作者math1209 (人到无求品自高)
看板Math
标题Re: [分析] 两题
时间Wed Jan 19 11:56:51 2011
※ 引述《cckk3333 (皓月)》之铭言：
: 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.
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
的数都可以，其目的在保证 f 为 K 上之连续函数。
--
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※ 编辑: math1209 来自: 140.113.25.169 (01/19 13:06)
1F：推 cckk3333 :不好意思可以问一下怎麽保证f连续吗？ 01/19 14:01
2F：推 cckk3333 :sorry 请忽略第一行 01/19 14:15
3F：→ cckk3333 :不好意思可以问一下 x'_n_j+1 跟 x'_n_j 有甚麽直接 01/19 15:02
4F：→ cckk3333 :关系吗? 01/19 15:02
5F：→ math1209 :如果你看得不习惯:命 x'_n_j = X_j. 则 01/19 15:13
6F：→ math1209 :x'_n_(j+1) = X_(j+1) 与 x'_n_j = X_j. 01/19 15:14
7F：→ math1209 :也就是说(对子列来说):x'_n_j+1 是 x'_n_j 下一项. 01/19 15:14
8F：推 cckk3333 :我只是觉得x'n_j只是x_n的preimage而已 01/19 15:16
9F：→ cckk3333 :x'_n_j+1 也同样只是 x_n_j+1的preimage 01/19 15:17
10F：→ cckk3333 :两者乍看之下好像跟f没有关系 01/19 15:17
11F：→ cckk3333 :如有误解请见谅 01/19 15:17
12F：→ math1209 :嗯~d(x,y) = d(f(x),f(y)) 中的证明有误. 01/19 16:26
13F：→ math1209 :我再想想...=.= 01/19 16:27