推 hcsoso :f^{-1} 的連續性? 10/26 01:25
→ VFresh :你要用矛盾法證明open 不能假設他是closed 10/26 06:50
→ VFresh :因為一個集合不是open不代表就是closed 10/26 06:50
推 hcsoso :忘記是誰說過, 集合不是一扇門... 10/26 08:13
→ jacky7987 :抱歉腦殘了 10/26 22:52
※ 引述《bineapple (パイナップル)》之銘言：
: Let f:R^n→R^m be one-to-one and continuous.
: If A is an open and disconnected subset of R^n,
: show that f(A) is open and disconnected in f(R^n).
: 可以請高手給一點提示嗎??
: 謝謝!!
Notation f^(-1) is the inverse image of f
獻醜了，有誤歡迎大家更正
不好意思今天很匆忙的思考這幾題，回到房間才有時間認真想
如果造成大家不便請見諒
1.proof of f(A) is open :
Suppose not,i.e f(A) is closed
Since f is conti. & f(A) is closed
by thm 4.24 in Apostol
f^(-1)(f(A)) is closed
→A = f^(-1)(f(A)) is closed (→←)
Note that the "=" hold because f is 1-1
if f is not 1-1 then we only can conclude that
A is contained in f^(-1)(f(A))
2.proof of f(A) is disconnected
Similary,suppose not i.e f(A) is connected
then f(A) and empty are the only two sets are both open and closed
by 1. and thm 4.23 in Apostol
We can also get A is also both open and closed
which implies A is connected (→←)
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