For each m ∈ ℕ let Yₘ = (xₘₙ) be a sequence in ℝᵖ converging to yₘ. The collection {Yₘ : m ∈ ℕ} is uniformly convergent in case, for each ε > 0, there is a natural number N(ε) — depending on ε but not on m — such that n ≥ N(ε) implies ‖xₘₙ − yₘ‖ < ε for all m ∈ ℕ.
If the double limit of the double sequence X = (xₘₙ) exists and if, for each m ∈ ℕ, the sequence Yₘ = (xₘₙ : n ∈ ℕ) is convergent, then this collection is uniformly convergent.
證明計畫 · 由所求想起 把列分成兩批。編號大的那批（列號本身已經夠大）靠雙重極限一次全部搞定：這些列上的每一項與該列的極限都離 x 不遠，繞經 x 即可。編號小的那批只有有限多列，各自的收斂各給一個起算點，取最大的那個。兩批的起算點再取一次最大。
Proof. Let ε > 0 and take N = N(ε) with ‖xₘₙ − x‖ < ε for all m, n ≥ N. As in the proof of 19.6, letting n → ∞ and applying 15.8 gives ‖yₘ − x‖ ≤ ε for every m ≥ N. Hence for m, n ≥ N, ‖xₘₙ − yₘ‖ ≤ ‖xₘₙ − x‖ + ‖x − yₘ‖ < 2ε.
這一步的所求：一口氣處理掉列號 ≥ N 的所有列。做法是繞經雙重極限 x：這些列上的項離 x 不遠（雙重極限直接給的），而它們的極限 yₘ 也離 x 不遠（19.6 證明裡那一步過極限的結論），兩段相加即得。關鍵在於這個估計裡的起算點 N 完全不看 m——它是從雙重極限來的，而雙重極限的條件對兩個下標一視同仁。
There remain the finitely many indices m = 1, 2, …, N − 1. For each such m the sequence Yₘ converges to yₘ, so there is Kₘ ∈ ℕ with ‖xₘₙ − yₘ‖ < 2ε for all n ≥ Kₘ. Put K = sup {K₁, …, K_{N−1}}, taking K = 1 when N = 1.
Let M = sup {N, K}. If n ≥ M, then ‖xₘₙ − yₘ‖ < 2ε holds for every m ∈ ℕ — for m ≥ N by the first estimate, for m < N by the second. Since ε > 0 was arbitrary, the collection is uniformly convergent.
Suppose that the single limits yₘ = limₙ (xₘₙ) and zₙ = limₘ (xₘₙ) exist for all m, n ∈ ℕ, and that the convergence of one of these collections is uniform. The two iterated limits then exist, the double limit exists as well, and the three coincide.
Proof. Let ε > 0. By uniformity there is N(ε) ∈ ℕ such that (∗) ‖xₘₙ − yₘ‖ < ε for all n ≥ N(ε) and all m ∈ ℕ. Fix the single column index q = N(ε). Since z_q = limₘ (x_{mq}) exists, the sequence (x_{mq} : m ∈ ℕ) is convergent and hence Cauchy by 16.7: there is R ∈ ℕ with ‖x_{kq} − x_{lq}‖ < ε for all k, l ≥ R.
For k, l ≥ R we then have ‖yₖ − yₗ‖ ≤ ‖yₖ − x_{kq}‖ + ‖x_{kq} − x_{lq}‖ + ‖x_{lq} − yₗ‖ < 3ε. Hence (yₘ) is a Cauchy sequence in ℝᵖ, and by 16.10 it converges to some y ∈ ℝᵖ. This is the row iterated limit y = limₘ limₙ (xₘₙ).
這一步是全篇的核心，值得慢讀。三段插入的頭尾兩段用 (∗)——注意它們用的是同一個 q，卻服務兩個不同的列 k 與 l，這件事只有在收斂均勻時才做得到；中間那段用第 q 行的 Cauchy 性。若各列的收斂不均勻，頭尾兩段就得各自挑自己的行號，插入的位置對不上，整條估計串不起來——例 8 失敗的地方正是這裡。得到 Cauchy 之後，完備性由 16.10 向 ℝᵖ 借，極限 y 就這樣造出來。
Since y = limₘ (yₘ), there is M ∈ ℕ with ‖yₘ − y‖ < ε for m ≥ M. Put K = sup {N(ε), M}. For m, n ≥ K, using (∗) again, ‖xₘₙ − y‖ ≤ ‖xₘₙ − yₘ‖ + ‖yₘ − y‖ < 2ε. Therefore the double limit exists and equals y.
這一步的所求：從剛造出的 y 反推雙重極限。兩段插入：第一段是「這一項離自己那列的極限多遠」（由均勻性，起算點不看 m），第二段是「那列的極限離 y 多遠」（由 yₘ → y，起算點不看 n）。兩個起算點各管一個下標，取較大者之後兩者同時生效——這正是雙重極限的定義要的形狀。
Finally, both collections of single limits exist by hypothesis and the double limit has just been shown to exist, so 19.7 applies: the column iterated limit exists and also equals y.