推 herstein:第一句話有點措...open connected => path connected 08/18 16:08
推 Lonson:Thanks a lot~ 08/18 21:29
: 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.
: 順便請問有沒有推薦這方面的入門書可以讓我研讀一下?
: (我只有修過大一微積分)
Normally, you don't need any background knowledge to learn
topology, but it will help you to develop the insight and
skills of writing proofs if you are familiar with the topology
of Euclidean space.
The textbooks written by Munkres, Kelley or Armstrong are
standard, but you could pick up any book which suits your needs
as long as it is rigorous enough as well as comfortable to you.
In case you need more examples other than the Euclidean space,
you may also check "Counterexamples in Topology"
by Lynn Steen and J. Arthur Seebach, Jr.
: 另外感謝前面LimSinE板友有提到用convex性質去證
: 但我還是想了解比較依照定義的方式
: 譬如說我畫一個沒有洞的星星狀, 就沒辦法套convex性質了
: ※ 編輯: Lonson 來自: 58.114.103.97 (08/17 22:58)
: → THEJOY:我記得APOSTOL第四章好像有提過? 08/17 22:59
: → WINDHEAD:請搜尋 : star-shaped 08/17 23:06
: → WINDHEAD:比較保險的進路 : 先證 star-shaped => 與 single point 08/17 23:06
: → WINDHEAD:homotopy equivalent, 再據此推出 \pi_1 = trivial 08/17 23:07
: 推 herstein:http://0rz.tw/6rqGc 08/18 08:40
※ 引述《hanabiz (死神的精準度)》之銘言：
: ※ 引述《Lonson ()》之銘言：
: : 我的問題是："沒有洞" 到底是否能保證 simply-connected?
: : 順便請問有沒有推薦這方面的入門書可以讓我研讀一下?
: : (我只有修過大一微積分)
: 印象中只有符合 open connected 的集合才會去討論 它是不是simply-connected耶
: 如果有一個open connected 的集合"沒有洞", 那它就是simply connected
: (從定義上來看就是如此)
: 舉幾個例子 像是open ball, open rectangle都是simply connected
: 想像一個被從上往下壓扁的甜甜圈 扁到融入xy平面
: 它 open connected, 但不是simply connected
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◆ From: 140.109.23.38
※ 編輯: xcycl 來自: 140.109.23.38 (08/18 13:33)