覆蓋、子覆蓋與定義
空間 X 的一個開覆蓋是一族開集，其並為整個 X——每個點都至少被某個集合捕獲。有限子覆蓋是從該族中選出的有限幾個，它們仍覆蓋 X。若每個開覆蓋都有有限子覆蓋，則稱 X 緊。可把它讀成一場挑戰—應答遊戲：對手交給你任意一個開覆蓋，無論多麼無限或巧妙，你都必須以其中有限多個成員作答，而它們已足以完成覆蓋。
Heine–Borel 與 [0,1] 的證明
Heine–Borel 定理是通往日常直覺的橋樑：Rⁿ 的子集緊當且僅當它有界且閉。僅有界不行——(0, 1) 有界卻不緊；僅閉也不行——整個 R 是閉的卻不緊。二者缺一不可。定理的核心是：任意有界閉區間都是緊的；其證明本質上用到了 R 的完備性。
Theorem. [0, 1] is compact.
Let C be an open cover of [0, 1]. Define
S = { x in [0,1] : [0, x] can be covered by FINITELY many sets of C }.
We show 1 ∈ S, which says [0,1] = [0,1] has a finite subcover.
Step 1 (S nonempty). 0 ∈ [0,1] lies in some open U0 ∈ C, so {U0} covers
[0,0] = {0}. Hence 0 ∈ S, and S is nonempty and bounded above by 1.
Step 2 (let s = sup S). By the least-upper-bound property s exists,
and 0 ≤ s ≤ 1. Some open U ∈ C contains s; openness gives a small
interval (s - d, s + d) ⊆ U for some d > 0.
Step 3 (s ∈ S). Pick a point t with s - d < t ≤ s and t ∈ S
(possible since s = sup S). Then [0, t] has a finite subcover F.
Adding U covers [0, t] ∪ (s - d, s + d) ⊇ [0, s].
So F ∪ {U} is a finite cover of [0, s], giving s ∈ S.
Step 4 (s = 1). If s < 1, the SAME finite set F ∪ {U} already covers
[0, s'] for some s' with s < s' < min(s + d, 1), so s' ∈ S — contradicting
that s is an upper bound of S. Hence s = 1, and 1 ∈ S by Step 3.
Therefore [0,1] has a finite subcover of C. Since C was arbitrary,
[0,1] is compact. ∎列緊性
還有第二個更具操作性的概念。若空間中每個序列都有收斂到空間內某點的子序列，則稱空間列緊。在 Rⁿ——以及每個度量空間——中，這與緊性等價。對 [0, 1] 而言，這一等價正是 Bolzano–Weierstrass 定理：每個有界序列都有收斂子序列，而閉性把極限留在內部。兩個定義觀感不同——覆蓋對序列——卻刻畫了完全相同的度量空間。