Open Sets

Open Ball

Definition 39.1. Let \( (M, d) \) be a metric space. Let \( \varepsilon > 0 \) and let \( x \in M \). We let \[ B_\varepsilon(x) = \{ y \in M \mid d(x, y) < \varepsilon \} \] \( B_\varepsilon(x) \) is called the open ball of radius \( \varepsilon \) centered at \( x \).

열린 집합 $(a, b)$를 일반화 시킨 개념이다. 특별히 어떤 점 $x$를 중심으로 $\varepsilon > 0$ 만큼 좌우로 떨어진 점들을 모아놓은 집합을 의미한다.

Open Sets

Definition 39.2. Let \( M \) be a metric space and let \( X \) be a subset of \( M \). We say that \( X \) is open if for every \( x \in X \), there exists an open ball \( B_\varepsilon(x) \) (centered at \( x \)) such that \( B_\varepsilon(x) \subset X \).

직관적으로 생각했을 때 만약 경계 부분을 포함하고 있는 집합이라고 하면 결코 open set이 될 수 없음을 알 수 있다. 경계 부분에서는 어떻게 해도 기존 집합에 완전히 포함되는 open ball을 잡아낼 수 없기 때문이다. 

Theorem 39.3

Theorem 39.3. Let \( M \) be a metric space. Then \( M \) and \( \emptyset \) are open subsets of \( M \).

Proof. Let $x \in M$. Then $\forall \varepsilon > 0$, clearly $B_{\varepsilon}(x) \subset M$. Thus $M$ is open. 
Suppose that $\emptyset$ is not open. Then $\exists x \in \emptyset$ such that $\forall \varepsilon > 0, B_{\varepsilon} \not \subset \emptyset$. But it is impossible, so that $\emptyset$ is open. $\blacksquare$

$M$ 자기 자신과 공집합 $\emptyset$은 closed이면서 open임을 알 수 있다. 두 정의는 상호배타적인 것 같지만, 정의만 놓고 봤을 때 closed가 아니면 open일 이유가 전혀 없다.

Theorem 39.4

Theorem 39.4. Let \( M \) be a metric space. Let \( x \in M \) and let \( \varepsilon > 0 \). Then the open ball \( B_\varepsilon(x) \) is an open subset of \( M \).

Proof. Let $y \in B_{\varepsilon}(x)$. Let $\delta = \varepsilon - d(y, x)$.
Let $z \in B_{\delta}(y)$. Then $$d(z, y) < \delta = \varepsilon - d(y, x) \\ \Longrightarrow d(x, z) \leq d(x, y) + d(y, z) < \varepsilon \\ \Longrightarrow z \in B_{\varepsilon}(x). $$ Then $B_{\delta}(y) \subset B_{\varepsilon}(x)$. Thus $B_{\varepsilon}(x)$ is open. $\blacksquare$

Theorem 39.5

Theorem 39.5. Let \( M \) be a metric space and let \( X \subset M \). Then \( X \) is open if and only if \( X^c \) is closed.

Proof. ($\Longrightarrow$) Suppose that $X$ is open. Let $x$ be a limit point of $X^c$. Then there exists $\{ x_n \}$ such that $x_n \in X^c, \forall n$ and $x_n \to x$. Then $x_n \notin X, \forall n$.
If $x \in X$, then $\exists B_{\varepsilon}(x) \subset X$. Since $x_n \to x$, $\exists N$ such that $d(x_n, x) < \varepsilon, \forall n \geq N$. Then $x_n \in B_{\varepsilon}(x) \subset X, \forall n \geq N. \bigotimes$
Thus $x \not X \iff X \in X^c$, which implies that $X^c$ is closed. 
($\Longleftarrow$) Suppose that $X^c$ is closed. If $X$ is not open, then $\exists x \in X$ such that $\forall \varepsilon > 0, B_{\varepsilon}(x) \not \subset X$. 
If we take $\varepsilon = \frac{1}{n}$, then $B_{\frac{1}{n}}(x) \not \subset X$. For any positive integers $n$, $\exists y_n \in B_{\frac{1}{n}}(x)$ such that $y_n \notin X \iff y_n \in X^c$.
Note that $d(y_n, x) < \frac{1}{n}, \forall n$. $\forall \varepsilon > 0, \exists N$ such that $\frac{1}{N} < \varepsilon$. Then $d(y_n, x) < \frac{1}{n} \leq \frac{1}{N} < \varepsilon, \forall n \geq N$. Thus $\lim_{n \to \infty} y_n = x$. 
Then $x$ is a limit point of $X^c$, which means that $x \in X^c \iff x \notin X. \bigotimes$ Thus $X$ is open. $\blacksquare$

Open과 closed는 정의 자체만 보면 상호배타적이지 않지만, 여집합의 관계에 있어서는 상호배타적이라고 할 수 있다. 직관적으로 생각했을 때 closed set의 여집합은 당연히 open이고 역도 성립함을 알 수 있다. 

Theorem 39.6

Theorem 39.6. Let \( M \) be a metric space.
(i) If \( U_1, \ldots, U_n \) are open subsets of \( M \), then \( U_1 \cap U_2 \cap \cdots \cap U_n \) is an open subset of \( M \).
(ii) If \( \mathscr{O} \) is a collection of open subsets of \( M \), then \( \bigcup \mathscr{O} \) is an open subset of \( M \).

Proof. (i) Let $x \in \bigcap_{i=1}^n U_i$. Then $x \in U_i, \forall i$. Since each $U_i$ is opne, $\exists \varepsilon_i > 0$ such that $B_{\varepsilon_i}(x) \subset U_i$.
Let $\varepsilon = \min \{ \varepsilon_1, ..., \varepsilon_n \}$. Then $B_{\varepsilon}(x) \subset U_i, \forall i$, which implies that $B_{\varepsilon}(x) \subset \bigcap_{i=1}^n U_i$. Thus $\bigcap_{i=1}^n U_i$ is open.
(ii) Let $x \in \bigcup \mathscr{O}$. Then $\exists O \in \mathscr{O}$ such that $x \in O$. Since $O$ is open, $\exists \varepsilon > 0$ such that $B_{\varepsilon}(x) \subset O \subset \bigcup \mathscr{O}$. Thus $\bigcup \mathscr{O}$ is open. $\blacksquare$

Topology

Definition. If \( X \) is a set and \( \mathscr{T} \) is a collection of subsets of \( X \) satisfying
(i) \( X, \emptyset \in \mathscr{T} \)
(ii) The union of a subcollection of \( \mathscr{T} \) is a member of \( \mathscr{T} \)
(iii) The intersection of a finite subcollection of \( \mathscr{T} \) is a member of \( \mathscr{T} \) then \( \mathscr{T} \) is called a topology for \( X \). By Theorems 39.3 and 39.6, the collection of open subsets of a metric space \( M \) is a topology for \( M \).