※ 引述《math1209 (人到无求品自高)》之铭言： : [我将题目的符号 M 改成 K, 然後 K 是 compact metric space.] : Proof. : First, we show that f is bijective. For 1-1, it is clear. So, it remains to : show that f is surjective as follows. If not, i.e., there exists a point p_0 : in K such that p_0 in K － f(K). Define p_1 ＝ f(p_0), and p_(n＋1) ＝ f(p_n) : for n ＝ 1,2,3,... . Since K is compact and d(f(x),f(y)) ≦ 2 d(x,y) for all : x, and y, it follows f(K) is compact. So, we have : d(p_0, f(K)) ＝ r ＞ 0. : Notice that : d(p_n, p_(n＋k)) ≧ d(p_0, p_k) ≧ r for all n, k. (1) : by the hypothesis: d(x,y) ≦ d(f(x),f(y)). (1) implies that {p_n} has no : convergent subsequence which is absurd since K is sequentially compact. From : above sayings, we have proved that f is bijective. (恕删) 试着接下去: n n 给定 K 中任两点 x y, 记 x_n = f (x), y_n = f (y), x_0 = x, y_0 = y 存在一组子足码 {n_j} 使得: (i) x , y 收敛(到K中某点) n_j n_j (ii) m_j := n_{j+1} - n_j 随 j 递增到无限大 (i) 是因为我们有 sequentially compact 的假设, 而因为 {n_j} 是一组(无限延伸的) 子足码, 因此必要时可以再取 {n_j} 的子足码满足 (ii). 因为 d(a,b)≦ d(f(a),f(b)), 所以 d(x_{n-1}, x_n) ≦ d(x_n, x_{n+1}), 所以 (i) d(x ,x ) ≦ d(x ,x ) ≦ ... ≦ d(x , x ) ──→ 0 as j→∞ 0 m_j 1 m_j+1 n_j n_{j+1} 同理我们有: d(y ,y ) → 0 as j→∞. 0 m_j 因此 d(x ,y ) → d(x_0, y_0) as j→∞. m_j m_j 但 d(x_n, y_n) 是一组随 n 递增的数列, 因此这会强迫 d(x_1,y_1) = d(x_2,y_2) = ... = d(x_0,y_0) d(f(x),f(y)) = ￣￣￣￣￣ ￣￣￣￣￣ = d(x,y) --

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