C[a,b] 作为完备度量空间
把 [a,b] 上所有连续函数收进一个集合，即连续函数空间 C[a,b]。赋予它来自上确界范数的距离 d(f,g) = ||f - g||。则该度量下的收敛恰是[[uniform-convergence|一致收敛]]——这正是我们一路构建的抽象。前面两条定理合在一起说明该空间性质良好。
- 一致 Cauchy 判据（第 3 篇）：C[a,b] 中一致 Cauchy 的序列有一致极限。
- 一致极限定理（第 4 篇）：该极限本身连续，故仍落在 C[a,b] 内。
- 结论：C[a,b] 是完备度量空间——Cauchy 序列收敛，且极限仍在空间内。
哪些族是紧的？
在有限维空间里，闭且有界即紧——每个序列有收敛子序列（Bolzano–Weierstrass）。但在无限维的 C[a,b]，有界并不够。我们需要两个条件。其一，一致有界：存在单一常数 M，使族中每个 f 与每个 x 满足 |f(x)| <= M。其二，等度连续：一个 delta 同时服务整个族。
Equicontinuity (the key new idea).
A family F of functions on [a,b] is EQUICONTINUOUS if:
for every epsilon > 0 there is ONE delta > 0 such that
for ALL f in F and all x, y with |x - y| < delta,
|f(x) - f(y)| < epsilon.
(Same delta works for every member f at once.)
Why mere boundedness fails: f_n(x) = sin(n x) on [0, 2 pi].
Uniformly bounded: |sin(n x)| <= 1.
But the slopes n -> infinity destroy equicontinuity, and NO
subsequence converges uniformly (the sines never settle).
=> a bounded sequence in C[0,2pi] with no convergent subsequence.
A clean sufficient condition for equicontinuity: a common
Lipschitz bound |f(x) - f(y)| <= L|x - y| for all f in F
(take delta = epsilon / L).Arzelà–Ascoli 定理
把它们合起来，得到分析中最有用的紧性结果之一。Arzelà–Ascoli 定理：C[a,b] 中一致有界且等度连续的序列 (f_n) 有一致收敛的子序列。等价地，C[a,b] 的子集紧当且仅当它闭、有界且等度连续。这是 Bolzano–Weierstrass 在无限维下的正确替身。
Sketch of the proof (diagonal argument).
Let (f_n) be uniformly bounded and equicontinuous on [a,b].
(1) Pick a countable DENSE set {q_1, q_2, ...} in [a,b] (rationals).
(2) At q_1 the reals f_n(q_1) are bounded; by Bolzano-Weierstrass
take a subsequence converging at q_1. From IT take a further
subsequence converging at q_2. Repeat.
(3) DIAGONAL subsequence g_k = (k-th term of k-th subsequence)
converges at EVERY q_j.
(4) EQUICONTINUITY upgrades pointwise-on-a-dense-set to
uniform-on-all-of-[a,b]: given epsilon, pick delta from
equicontinuity, cover [a,b] by finitely many delta-balls
centered at q's; closeness at those q's + the common delta
force ||g_k - g_l|| < epsilon for large k, l.
(5) g_k is uniformly Cauchy => converges uniformly (C[a,b] complete).