法庭上要判一個人有罪，得舉證；要判無罪，卻不能只說「找不到證據」——得說清楚無罪的樣子是什麼。數學裡的處境類似：證明 X 收斂到 x，流程 §14-2 的 14.4 已經給了；證明 X不收斂到 x，卻常常有人卡住——因為那是一句全稱敘述的否定，量詞要整排翻面。
翻面的規則是機械的。「對每個 neighborhood V，存在K，使對每個n ≥ K 都有 xₙ ∈ V」的否定，是「存在某個 V，對每個K，存在n ≥ K 使 xₙ ∉ V」。每個量詞換一次身分、最裡層的結論取反，其餘一字不動。翻完之後那句話還可以再包裝一次，變成更好用的形式。
15.4 THEOREM
For a sequence X = (xₙ) in ℝᵖ and a point x, the following are equivalent. (a) X does not converge to x. (b) Some neighborhood V of x has this property: to every natural number n there corresponds a natural number m = m(n) ≥ n with x_m ∉ V. (c) Some neighborhood V of x and some subsequence X′ of X are such that no term of X′ lies in V.
下列三句話等價：(a) X 不收斂到 x；(b) 存在 x 的某個 neighborhood V，使得不論 n 取多大，總還找得到某個 m ≥ n 使 x_m 落在 V 外；(c) 存在 x 的某個 neighborhood V 與某個子數列 X′，使 X′ 的每一項都在 V 外。
正例：X = ((−1)ⁿ) 與 x = −1 合乎 (c)——取 V = (−2, 0) 與偶數項子數列 (1, 1, 1, ⋯)，整條在 V 外。反例：不能只找一項在外面就結案——(1/n) 與 x = 0 之間，x₁ = 1 確實落在 V = (−1/2, 1/2) 外，可是 (b) 要的是「不論 n 多大都還有」，而這裡 n ≥ 3 之後一項都沒有了。
Proof. (a) ⟹ (b). By Definition 14.3, convergence to x asserts: for every neighborhood V of x there is K such that xₙ ∈ V for all n ≥ K. Its denial reads: there is a neighborhood V such that for every natural number n the choice K = n fails, that is, some m ≥ n has x_m ∉ V. This is precisely statement (b).
第一段是純粹的邏輯操作。收斂的敘述有三層量詞：對每個V、存在K、對每個n ≥ K。否定時三層依序翻成存在V、對每個K、存在n ≥ K，最裡層的 xₙ ∈ V 變成 xₙ ∉ V。翻出來的意思要讀懂：不是「有一項在外面」，而是「例外項要多後面有多後面」——不論你把起算點訂在哪裡，它後面都還有例外。(b) 只是把翻面結果裡的 K 改名成 n、把那個例外項的編號記成 m(n)，強調它是隨 n 變動的。
(b) ⟹ (c). Take the neighborhood V furnished by (b) and build indices recursively: put r₁ = m(1), and once rₙ has been chosen put r_{n+1} = m(rₙ + 1). By the property in (b) each rₙ satisfies x_{rₙ} ∉ V, and r_{n+1} ≥ rₙ + 1 > rₙ, so the indices increase strictly. The resulting X′ = (x_{rₙ}) is a subsequence none of whose terms lies in V.
(c) ⟹ (a). Suppose, to the contrary, that X converged to x. Lemma 15.2 would make the subsequence X′ converge to x as well, so for the neighborhood V all terms of X′ from some index onwards would lie in V. That contradicts the assumption that no term of X′ lies in V. Hence X does not converge to x.
第三段是 §15-1 的 15.2（收斂的數列，每個子數列也收斂到同一點）反過來用。反證：假設 X 真的收斂到 x，那麼 (c) 裡那條子數列也得收斂到 x，於是它從某項起全部落進 V。可是 (c) 說它一項都不在V 裡。矛盾。注意這裡連 V 有多小都不必計較——只要 V 是 neighborhood，收斂就得把尾巴送進去。