连通意味着什么
若空间 X 不能被分割为两个非空、不相交且并为整个 X 的开集，则称 X 连通。这样的分割称为一个分离；连通即不存在分离。等价地说，同时既开又闭（“开闭集”）的子集只有 ∅ 与 X 本身。直观上，连通空间是一整块——你无法沿一道开的缝把它干净地切开。
关于 R 的基本事实是：它的区间恰好就是它的连通子集。像 {1} ∪ {2} 这样的集合是不连通的：开集 (0, 1.5) 与 (1.5, 3) 把它分离。一个相关而更强的概念是道路连通：空间内任意两点都可被一条连续道路相连。道路连通总蕴含连通；对 Rⁿ 的开子集二者一致。
连续像与介值定理
起作用的就这一条事实：连续映射把连通集送成连通集。证明是干净的逆否——若像分裂为两个开块，把它们经 f⁻¹ 拉回会使定义域分裂，与其连通性矛盾。于是介值定理立刻得到，它不再是另一桩奇迹，而是一条推论。
Lemma. If f : X -> Y is continuous and X is connected, then f(X) is connected.
Proof (contrapositive). Suppose f(X) is NOT connected: there are open sets
A, B in Y with
f(X) ⊆ A ∪ B, A ∩ f(X) ≠ ∅, B ∩ f(X) ≠ ∅, A ∩ B ∩ f(X) = ∅.
Then U = f^{-1}(A) and V = f^{-1}(B) are open in X (continuity), nonempty
(each part of the image is hit), disjoint, and U ∪ V = X.
That is a separation of X — contradicting that X is connected.
Hence f(X) is connected. ∎
Intermediate Value Theorem. Let f : [a, b] -> R be continuous and suppose
f(a) < y < f(b). Then f(c) = y for some c in [a, b].
Proof. [a, b] is connected (an interval), so by the Lemma f([a,b]) is a
connected subset of R, hence an interval. It contains f(a) and f(b),
so it contains every value between them, in particular y.
Thus y = f(c) for some c in [a, b]. ∎紧性与最值定理
紧性扮演孪生角色。连续映射把紧集送成紧集——给定像的一个开覆盖，把它拉回，对（紧的）定义域取有限子覆盖，再推回去即可。把它与 Heine–Borel 结合便得最值定理：紧集上的连续函数取得其最大值与最小值。
Extreme Value Theorem. A continuous f : [a, b] -> R attains a max and a min. Proof. 1. [a, b] is compact (Heine-Borel: closed and bounded). 2. The continuous image f([a,b]) is compact (compactness is preserved), so by Heine-Borel f([a,b]) is closed and bounded in R. 3. Bounded above => M = sup f([a,b]) exists and is finite. 4. M is a limit of values of f, so M ∈ closure of f([a,b]) = f([a,b]) (the set is closed). Hence M = f(c) for some c in [a, b]: the maximum is ATTAINED. 5. The same argument on -f, or using inf, gives the minimum. ∎ Where each hypothesis is used: - boundedness of [a,b] -> f is bounded (sup is finite); - closedness of [a,b] -> the sup is achieved, not just approached. Drop either and it fails: f(x)=x on (0,1) has sup 1 but no maximum.