※ 引述《cckk3333 (皓月)》之銘言： : 1. Let (M,d) be a metric space and f: M -> M satisfy : d(x,y) <= d(f(x),f(y))<= 2*d(x,y) for all x,y : Show that d(f(x),f(y)) = d(x,y) for all x,y and f is bijection [我將題目的符號 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. Next, check d(f(x),f(y)) ＝ d(x,y) for all x, and y. Given x and y in K, there exist two convergent sequences {x_n} and {y_n} in K with x_n →x and y_n →y. Since f is onto, we can write f(x'_n) ＝ x_n and f(y'_n) ＝ y_n. Hence, there exists two convergent subsequences {x'_n_j} and {y'_n_j} with x'_n_j → x' and y'_n_j → y',(where n_(j＋1) － n_j ≧ 2) since K is sequentially compact. Thus, we obtain x_n_j ＝ f(x'_n_j) and y_n_j ＝ f(y'_n_j). By uniqueness of limit and continuity of f, we get x ＝ f(x') and y ＝ f(y'). Consider d(x,y) ＝ d(f(x'),f(y')) ≧ d(x',y') by d(x,y) ≦ d(f(x),f(y)) for all x, and y, ＝ d(lim x'_n_(j＋1), lim y'_n_(j＋1)) j→∞ j→∞ ＝ d( lim f^(m) (x'_n_j), lim f^(m) (y'_n_j) ) for some m≧2, j→∞ j→∞ ≧ d( lim f^(2) (x'_n_j), lim f^(2) (y'_n_j)). j→∞ j→∞ [use d(x,y) ≦ d(f(x),f(y)) for all x and y again] ＝ d(f^(2) (x'), f^(2) (y')) by continuity of f^(2) ＝ d(f(x), f(y)). Therefore, d(x,y) ＝ d(f(x),f(y)) for all x and y and the hypothesis: d(x,y) ≦ d(f(x),f(y)) for all x, and y. NOTE. d(x,y) ≦ d(f(x),f(y)) ≦ 2 d(x,y) for all x,y 中的 "2" 改成大於等於 1 的數都可以，其目的在保證 f 為 K 上之連續函數。 -- Good taste, bad taste are fine, but you can't have no taste. -- ※ 編輯: math1209 來自: 140.113.25.169 (01/19 13:06)