Two equivalent versions of Arzelà–Ascoli theorem Disclaimer: This thread is meant to record. See: SE blog: Answer own Question and MSE meta: Answer own Question. Anyway, it is written as problem. Have fun! :)

Let $(X, d)$ be a metric space, $C(X)$ the space of real-valued continuous functions on $Y$, and $\mathcal{F} \subset C(X)$.

*

*We say that $\mathcal{F}$ is pointwise bounded if, for each $x \in X$, the set $\{f(x) \mid f \in \mathcal{F}\}$ is bounded.


*We say that $\mathcal{F}$ is uniformly bounded if the set $\{f(x) \mid f \in \mathcal{F}, x\in X\}$ is bounded.


*We say that $\mathcal F$ is pointwise equi-continuous if, for each $x\in X$, for each $\varepsilon>0$, there exists $\delta>0$ such that for all $f\in \mathcal F$, we get $d(x,y) < \delta \implies |f(y)-f(x)| < \varepsilon$.


*We say that $\mathcal F$ is uniformly equi-continuous if, for each $\varepsilon>0$, there exists $\delta>0$ such that for all $f\in \mathcal F$, we get $d(x,y)< \delta \implies |f(y)-f(x)| < \varepsilon$.
Wikipedia's ver 1: Assume $X$ is compact. A subset $\mathcal F$ of $C(X)$ is relatively compact in $\| \cdot\|_\infty$ if and only if $\mathcal F$ is pointwise bounded and pointwise equi-continuous.
Carothers' ver 2: Assume $X$ is a compact. A subset $\mathcal{F}$ of $C(X)$ is compact in $\| \cdot\|_\infty$ if and only if  $\mathcal{F}$ is closed, uniformly bounded, and uniformly equi-continuous.

Then the two above versions are indeed equivalent.

 A: 

*

*Let's assume ver 1 and prove ver 2.



*

*$\Rightarrow$ Assume that $\mathcal F$ is compact.

Certainly, $\mathcal F$ is closed  and bounded in $\|\cdot\|_\infty$, so $\mathcal F$ is uniformly bounded. It remains to prove that $\mathcal F$ is uniformly equi-continuous. We fix $\varepsilon>0$.
Clearly, $\mathcal F$ is relatively compact. By ver 1, $\mathcal F$ is pointwise equi-continuous, i.e., for each $x\in X$, there exists $\delta_x>0$ such that for all $f\in \mathcal F$, we get $d(x,y) < \delta_x \implies |f(y)-f(x)| < \varepsilon/2$.
We have $\{B(x, \delta_x/2) \mid x\in X\}$ is an open cover of $X$. There is a finite subset $\{x_1, \ldots, x_n\} \subset X$ such that $\{B(x_1, \delta_{1}/2), \ldots, B(x_n, \delta_{n}/2)\}$ covers $X$. Let
$$
\delta := \min \{\delta_{1}/2, \ldots, \delta_{n}/2\}.
$$
Fix $y,z\in X$ such that $d(y,z) <\delta$. Assume $y \in B (x_i, \delta_{i}/2)$ for some $i \in \{1, \ldots,m\}$. Then
$$
d(z, x_i) \le d(z,y)+d(y, x_i) <\delta +\delta_i/2 \le \delta_i.
$$
This implies $y,z \in B(x_i, \delta_i)$. Hence $|f(y)-f(z)| \le |f(y)-f(x_i)| + |f(z)-f(x_i)| < \varepsilon/2 + \varepsilon/2 =\varepsilon$.

*

*$\Rightarrow$ Assume that $\mathcal{F}$ is closed, uniformly bounded, and uniformly equi-continuous.

By ver 1, $\mathcal F$ is relatively compact. Because $\mathcal F$ is closed, it is compact.



*Let's assume ver 2 and prove ver 1.



*

*$\Rightarrow$ Assume that $\mathcal F$ is relatively compact.

Then $\overline{\mathcal F}$ is compact. By ver 2, $\overline{\mathcal F}$ is uniformly bounded and uniformly equi-continuous. Then $\overline{\mathcal F}$ is pointwise bounded and pointwise equi-continuous. Obviously, $\mathcal F \subset \overline{\mathcal F}$, so $\mathcal F$ is pointwise bounded and pointwise equi-continuous.

*

*$\Rightarrow$ Assume that $\mathcal{F}$ is pointwise bounded and pointwise equi-continuous.

Just as we proved above, $\mathcal{F}$ is uniformly equi-continuous by compactness of $X$. This means for each $\varepsilon>0$, there exists $\delta>0$ such that for all $f\in \mathcal F$, we get $d(x,y)< \delta \implies |f(y)-f(x)| < \varepsilon$. Fix $g \in \overline{\mathcal F}$. Then there is a sequence $(f_n) \subset \mathcal F$ such that $f_n \to f$ in $\| \cdot \|_\infty$. Obviously, $d(x,y)< \delta \implies |f_n(y)-f_n(x)| < \varepsilon$. Convergence in $\| \cdot \|_\infty$ implies pointwise convergence, so $d(x,y)< \delta \implies |g(y)-g(x)| < \varepsilon$. It follows that $\overline{\mathcal{F}}$ is uniformly equi-continuous.
Next we prove that $\overline{\mathcal{F}}$ is uniformly bounded. Fix $\varepsilon>0$. By uniform equi-continuity of $\overline{\mathcal{F}}$, there exists $\delta>0$ such that for all $f\in \overline{\mathcal{F}}$, we get $d(x,y)< \delta \implies |f(y)-f(x)| < \varepsilon$. Clearly, $B\{B(x, \delta) \mid x\in X\}$ is an open cover of $X$, so there is a finite subset $\{x_1, \ldots, x_n\} \subset X$ such that $\{B(x_1, \delta), \ldots, B(x_n, \delta)\}$ covers $X$. For each $y \in X$, there is $x_i$ such that $y \in B(x_i, \delta)$, so $|f(x)| \le |f(x_i) \pm\varepsilon|$ for all $f \in \overline{\mathcal{F}}$. This means
$$
|f(x)| \le \max_{i = 1, \ldots,m} |f(x_i) \pm\varepsilon| \quad \forall f \in \overline{\mathcal{F}}, x\in X.
$$
So
$$
\sup\{|f(x)| \mid f\in \overline{\mathcal{F}}, x\in X\} \le \max_{i = 1, \ldots,m} \sup_{f\in \overline{\mathcal{F}}}|f(x_i) \pm\varepsilon| < +\infty.
$$
It follows that $\overline{\mathcal{F}}$ is uniformly bounded. Clearly, $\overline{\mathcal{F}}$ is closed. By ver 2, $\overline{\mathcal{F}}$ is compact. Hence ${\mathcal{F}}$ is relatively compact.
