Arzelà–Ascoli Theorem

Relatively compact implies totally bounded: for any $\varepsilon > 0$ there is a finite $\varepsilon/3$-net $\{g_1, \dots, g_m\}$ of the closure $\overline{F}$.

That is, every $f \in F$ has some $g_j$ with $\lVert f - g_j \rVert_\infty < \varepsilon/3$.

The $g_j$ need not lie in $F$ — continuity is all we need.

Each $g_j$ is continuous on the compact metric space $K$, hence uniformly continuous (Heine–Cantor): there is $\delta_j > 0$ with $\lvert g_j(x) - g_j(y) \rvert < \varepsilon/3$ whenever $d(x, y) < \delta_j$. Set $\delta = \min_j \delta_j > 0$ — this $\delta$ does not depend on $f$.

Now take $f \in F$ and $x, y$ with $d(x, y) < \delta$.

Picking $g_j$ with $\lVert f - g_j \rVert_\infty < \varepsilon/3$, the triangle inequality gives $\lvert f(x) - f(y) \rvert \le \lvert f(x) - g_j(x) \rvert + \lvert g_j(x) - g_j(y) \rvert + \lvert g_j(y) - f(y) \rvert < \varepsilon/3 + \varepsilon/3 + \varepsilon/3 = \varepsilon$.

Since $\delta$ works for every $f \in F$, the family $F$ is equicontinuous.

In addition, relative compactness gives uniform boundedness (a bounded set has bounded closure): $M = \sup_{f \in F} \lVert f \rVert_\infty < \infty$. Conversely, on compact $K$, pointwise bounded + equicontinuous implies uniformly bounded: cover $K$ by a finite $\delta$-net $\{y_1, \dots, y_r\}$ for the equicontinuity modulus at $\varepsilon = 1$; each $x$ is $\delta$-close to some $y_i$, so $\lvert f(x) \rvert \le \lvert f(y_i) \rvert + 1 \le \max_i \sup_f \lvert f(y_i) \rvert + 1$.

A compact metric space is separable (take the union of finite $1/r$-nets).

Now look at $(f_n)$.

For each $k$, the scalar sequence $(f_n(x_k))_n$ is trapped in $[-M, M]$, so by Bolzano–Weierstrass it has a convergent subsequence.

Build nested index sequences: choose $n^{(1)}_1 < n^{(1)}_2 < \cdots$ so that $f_{n^{(1)}_j}(x_1)$ converges, and inductively choose $n^{(k)}$ as a subsequence of $n^{(k-1)}$ with $f_{n^{(k)}_j}(x_k)$ convergent. Fixing $k$, every later stage refines $n^{(k)}$, so convergence at $x_1, \dots, x_k$ is preserved.

Set the diagonal $m_j = n^{(j)}_j$. First, $(m_j)$ is strictly increasing: writing $n^{(j+1)}_i = n^{(j)}_{\sigma^{(j)}(i)}$ with the strictly increasing refinement map $\sigma^{(j)}: \mathbb{N} \to \mathbb{N}$, we get $m_{j+1} = n^{(j+1)}_{j+1} = n^{(j)}_{\sigma^{(j)}(j+1)} > n^{(j)}_j = m_j$ (since $\sigma^{(j)}(j+1) \ge j+1 > j$ — a strictly increasing map satisfies $\sigma(i) \ge i$, with equality for the identity refinement).

Fix $k$.

For each $j \ge k$, $m_j = n^{(j)}_j$ is a term of $(n^{(k)}_i)$, so write $m_j = n^{(k)}_{i_j}$.

The nested composition $\psi_k(j) = \sigma^{(k)} \circ \cdots \circ \sigma^{(j-1)}(j)$ gives $i_j = \psi_k(j)$, and strictly increasing composition preserves strict increase, yielding the index inequality $i_{j+1} = \psi_k(j+1) > \psi_k(j) = i_j$.

Hence the tail $(m_j)_{j \ge k} = (n^{(k)}_{i_j})_{j \ge k}$ is a subsequence of $(n^{(k)}_i)$, and $f_{m_j}(x_k)$ is a subsequence of the convergent sequence $(f_{n^{(k)}_i}(x_k))_i$, so it converges.

Since $k$ was arbitrary, $(f_{m_j})$ converges at every $x_k$ — pointwise convergence on the dense set.

Note: only pointwise boundedness was used here. Upgrading this pointwise convergence to uniform convergence on all of $K$ is the job of equicontinuity, carried out by the $\varepsilon/3$ argument in the next problem.

Given $\varepsilon > 0$, equicontinuity yields $\delta > 0$: $d(x, y) < \delta$ implies $\lvert f_n(x) - f_n(y) \rvert < \varepsilon/3$ for every $n$. Since $K$ is compact, finitely many points $x_{k_1}, \dots, x_{k_r}$ from the dense set $\{x_k\}$ form a $\delta$-net of $K$.

At the finitely many net points, the pointwise convergence of $(f_n)$ itself becomes "uniform": each $(f_n(x_{k_i}))$ converges, hence is Cauchy, so choose $N$ so large that $p, q \ge N$ gives $\lvert f_p(x_{k_i}) - f_q(x_{k_i}) \rvert < \varepsilon/3$ for every $i$ (a finite intersection of tails).

For arbitrary $x \in K$, pick $x_{k_i}$ with $d(x, x_{k_i}) < \delta$.

By the triangle inequality, for $p, q \ge N$, $\lvert f_p(x) - f_q(x) \rvert \le \lvert f_p(x) - f_p(x_{k_i}) \rvert + \lvert f_p(x_{k_i}) - f_q(x_{k_i}) \rvert + \lvert f_q(x_{k_i}) - f_q(x) \rvert < \varepsilon/3 + \varepsilon/3 + \varepsilon/3 = \varepsilon$.

Hence $(f_n)$ itself is uniformly Cauchy, so $(f_n)$ converges uniformly by completeness of $C(K)$.

This completes ($\Longleftarrow$): pointwise bounded + equicontinuous implies uniformly bounded (as noted in the first problem).

Also, applying the diagonal extraction of the previous problem to any sequence gives a subsequence converging on the dense set, and applying steps 1–3 to that subsequence shows it converges uniformly.

Hence every sequence has a uniformly convergent subsequence, so by sequential compactness in metric spaces the set is relatively compact.

Corollary: $(x^n)$ has no uniformly convergent subsequence on $[0,1]$.

Its pointwise limit is discontinuous — 0 on $[0,1)$, 1 at $x = 1$. If some subsequence converged uniformly, its limit would be continuous (a uniform limit of continuous functions) and would have to equal the pointwise limit — contradiction.

So $\{x^n\}$ is not equicontinuous at $x = 1$, and not totally bounded either: the subsequence $x^{4^k}$ is $7/16$-separated — for $a = 4^k < 4^\ell = b$, at the witness point $x = 2^{-1/a}$ we have $x^a = 1/2$ and $x^b = 2^{-b/a} \le 2^{-4} = 1/16$, so $\lVert x^a - x^b \rVert_\infty \ge 1/2 - 1/16 = 7/16 = 0.4375 \ge 0.4$.

(Not every distinct pair is this far apart: $\lVert x - x^2 \rVert_\infty = 1/4$.)