推 math1209 :-) 01/19 22:49
※ 引述《math1209 (人到無求品自高)》之銘言：
: [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.
(恕刪)
試著接下去:
n n
給定 K 中任兩點 x y, 記 x_n = f (x), y_n = f (y), x_0 = x, y_0 = y
存在一組子足碼 {n_j} 使得:
(i) x , y 收斂(到K中某點)
n_j n_j
(ii) m_j := n_{j+1} - n_j 隨 j 遞增到無限大
(i) 是因為我們有 sequentially compact 的假設, 而因為 {n_j} 是一組(無限延伸的)
子足碼, 因此必要時可以再取 {n_j} 的子足碼滿足 (ii). 因為 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}
同理我們有: 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) 是一組隨 n 遞增的數列, 因此這會強迫
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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◆ From: 114.32.4.99
※ 編輯: ppia 來自: 114.32.4.99 (01/19 22:18)