覆盖、子覆盖与定义
空间 X 的一个开覆盖是一族开集，其并为整个 X——每个点都至少被某个集合捕获。有限子覆盖是从该族中选出的有限几个，它们仍覆盖 X。若每个开覆盖都有有限子覆盖，则称 X 紧。可把它读成一场挑战—应答游戏：对手交给你任意一个开覆盖，无论多么无限或巧妙，你都必须以其中有限多个成员作答，而它们已足以完成覆盖。
Heine–Borel 与 [0,1] 的证明
Heine–Borel 定理是通往日常直觉的桥梁：Rⁿ 的子集紧当且仅当它有界且闭。仅有界不行——(0, 1) 有界却不紧；仅闭也不行——整个 R 是闭的却不紧。二者缺一不可。定理的核心是：任意有界闭区间都是紧的；其证明本质上用到了 R 的完备性。
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. ∎列紧性
还有第二个更具操作性的概念。若空间中每个序列都有收敛到空间内某点的子序列，则称空间列紧。在 Rⁿ——以及每个度量空间——中，这与紧性等价。对 [0, 1] 而言，这一等价正是 Bolzano–Weierstrass 定理：每个有界序列都有收敛子序列，而闭性把极限留在内部。两个定义观感不同——覆盖对序列——却刻画了完全相同的度量空间。