C[a,b] 作為完備度量空間
把 [a,b] 上所有連續函數收進一個集合，即連續函數空間 C[a,b]。賦予它來自上確界範數的距離 d(f,g) = ||f - g||。則該度量下的收斂恰是[[uniform-convergence|一致收斂]]——這正是我們一路建構的抽象。前面兩條定理合在一起說明該空間性質良好。
- 一致 Cauchy 判據（第 3 篇）：C[a,b] 中一致 Cauchy 的序列有一致極限。
- 一致極限定理（第 4 篇）：該極限本身連續，故仍落在 C[a,b] 內。
- 結論：C[a,b] 是完備度量空間——Cauchy 序列收斂，且極限仍在空間內。
哪些族是緊的？
在有限維空間裡，閉且有界即緊——每個序列有收斂子序列（Bolzano–Weierstrass）。但在無限維的 C[a,b]，有界並不夠。我們需要兩個條件。其一，一致有界：存在單一常數 M，使族中每個 f 與每個 x 滿足 |f(x)| <= M。其二，等度連續：一個 delta 同時服務整個族。
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
把它們合起來，得到分析中最有用的緊性結果之一。Arzelà–Ascoli 定理：C[a,b] 中一致有界且等度連續的序列 (f_n) 有一致收斂的子序列。等價地，C[a,b] 的子集緊當且僅當它閉、有界且等度連續。這是 Bolzano–Weierstrass 在無限維下的正確替身。
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).