Let a be a point in the domain D(f). The following statements are equivalent. (a) f is continuous at a. (b) For each ε > 0 there is a number δ(ε) > 0 such that ‖f(x) − f(a)‖ < ε holds whenever x ∈ D(f) and ‖x − a‖ < δ(ε). (c) If (xₙ) is any sequence in D(f) converging to a, then (f(xₙ)) converges to f(a).
Proof. (a) ⟹ (b). Let ε > 0 and put V = {y ∈ ℝ^q : ‖y − f(a)‖ < ε}, a neighborhood of f(a). By 20.1 there is a neighborhood U of a with f(x) ∈ V for all x ∈ U ∩ D(f). Since U is a neighborhood of a, some open ball of radius δ(ε) > 0 about a lies in U. Hence x ∈ D(f) and ‖x − a‖ < δ(ε) imply ‖f(x) − f(a)‖ < ε.
這一步的所求：把兩端的 neighborhood 都換成球。輸出端是我們自己選的，直接把 V 取成半徑 ε 的 open ball（§9-1 例 7 證過 open ball 是 open set，所以它是自己的 neighborhood）；輸入端不是我們選的，只能從 20.1 交來的 U 裡面挖——依 9.7（neighborhood 藏著含該點的 open set），U 裡藏著一個含 a 的 open set，而 open set 的定義又保證那裡面有一顆以 a 為心的球，半徑就取它。舉個數字：若 U = (a − 0.4, a + 0.7)，挖出來的球是 (a − 0.4, a + 0.4)，半徑 0.4。
這張圖在說第一步做了什麼：右邊的球是我們主動取的 V，左邊那個形狀不規則的才是 20.1 交回來的 U，藍色的球是從它裡面挖出來的。挖出來的球比 U 小，所以要求只會更嚴，結論照樣成立。
(b) ⟹ (c). Let (xₙ) be a sequence in D(f) converging to a, and let ε > 0. Take δ(ε) > 0 as in (b). Because (xₙ) converges to a, there is N ∈ ℕ such that ‖xₙ − a‖ < δ(ε) for all n ≥ N. Since every xₙ lies in D(f), statement (b) gives ‖f(xₙ) − f(a)‖ < ε for all n ≥ N. Hence (f(xₙ)) converges to f(a).
(c) ⟹ (a). We argue indirectly. Suppose (a) fails: there is a neighborhood V₀ of f(a) such that for every neighborhood U of a some point of U ∩ D(f) is sent outside V₀. For each n ∈ ℕ apply this to Uₙ = {x ∈ ℝᵖ : ‖x − a‖ < 1/n} and pick xₙ ∈ Uₙ ∩ D(f) with f(xₙ) ∉ V₀.
Then ‖xₙ − a‖ < 1/n for every n, so (xₙ) is a sequence in D(f) converging to a. If (f(xₙ)) converged to f(a), then all but finitely many of its terms would lie in the neighborhood V₀ of f(a); but no term does. Hence (c) fails as well, and the contrapositive gives (c) ⟹ (a).
三個版本裡最容易被低估的是 (c)。它把「連續」與「取極限」串成一句話：f 在 a 連續，等於說極限符號可以搬進函數裡，
lim (f(xₙ)) = f(lim (xₙ))。
而它還有一個更省力的用途——反過來用。
20.3 DISCONTINUITY CRITERION
The function f fails to be continuous at a point a ∈ D(f) if and only if there is a sequence (xₙ) in D(f) which converges to a but for which (f(xₙ)) does not converge to f(a).
Proof. By 20.2, f is continuous at a if and only if (c) holds. Therefore f fails to be continuous at a if and only if (c) fails, that is, if and only if not every sequence in D(f) converging to a has (f(xₙ)) converging to f(a) — which says exactly that some such sequence does not.
整段只用了一次「否定全稱句就得到存在句」。(c) 說的是「每一條收斂到 a 的數列，像數列都收斂到 f(a)」，它的否定是「存在一條收斂到 a 的數列，像數列不收斂到 f(a)」——量詞從「每一條」翻成「存在一條」，內層的結論跟著加上否定。值得留意的是這個判準的實用價值全在存在句上：要找到一個東西，遠比要驗證一整族東西容易。