Proof. Assume f is injective. By 22.7 applied to f on the compact set S there is a point x_* ∈ S with ‖f(x_*)‖ = m = inf {‖f(x)‖ : x ∈ S}. Since x_* ≠ 0 and f is injective with f(0) = 0, we have f(x_*) ≠ 0, so m > 0.
這一步的所求：讓 inf 離開零。光說「每個 x ∈ S 都有 ‖f(x)‖ > 0」推不出 inf 大於零——無限多個正數的 inf 可以是零。真正把事情關上的是 22.7：inf 被 S 上某一個點取到，於是它等於一個具體的長度，而那個長度不是零。用到 f(0) = 0（§21-1 的兩條規則附帶給出）加單射：只有原點被送到原點，而 x_* 長度是 1，不是原點。
Now let u ∈ ℝᵖ. If u = 0 both sides vanish. If u ≠ 0, then u/‖u‖ ∈ S, so ‖f(u/‖u‖)‖ ≥ m; by homogeneity the left side equals ‖f(u)‖ / ‖u‖, and multiplying by ‖u‖ gives ‖f(u)‖ ≥ m‖u‖.
這一步把球面上的結論推到全空間，手法與 §21-4 算 ‖f‖ 時完全一樣：先把 u 縮成單位長度送進去，再把縮放倍數提出來。提得出來是因為 f 齊次而 norm 對正數的純量倍也齊次。零向量另案處理，因為除法要求分母不為零，這一格是邊界簿記。
Conversely, suppose ‖f(x)‖ ≥ m‖x‖ holds for all x with some m > 0. If f(x₁) = f(x₂), then linearity gives 0 = ‖f(x₁) − f(x₂)‖ = ‖f(x₁ − x₂)‖ ≥ m‖x₁ − x₂‖, so ‖x₁ − x₂‖ = 0 and x₁ = x₂. Hence f is injective.
Let K ⊆ ℝᵖ be compact and let f be continuous and injective with domain K and range f(K) ⊆ ℝ^q. Then the inverse function g = f⁻¹, with domain f(K) and range K, is continuous.
定義域 compact 時，連續加單射就自動換來反函數連續——不必對 g 做任何額外檢查。這裡的 f⁻¹ 是真正的反函數（§2-2 的 2.6），單射保證它存在；與前幾篇當成「把集合拉回去」的同名記號不同。
單射保證 g = f⁻¹ 存在，定義域是 f(K)、值域是 K。要證 g 連續，可用的判準有三個，選 22.1(c)——因為它要的是 closed set 的逆像，而 compact 集合的像正好 closed。
證明計畫 · 由所求想起 所求是：任給 ℝᵖ 裡的 closed set H，找一個 closed set 把 g⁻¹(H) 從 g 的定義域裡切出來。關鍵的辨認是 g⁻¹(H) 其實就是 f(H ∩ K)——「反函數把它送進 H 的那些點」與「H 裡屬於 K 的點的像」是同一批。這一步把問題從反函數翻回正函數，而正函數這邊 22.5 立刻適用：H ∩ K compact，所以像 compact，所以 closed。
Proof. Let H be closed in ℝᵖ. Then H ∩ K is closed by 9.6(c) and bounded because K is, so it is compact by 11.3. By 22.5 the set H₁ = f(H ∩ K) is compact in ℝ^q, hence closed.
We claim H₁ = g⁻¹(H). Indeed, a point y lies in g⁻¹(H) exactly when y ∈ f(K) and its unique preimage x = g(y) lies in H; that is, exactly when y = f(x) for some x ∈ H ∩ K.
這一步是全證明的樞紐，值得慢讀。g⁻¹(H) 依定義是「g 把它送進 H 的那些 y」，而 g(y) 就是 y 在 f 之下那個唯一的原像——所以條件等於說 y 的原像住在 H 裡，也就是 y 是 H ∩ K 中某點的像。單射在這裡不可省：原像若不唯一，「g(y) 落在 H」這句話就沒有意義。要留意兩個記號雖然長得一樣，這裡出現的 f(H ∩ K) 是正像、g⁻¹(H) 是逆像，而等式把兩者接了起來。
Since H₁ ⊆ f(K) = D(g), the equality can be written as H₁ ∩ D(g) = g⁻¹(H) with H₁ closed. As H was an arbitrary closed set, 22.1(c) shows that g is continuous.
Let D ⊆ ℝᵖ. The collection of all continuous functions on D to ℝ^q is written C_{pq}(D). The collection of those which are in addition bounded is written BC_{pq}(D). When p and q are clear from the context we abbreviate these to C(D) and BC(D).
兩個記號差一個 B，差別就是有沒有額外要求有界。一般而言 BC(D) 真的比 C(D) 小——f(x) = x 在 D = ℝ 上連續卻無界。可是 D 一旦 compact，22.7 說連續必定有界，兩者就重合。
(a) C_{pq}(D) and BC_{pq}(D) are vector spaces under the pointwise operations (f + g)(x) = f(x) + g(x), (cf)(x) = c f(x). (b) BC_{pq}(D) is a normed space under ‖f‖_D = sup {‖f(x)‖ : x ∈ D}.
Proof. (a) If f, g are continuous at each point of D, so are f + g and cf by 20.6; the remaining vector space axioms hold because they are checked value by value in ℝ^q. If moreover f and g are bounded, say by M and N, then f + g is bounded by M + N and cf by |c|M, so BC_{pq}(D) is closed under the operations too.
(b) The set BC_{pq}(D) is a subset of the space of all bounded functions on D to ℝ^q, on which ‖·‖_D was shown to be a norm in 17.8. By (a) it is closed under the operations, so it is a subspace, and a norm restricted to a subspace is again a norm. Finally, if D is compact then every continuous f on D is bounded by 22.7, so C_{pq}(D) = BC_{pq}(D).