Re: [分析] 兩題

[math1209 (人到無求品自高)]

※ 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。

--
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--
※ 編輯: math1209        來自: 140.113.25.169       (01/19 13:06)
推 cckk3333    :不好意思可以問一下怎麼保證f連續嗎？
推 cckk3333    :sorry 請忽略第一行
→ cckk3333    :不好意思可以問一下 x'_n_j+1 跟 x'_n_j 有甚麼直接
→ cckk3333    :關係嗎?
→ math1209    :如果你看得不習慣:命 x'_n_j = X_j. 則
→ math1209    :x'_n_(j+1) = X_(j+1) 與 x'_n_j = X_j.
→ math1209    :也就是說(對子列來說):x'_n_j+1 是 x'_n_j 下一項.
推 cckk3333    :我只是覺得x'n_j只是x_n的preimage而已
→ cckk3333    :x'_n_j+1 也同樣只是 x_n_j+1的preimage
→ cckk3333    :兩者乍看之下好像跟f沒有關係
→ cckk3333    :如有誤解請見諒
→ math1209    :嗯~d(x,y) = d(f(x),f(y)) 中的證明有誤.
→ math1209    :我再想想...=.=