Let X = (xₙ) be a convergent sequence in ℝᵖ with limit x. If there are c ∈ ℝᵖ and r > 0 with ‖xₙ − c‖ ≤ r for all sufficiently large n, then ‖x − c‖ ≤ r.
設 X = (xₙ) 收斂到 x。若有 c ∈ ℝᵖ 與 r > 0 使得夠後面的每一項都滿足 ‖xₙ − c‖ ≤ r，則極限也滿足 ‖x − c‖ ≤ r。
證明計畫 · 由所求想起
⇢ 把「不滿足結論的位置」整片圈起來：V = {y : ‖y − c‖ > r}。
⇢ 這片區域是 open set，所以只要 x 落在裡面，它就是 x 的 neighborhood。
⇢ 收斂於是逼得夠後面的項全部進來——與假設「夠後面的項都滿足 ‖xₙ − c‖ ≤ r」正面衝突。
Proof. The set V = {y ∈ ℝᵖ : ‖y − c‖ > r} is the complement of the closed ball of centre c and radius r, which is a closed set; hence V is open. Suppose x ∈ V. Then V is a neighborhood of x, so by Definition 14.3 all terms xₙ from some index onwards lie in V, that is, satisfy ‖xₙ − c‖ > r. This contradicts the hypothesis, which places all sufficiently large n under ‖xₙ − c‖ ≤ r. Therefore x ∉ V, which is to say ‖x − c‖ ≤ r.
整份證明只有一個動作，關鍵在把結論的否定畫成一個集合。所求：排除 ‖x − c‖ > r 這種情況；把所有這樣的點收成一片 V，問題就變成「x 能不能落在 V 裡」。V 正是閉球 {y : ‖y − c‖ ≤ r} 的補集，而 §9-2 例 6 驗過閉球是 closed set，所以由 9.4（closed set 就是 open set 的補集）得知 V 是 open。V 一旦是 open 又含著 x，它就是 x 的 neighborhood，於是收斂的定義把夠後面的項全部趕進 V——但那些項按假設應該待在閉球裡。兩邊互斥，x 進不了 V。