推 herstein :那麼你必須要定義甚麼叫"洞" 08/20 15:55
→ Lonson :很有道理....XD 08/20 16:04
※ 編輯: Lonson 來自: 210.69.40.253 (08/20 17:22)
※ 引述《xcycl (XOO)》之銘言：
: : 坦白說我看不是很懂
: : 假設題目換成這樣好了
: : S={(x,y)| x^2+y^2 < 1} 這樣的圖
: : unit circle不在S裡
: : (問題就在邊界)
: : S可以算是simply-connected嗎
: : 如果是,
: : 我需要證明些什麼東西?
: : copied from wiki:
: : A topological space X is called simply connected if
: : (1)it is path-connected and
: : (2) any continuous map f : S1 → X (where S1 denotes the unit circle in
: : Euclidean 2-space) can be contracted to a point in the following sense:
: : there exists a continuous map F : D2 → X (where D2 denotes the unit disk
: : in Euclidean 2-space) such that F restricted to S1 is f.
: : 我想path-connected的部分沒什麼問題
: : 可是請問"縮成一點"這個動作
: : 我要怎麼去"寫"出來??
: : 我也另外找到一種定義:
: : http://ppt.cc/jsFY 第三頁
: : A two-dimensional regi
: : 我的問題是："沒有洞" 到底是否能保證 simply-connected?
: Not exactly. In R^n, connected is equivalent path-connected, but
: in general path-connected implies connected only. Even if there is
: no hole, it cannot be simply-connected simply due to that it
: is not path-connected.
: For example, topologist's sine curve is a topological space defined by
: the union of (0, 0) and A = {(x, sin(1/x)) : x in (0, 1]} with the topology
: induced from R^2.
: It is connected because it is the closure of A and A is connected dut to that
: (0, 1] is connected and (0, 1] -> A is a continuous map.
: However there is no path between (0,0) to any point in A
: which means it is not path-connected. Also, there is no "hole" instinctively.
Thanks for your reply. :)
I did neglect the requirement of path-connectedness.
How about this:
Supose there is an open region C in R^2.
If C contains no holes `and' it is path-connected,
C is simply connected.
Am I right? (Just my guess. I dont know how to prove the statement above.)
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