A set ℱ of functions on K to ℝ^q is uniformly equicontinuous on K in case for each ε > 0 there is δ(ε) > 0 such that ‖f(x) − f(y)‖ < ε whenever x, y ∈ K, ‖x − y‖ < δ(ε) and f ∈ ℱ.
一個會用到的預備事實：ℝᵖ 的每個子集 K 都含有一個可數的稠密子集C——意思是任給 y ∈ K 與 ε > 0，C 中都有一點離 y 不到 ε。作法：對每一組「有理座標的點 r 與自然數 n」，若以 r 為心、1/n 為半徑的球碰得到 K，就從交集裡挑一個點收進 C。這樣的組合只有可數多個（3.4），所以 C 可數；而任給 y 與 ε，取 n 使 2/n < ε，再由稠密性（6.10 逐座標用一次）取 r 離 y 不到 1/n，該球必定碰得到 K（它含 y），於是 C 裡那一點離 y 不到 2/n。
26.7 ARZELÀ-ASCOLI THEOREM
Take K compact in ℝᵖ and let ℱ consist of functions that are continuous on K and take values in ℝ^q. The following are equivalent. (a) ℱ is bounded and uniformly equicontinuous on K. (b) Every sequence from ℱ has a subsequence which converges uniformly on K.
Proof. If ℱ is not bounded, choose fₙ ∈ ℱ with ‖fₙ‖_K ≥ n. Every subsequence is again unbounded, and a uniformly convergent sequence is bounded (the argument of 24.2), so no subsequence converges uniformly.
If ℱ is not uniformly equicontinuous, there is ε₀ > 0 such that for each n the choice δ = 1/n fails: there are fₙ ∈ ℱ and xₙ, yₙ ∈ K with ‖xₙ − yₙ‖ < 1/n and ‖fₙ(xₙ) − fₙ(yₙ)‖ ≥ ε₀.
Suppose some subsequence (f_{n(k)}) converged uniformly to g. Then g is continuous by 24.1 and uniformly continuous by 23.3, so there is δ with ‖g(x) − g(y)‖ < ε₀/3 whenever ‖x − y‖ < δ. Taking k large enough that ‖f_{n(k)} − g‖_K < ε₀/3 and 1/n(k) < δ gives ‖f_{n(k)}(x_{n(k)}) − f_{n(k)}(y_{n(k)})‖ < ε₀, contradicting the choice above.
這一步是矛盾的核心。三段拆解與 24.1 的證明逐句同構：從 f_{n(k)} 跳到 g、在 g 上走過去、再跳回來，頭尾兩段由均勻收斂統一壓住，中間那段由 g 的均勻連續負責。差別在於這次中間那段需要的是均勻連續而非逐點連續，因為 x_{n(k)} 與 y_{n(k)} 的位置隨 k 移動，沒有固定的基準點。
證明計畫 · 由所求想起 所求是一條在整個K 上均勻收斂的子列，而 K 有無限多個點，一次處理不完。先只要求在可數多個點上收斂：在 x₁ 上用 Bolzano-Weierstrass 挑一次子列，再在 x₂ 上從剛才的子列裡再挑一次⋯每一輪保住前面所有輪的成果。可是這樣挑不完，所以取對角線——第 n 條子列的第 n 項。它從第 n 項起是第 n 條的子列，於是在每個 x_k 上都收斂。 最後把「可數多個點上收斂」升級成「整個 K 上均勻收斂」：等度連續讓每個 x ∈ K 借用附近某個 x_k 的資訊，而 compact 保證借用的對象只需有限多個。
Proof. The sequence (fₙ(x₁)) is bounded in ℝ^q, so by 16.4 it has a convergent subsequence; write the corresponding functions as (f¹ₙ). From (f¹ₙ(x₂)) extract a convergent subsequence (f²ₙ), and so on. Put gₙ = fⁿₙ. For each k, the terms of (gₙ) from the k-th onwards form a subsequence of (f^kₙ), so (gₙ(x_k)) converges.
這一步是對角線手法，值得慢讀。每一輪的子列都是前一輪的子列，所以越後面的子列越乖——它在前面所有的取樣點上都收斂。對角線那一條之所以全部繼承，是因為它從第 k 項起整條住在第 k 條裡面，而 15.3 說極限只認尾巴，前面幾項不影響收斂。這與 §3-3 證明實數不可數時用的是同一個對角線構造，只是那裡用來製造差異，這裡用來收攏。
Let ε > 0 and take δ(ε) from equicontinuity. The balls of radius δ(ε) about points of C cover K, so by compactness finitely many suffice; let their centres be y₁, ⋯, y_k ∈ C. Since each (gₙ(y_i)) converges, there is M with ‖gₙ(y_i) − g_m(y_i)‖ < ε for all n, m ≥ M and all i.
這一步把可數降成有限。稠密性保證那些球蓋得住 K，compact 把它們裁成有限多個——而「有限」正是取得到共同編號 M 的理由：k 個收斂的數列各給一個編號，取最大值即可；無限多個就取不到。這與 23.3 的覆蓋路線把無限多個 δ 裁成有限多個是同一個動作。
Given x ∈ K, choose y_i with ‖x − y_i‖ < δ(ε). For n, m ≥ M, ‖gₙ(x) − g_m(x)‖ ≤ ‖gₙ(x) − gₙ(y_i)‖ + ‖gₙ(y_i) − g_m(y_i)‖ + ‖g_m(y_i) − g_m(x)‖ < 3ε. Hence ‖gₙ − g_m‖_K ≤ 3ε for n, m ≥ M, and 17.11 gives uniform convergence of (gₙ) on K.