어릴때 썼던거라.. 다소 과한 표현이 많을 것 같습니다.. 부디 귀엽게 봐주셔요..
$\def\m#1{\mbox{#1}}$
Chapter 3. Compact and Connected Set
3-3. Nested Set Property
$\ (m,d) \m{: metric space} \\ \m{If Compact set }F_k \m{ is }F_k \neq \phi \m{ and } F_k \supset F_{k+1} \m{ for all k}\in \N ,\m{then } \overset{\infty}{\underset{k=1}{\bigcap}} F_k \neq \phi$
pf)
$\ \m{Let } \bigcap F_k = \phi \\ \Rightarrow M = \phi ^c = (\bigcap F_k)^c = \bigcup F_k^c \\ {F_k^c \m{ is open because }F_k \m{ is compact and closed}} \\ \Rightarrow {F_k^c} \m{ : open cover of }F_1 \\ \Rightarrow \exists F_{k_1}^c,F_{k_2}^c,…,F_{k_N}^c \m{ such that } F_1 \subset \bigcup F_{k_i}^c = F_{k_N}^c {\because F_1 \m{ is compact and }F_k \supset F_{k+1}} \\ \m{At the same time, } F_{k_N} \subset F_1 \m{ so } F_{k_N} \subset F_{k_N}^c \Rightarrow F_{k_N} =\phi \ {\m{Contradiction!}} $
- $d(A) := \underset{x,y \in A}{max} \ d(x,y) \\ d(\bigcap F_k) = 0 \m{ but } \bigcap F_k \m{ is not empty set} \\ \Rightarrow F_k = {x} \m{: singleton set}$
여기서 $F_k$는 공집합이 아니지만, distance는 0이다. 공집합이 아니라는 것에서 존재성이, 0이라는 점에서 유일성이 입증되어, Nested Set이 하나의 원소로 수렴한다라는 개념으로 이어질 수 있다.
다음으로 다룰 내용들은 connected set에 관한 것이다. connected 개념 또한 Compact 개념과 유사하게 path-connected라는 하위개념이 존재 한다. 다른 점이 있다면, path- connected 할시 connected set이지만, connected set이라고 path-connected set이 되지는 않는다는 점이다.
3-4. Path-Connected
path connected를 정의하기 위해서는 여러가지 개념들이 선행해서 정의해야 한다.
Continuous
$\ \m{For (m,d) : metric space} \\ \psi : [0,1] \rightarrow M \m{: continuous at } t \in [0,1] \\ \iff \forall {t_k} \subset [0,1], \m{as }t_k \m{ goes to } t , \psi(t_k) \m{ converges to } \psi(t)$
Path Joining
$\psi:[0,1] \rightarrow M \m{ : Path Joining x & y } \\ \iff \psi \m{: continuous in }[0,1] \m{ and } \psi(0)=x,\psi(1)=y$
Path Lies
$\m{For }A \subset M \\ \psi:[0,1] \rightarrow M \m{ : Path lies in }A \\ \iff \psi(t) \in A,\forall t \in [0,1] $
Path Connected
$A \subset M \m{: path connected } \\ \iff \forall x,y \in A , \exists \m{ contiuous path } \psi:[0,1] \rightarrow M \m{ joining x & y and lies in }A$
Convex Set
$\ A \subset \R^n \m{ is convex } \\ \iff \forall x,y \in A, \forall t \in [0,1], tx +(1-t)y \in A$
convex set이면 path-connected하다
ex)
$A = $${(0,y)\in \R^2 \mid -1 \leq y \leq 1} \\ \qquad \cup {(x,y) \in \R^2 \mid y = sin \frac{1}{x},0<x\leq 1} \\ = A_1 \cup A_2$
이 A는 Not Path - Connected이다.
pf)
$A_1,A_2 \m{ : closed in } \R^2 \\ x= (0,0) \in A_1 ,y = (\frac{1}{2\pi},0) \in A_2 \\ \psi :[0,1] \rightarrow A \m{ : Continuous Path Joining x & y such that } \psi(t) \in A , \forall t \in [0,1] \\ B_1 = \psi^{-1}(A_1) = {t \in [0,1] \mid \psi (t) \in A_1}\\ B_2 = \psi^{-1}(A_2) = {t \in [0,1] \mid \psi (t) \in A_2} $
$\m{If }A_1 \m{ is closed, then } B_1 \m{ is closed}$
pf) $t \in cl(B_1)$$\ \Rightarrow \exists t_k \rightarrow t, t_k \in B_1 \\ \Rightarrow \psi(t_k) \in A_1 , \psi(t_k) \rightarrow \psi(t) \\ \Rightarrow \psi(t) \in cl(A_1) = A_1 \\ \Rightarrow t \in B_1 \\ \Rightarrow cl(B_1) \subset B_1 \Rightarrow B_1 \m{ is closed }$
Path $ \psi$를 다음과 같이 설정하자
$\psi (0) = x = (0,0) \in A_1 , \psi(1) = y = (\frac{1}{2 \pi},0) \in A_2$
그러면 $0 \in B_1 \neq \phi , 1 \in B_2 \neq \phi$이다.
$\Rightarrow \sup B_1 = t_0 \in[0,1] \\ A_1 \cap A_2 = \phi \Rightarrow \phi = \psi^{-1}(A_1 \cap A_2 ) = \psi^{-1}(A_1) \cap \psi^{-1}(A_2) =B_1 \cap B_2 \\ \Rightarrow t_0 - \frac{1}{n} < x_n < t_0 , \exists x_n \in B_1 ,\forall n \in \N \\ \Rightarrow x_n \in B_1 \m{&} x_n \rightarrow t_0 , t_0 \in cl(B_1) = B_1 \Rightarrow t_0 \in B_1$
$\m{If } t_0 =b \in B_2 \Rightarrow (\m{contradiction to } B_1 \cap B_2 = \phi) \\ \m{so } t_0 <b , (t_0,t_0+\frac{1}{n}) \not\subset B_1 (t_0 = \sup B_1) \\ B_1 \cup B_2 = \psi^{-1}(A_1 \cup A_2 ) = \psi^{-1}(A) = [0,1] \\ \m{So, } \exists t_n \in (t_0,t_0+\frac{1}{n}) \cap B_2 \\ \Rightarrow t_n \in B_2 , t_n \rightarrow t_0 \Rightarrow t_0 \in cl(B_2) = B_2 \\ \Rightarrow t_0 \in B_2 \Rightarrow t_0 \in B_1 \cap B_2 \neq \phi \Rightarrow (\m{contradiction to } B_1 \cap B_2 = \phi)$
3-5. Connected set
Connected Set을 정의하기 위해서 또 다른 개념을 선행해서 정의하자.
Seperation
$\m{For (m,d) : Metric Space} ,A,U,V \subset M \m{ : Non-Empty Open Sets } \\ U,V \m{ : Separation of A} \\ \iff \begin{cases} \ 1) U \cap V \cap A = \phi \\ 2) U \cap A \neq \phi \\ 3) V \cap A \neq \phi \\ 4) A \subset U \cup V\end{cases}$
Disconnected
$\m{A is disconnected } \\ \quad \iff \m{There exists Separation of }A$
Connected
$\m{A is Connected } \\ \quad \iff A \m{ is not Disconnected}$
A is Path-Connected,then A is Connected
pf)
$A \m{: path connected } \rightarrow A \m{: connected}$
$A \m{ is path - connected } \\ \m{Suppose that }A \m{ is not connected, so there is separation } U,V \\ \m{By 2),3)} \Rightarrow x \in U \cap A \neq \phi , V \cap A \neq \phi \\ \psi : [0,1] \rightarrow A \m{ : continuous path such that } \forall t \in [0,1] , \psi(t) \in A \m{ and } \psi(0) =x , \psi(1)=y’ \\ C = \psi^{-1}(U \cap A) \neq \phi \\ D =\psi^{-1}(V \cap A) \neq \phi \\ \\ \m{Claim 1 : }C,D \m{ is closed} \\ t_k \rightarrow t, t_k \in C \\ \quad \Rightarrow \psi(t_k) \rightarrow \psi(t) \m{ & } \psi(t_k) \in U \cap A \\ \quad \Rightarrow \m{If } \psi(t) \notin U \cap A, \m{then }\psi(t) \in V \cap A \\ \qquad \Rightarrow D(\psi(t),\epsilon) \subset V,\exists \epsilon >0 \\ \qquad \Rightarrow \psi(t_k) \rightarrow \psi(t) \Rightarrow \psi(t_k) \in D(\psi(t),\epsilon) \subset V \m{(Contradiction to 1st condition)} \\ \quad \Rightarrow \psi(t) \in U \cap A \m{ so C is closed and simillary D is closed}$
$\m{Claim 2 : } C \cap D = \phi , C \cup D = [0,1] \\ C \cap D = \psi^{-1}(U \cap A) \cap \psi^{-1}(V \cap A) \\ \qquad = \psi^{-1}((U \cap A) \cap (V \cap A)) \\ \qquad = \psi^{-1}(\phi) = \phi \\ C \cup D = \psi^{-1}(U \cap A) \cup \psi^{-1}(V \cap A) \\ \qquad = \psi^{-1}((U \cap A) \cup (V \cap A)) \\ \qquad = \psi^{-1}((U \cup V) \cap A) \\ \qquad = \psi^{-1}(A) =[0,1] $
$E := C^c , F:= D^c , 1 \in E ,0 \in F \\ C,D \m{ is closed } \Rightarrow E,F \m{ is open} \\ 1) E \cap F \cap [0,1] =(C \cup D)^c \cap [0,1] = \phi \\ 2) E \cap [0,1] \neq \phi \quad 3) F \cap [0,1] \neq \phi \\ 4) C \cap D = \phi \Rightarrow \R = (C\cap D)^c = E \cup F \supset [0,1] \\ \therefore E,F \m{ is separation so A is connected so contradiction } \\ \therefore A \m{ is connected}$