連通意味著什麼
若空間 X 不能被分割為兩個非空、不相交且並為整個 X 的開集，則稱 X 連通。這樣的分割稱為一個分離；連通即不存在分離。等價地說，同時既開又閉（「開閉集」）的子集只有 ∅ 與 X 本身。直觀上，連通空間是一整塊——你無法沿一道開的縫把它乾淨地切開。
關於 R 的基本事實是：它的區間恰好就是它的連通子集。像 {1} ∪ {2} 這樣的集合是不連通的：開集 (0, 1.5) 與 (1.5, 3) 把它分離。一個相關而更強的概念是道路連通：空間內任意兩點都可被一條連續道路相連。道路連通總蘊含連通；對 Rⁿ 的開子集二者一致。
連續像與介值定理
起作用的就這一條事實：連續映射把連通集送成連通集。證明是乾淨的逆否——若像分裂為兩個開塊，把它們經 f⁻¹ 拉回會使定義域分裂，與其連通性矛盾。於是介值定理立刻得到，它不再是另一樁奇蹟，而是一條推論。
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]. ∎緊性與最值定理
緊性扮演孿生角色。連續映射把緊集送成緊集——給定像的一個開覆蓋，把它拉回，對（緊的）定義域取有限子覆蓋，再推回去即可。把它與 Heine–Borel 結合便得最值定理：緊集上的連續函數取得其最大值與最小值。
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.