N_i$ each open set $U_{lj}$, $1 \leq j \leq N_l$, contains at-least one point $x_k$ with $1 \leq k \leq K$

Step 4: For any $x \in X$, there exist $j,k$ such that $x \in U_{lj}, x_k \in U_{lj}$ where $1 \leq j \leq N_l$. For this $k$, \[\|f_n(x) - f_m(x)\| \leq \|f_n(x)-f_n(x_k)\| + \|f_n(x_k) - f_m(x_k)\| + \|f_m(x_k) - f_m(x)\| < 3\epsilon_l\] for all $n,m > \max(N(\epsilon,x_1),N(\epsilon,x_2),\cdots,N(\epsilon,x_K))$. Therefore, the sequence is uniformly-cauchy and hence converges to a continuous function $g \in C(X;\mathbb{R}^n)$. It is obvious that $g \in \bar{\mathcal{F}}$