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$\def\m#1{\mbox{#1}}$
Chapter 4. Continuous Mapping
4-1. Continuity
$(M,d),(N,p) : \m{metric space}$
$f : A \rightarrow N \m{ : function} \\ x_0 \in A’ = {x \in M \mid x \m{: accumulation point of A}} \\ (x \in A’ \iff \forall \epsilon >0 , \rho(x,\epsilon)\setminus {x} \cap A \neq \phi) $
에 대해
Limit of Function
$b \in N \m{: limit of f at }x_0 \\ \iff \forall \epsilon>0 , \exists \delta>0 \m{ such that if } 0< d(x,x_0) < \delta \m{, then } \rho (f(x),b) < \epsilon \\ \iff \underset{x \rightarrow x_0}{\lim} f(x) =b $
이의 역은 다음과 같다.
Not Limit of Function
$b \in N \m{ is not limit of f at }x_0$
$\ \iff \exists \epsilon_0 > 0 , \forall \delta >0, \exists x_\delta , 0 < d(x_\delta , x_0) < \delta \m{ but } \rho(f(x_\delta),b) \geq \epsilon_0 \\ \iff \exists \epsilon_0 >0 , \forall n \in \N ,\exists x_n \neq x_0 , x_n \in D(x_0,\frac{1}{n}) \m{ but } \rho(f(x_n),b) \geq \epsilon_0 \\ \iff \underset{x \rightarrow x_0}{\lim} f(x) \neq b$
2장을 할때 사용했던 개념을 다시 생각해보자.
-
$x_0 \in A’ $$\ \iff \forall \epsilon>0 , D(x_0,\epsilon)\setminus {x_0} \cap A \neq \phi \\ \iff \exists {x_n} \subset A \setminus{x_0} \m{ such that } x_n \rightarrow x_0$
-
$x_0 \in cl(A) $ $\ \iff \forall \epsilon>0 , D(x_0,\epsilon) \cap A \neq \phi \\ \iff \exists {x_n} \subset A \m{ such that } x_n \rightarrow x_0$
그러면 여기에 대해서 함수의 continuous를 다음과 같이 정의될 수 있다.
$\m{For }f: A \rightarrow N , x_0 \in A \cup A’ \\ f\m{: continuous at }x_0 \\ \iff \begin{cases} x_0 \notin A’ \\ or \\ x_0 \in A’ \m{ then } \underset{x \rightarrow x_0}{\lim} f(x) = f(x_0)\end{cases}$
$x_0$가 A에는 존재하지만 Accumulation Point에 존재하지 않는다면 자동으로 continuous하다.
Continuous Function
$f:A \rightarrow N \m{ is continuous function } \iff f \m{ is continuous } \forall x \in A$
이러한 continuous function에는 여러가지 성질이 숨어있다. 우선 다음과 같은 개념을 새로이 정의해보자.
Open Relative and Closed Relative
$\m{For }f : A \rightarrow N \m{ : continuous function & } F,G \subset N, F \m{: closed } ,G \m{: open}$
$f^{-1}(F) \m{ is closed relative to A} \\ \iff f^{-1}(F) = A \cap V \m{ for some closed set } V \subset M $
$f^{-1}(G) \m{ is open relative to } A \\ \iff f^{-1}(G) = A \cap U \m{ for some open set }U \subset M$
If B is closed relative to A, following proposition holds
- $x_n \in B = A \cap V ,x_n \rightarrow x \in A \Rightarrow x \in B$
- $B = A \cap cl(B)$
이를 이용하면 다음과 같은 notion들을 정리할 수 있다.
$f \m{ : continuous on }A \\ \iff \forall x_k \rightarrow x ,x_m \in A \Rightarrow f(x_k) \rightarrow f(x) \\ \iff \forall \m{ closed set F } \subset N , f^{-1}(F) \m{ : closed relative to A} \\ \iff \forall \m{ open set G} \subset N , f^{-1}(G) \m{ : open relative to }A$
4-2. Image of compact and connected
Image는 함수값 집합으로 치역으로 해석된다. 이 image와 관련해서는 아래의 중요한 Theorem이 알려져 있다.
Theorem 4.2.1
$\m{For } f : M \rightarrow N \m{: continuous} \\ K \subset M, K \m{: connected or path-connected } \Rightarrow f(K) \m{ is connected or path-connected respectively}$
pf)
$f: M \rightarrow N \m{: conti} , K \subset M \m{ : connected}$
A) $K: connected $
Suppose that f(K) is not connected
$\Rightarrow \exists G,F \m{: seperation of }f(K) \\ G,F \m{ is nonempty open set} \\ 1. G \cap F \cap f(K) = \phi \\ 2. G \cap f(K) \neq \phi , F \cap f(K) \neq \phi \\ 3.f(K) \subset G \cup F $
$U := f^{-1}(G) , V := f^{-1}(F) ; \m{ open } \\ G \cap f(K)\neq \phi \Rightarrow \exists y \in G \m{ such that }y \in f(K) = {y \mid y =f(x) , x\in K } \\ \qquad \qquad \qquad \Rightarrow \exists x \in f^{-1}(G) =U ,x \in K\\ \qquad \qquad \qquad \Rightarrow x\in U \cap K \neq \phi {\m{ similar to V }} \\ F\cap G\cap f(K) = \phi \Rightarrow \phi = f^{-1}(F \cap G \cap f(K)) =f^{-1}(F) \cap f^{-1}(G) \cap f^{-1}(f(K)) \\ \qquad\qquad\qquad\qquad \quad = V \cap U \cap K = \phi \\ f(K) \subset F \cup G \Rightarrow f^{-1}(f(K)) \subset f^{-1}(F)\cup f^{-1}(G) \\ \qquad \qquad \qquad \Rightarrow K \subset V \cup U $
즉 U와 V는 K의 Separation이 된다. 이 경우 K는 dis-connected이므로 contradiction이 발생한다. 따라서 f(K)도 connected set이다.
B) $K : path-connected$
$y_1 \m{ & } y_2 \in f(K) = {y \in N \mid f(x) = y , \exists x \in K} \\ \Rightarrow \exists x_1,x_2 \in K \m{ such that } y_1 = f(x_1), y_2=f(x_2) \\ \exists \phi : [0,1] \rightarrow K \m{: continuous path such that } \phi(0)=x_1,\phi(1)=x_2 \\ \psi : [0,1] \rightarrow N \m{ such that } \psi(t) = f(\phi(t)) = f \cdot \phi (t) \\ \psi (0) = f(\phi(0))=f(x_1)=y_1 , \psi(1)=f(\phi(1))=f(x_2)=y_2 \\ \m{So } \psi \m{ is path joining }y_1 \m{ & }y_2 \\ \Rightarrow f(K) \m{: path-connected }$
- If $f ,g $ is continuous,then $f \cdot g$ is continuous
Compact의 경우에도 마찬가지로 위의 정리가 성립한다.
Theorem 4.2.1
$\m{For } f : M \rightarrow N \m{: continuous} \\ K \subset M, K \m{: compact } \Rightarrow f(K) \m{ is comapct}$
pf 1)
${F_k} \m{: open conver of }f(K) \\ \Rightarrow U_n = f^{-1}(F_k) \m{: open cover in M} \\ f(K) \subset \bigcup F_k \Rightarrow K\subset f^{-1}(f(K)) \subset f^{-1}(\bigcup F_k) = \bigcup f^{-1}(F_k) =\bigcup U_k$
따라서 ${U_k}$는 K의 open cover이다. K는 compact set이므로 다음과 같은 논리를 전개할 수 있다.
$K \subset U_1 \cup U_2 \cup … \cup U_N \\ \Rightarrow f(K) \subset f(U_1) \cup f(U_2) \cup … \cup f(U_N) \\ \qquad \qquad \quad = f(f^{-1}(U_1)) \cup f(f^{-1}(U_2))\cup … \cup f(f^{-1}(U_N)) \\ \qquad \qquad \quad \subset F_1 \cup F_2 \cup … \cup F_N$
따라서 $f(K)$도 compact set이다.
pf 2)
$y_n \in f(K) \Rightarrow y_n = f(x_n) ,\exists x_n \in K \\ \Rightarrow \exists {x_{n_k}} \subset {x_n} \m{ such that } x_{n_k} \rightarrow x \in K \\ \Rightarrow f(x_{n_k})=y_{n_k} \rightarrow f(x)=y \in f(K) {\m{ f is continuous function}}$
$\therefore \exists {y_{n_k}} \subset {y_n} , y= f(k) \m{ such that } y_{n_k} \rightarrow y \in f(K)$
$\therefore f(K) \m{ is sequentially compact and so it is compact}$
4-3. Operation of Continuous Mapping
연속함수의 연산 결과의 연속성은 다음과 같이 정의된다.
$(M,d) , (N ,\rho) , (P,\gamma) : \m{ metric space }\\ \m{For }f:M\rightarrow N, g: N \rightarrow P , \m{f and g is continuous function}$
다음 함수들은 모두 연속함수이다.
- $f \cdot g$
- $f+g$
- $f\times g$
- $f / g \quad {g \m{ is not 0 at concerned point}} $
4-4. The Boundedness of Continuous Function on Compact Set
아래의 정리는 장통근 교수님께서 매우 중요하다고 한 정리이다.
Max - Mini Theory
$f: M \rightarrow \R \m{: continuous at }K \subset M \m{: compact} \\ \Rightarrow \exists x \in K \m{ such that } f(x) = \sup f(K)$
pf)
$f(K) \m{: bounded } \Rightarrow y_1 = \sup f(K) , y_2 = \inf f(K) \\ \qquad \qquad \qquad \qquad y_1,y_2 \in \R\m{(by completeness, LUBP)} \\ f(K) \m{: closed } \iff \exists y_n \in f(K) \m{ such that } y_n \rightarrow y_1 ,\exists x_n \in K \m{ such that } f(x_n) = y_n \\ \qquad \qquad \qquad \qquad \Rightarrow \exists {x_{n_k}} \subset {x_n},x \in K \m{ such that }x_{n_k} \rightarrow x \\ \qquad \qquad \qquad \qquad \Rightarrow f(x_{n_k})=y_{n_k} \rightarrow f(x)=y,x \in K \\ \qquad \qquad \qquad \qquad \Rightarrow f(x) = y_1 = \sup f(K)\quad \exists x \in K$
4-5. The Intermediate Value Theorem
$f:M \rightarrow \R \m{ : continuous }, K \subset M \m{ : connected } x,y \in K \\ \forall c \in (f(x),f(y)) \Rightarrow \exists z \in K \m{ such that } f(z)=c$
pf)
If not $\exists c \in (f(x),f(y)) \m{ such that }f(z)\neq c , \forall z \in K $
$G = ( - \infty,c),F=(c,\infty) \m{: open in } \R \\ G \cup F = \R \setminus {c} \Rightarrow f(K) \subset G \cup F \\ f(x) < c< f(y) \Rightarrow f(x) \in G \neq \phi,f(y)\in F \neq \phi \\ \qquad \qquad \qquad f(x) \in G \cap f(K) \neq \phi , f(y) \in F \cap f(K)\neq \phi \\ G \cap F = \phi \Rightarrow G \cap F \cap f(K) = \phi$
따라서 G와 F는 K의 separation이 되므로 K는 disconnected하다. (Contradiction!)