Point topology 定義與性質

(集合)直徑 (Diameter)

定義： $S \subseteq X$ is bounded if $\exists M \in \mathbb{R}$ and $y \in X$ $\ni$ $d(x,y) < M, \ \forall x \in X$ .

定義：Let $a \in X$ and $r>0$ , the open ball $N_a(r) = \{b\in X \vert \ d(a,b) < r\}$ .
$N_a(r)$ 包含了以 $a$ 為圓心，半徑小於 $r$ 的所有屬於集合 $X$ 中的點，但不含圓周上的點。

定義：Let $a \in S$ , $a$ is called an interior point of $S$ $\Leftrightarrow$ $\exists r > 0$ $\ni$ $N_r(a) \subseteq S$ .
內點會如此定義的原因是因為若點恰好在集合 $S$ 的邊界上中，會找不到符合上述定義的半徑 $r$ ；反之若點在集合 $S$ 內，則一定可以找到滿足上述定義的半徑 $r$ 。
$S$ 中所有interior points所形成的集合記為 $\text{int} S$ .

定義： $S \subset X$ is an open set if all its points are interior points (此定義中，排除了boundary points).
$\forall a \in S$ , $a \in \text{int} S$ . (開集合中所有的點都是內點)
- Proof: Given $p \in S$ , $r > 0$ and $q \in N_r(p)$ .
- Let $h = d(p,q)$ , then $r_1 = r-d > 0$ , thus $N_{r_1}(q) \subseteq N_r(p)$ . (QED).
S is an open set $\Leftrightarrow$ $S = \text{int} S$ .
S in not an open set $\Leftrightarrow$ $\exists a \in S$ , $\forall r > 0$ , $N_r(a) \nsubseteq S$ .
任意(無限)個開集點的聯集仍是開集合。 $A_n$ is an open set, then $\cup_{n=1}^{\infty}A_n$ is an open set.
有限個開集點的交集仍是開集合。 $A_n$ is an open set, then $\cap_{n=1}^{N} A_n$ is an open set.
任意(無限)個開集合的交集不一定為開集合。
- E.g. $A_n=\left(\frac{-1}{n}, \frac{1}{n} \right)$ , then $\cap_{n=1}^{\infty} A_n =0$ is a point, not an open set.

$S \subseteq X$ , and $x \in X$
定義： $x$ is called a limit point of $S$ if
- $\forall r > 0$ , $N_r(x) \cap S - \lbrace x \rbrace \neq \phi$ .
- (反) $\exists r > 0$ , $N_r(x) \cap S - \lbrace x \rbrace = \phi$ .
- $x$ 的任意鄰域 $N_r(x)$ 必定與集合 $S$ 中不等於 $x$ 的元素之交集不為空集合。
可用limit point的概念得到以下序列
- 取 $r_1 =1$ ，可得 $x_1 \in N_{r_1} (x) \cap S \neq x_1$ .
- 取 $r_2 = \frac{1}{2} \begin{vmatrix} x_1 - x\end{vmatrix} > \frac{1}{2}$ , 得 $x_2 \in N_{r_2}(x) \cap S - \lbrace x \rbrace$ ，則 $x_2 \neq x_1, x$ 。
- 以此類推，得相異點 $\lbrace x_1, x_2, \cdots \rbrace \subset S$ , 且 $\lim_{ n \rightarrow \infty} x_n = x$ .
E.g. $S=\lbrace \frac{1}{n},\ \forall n \in \mathbb{N}\rbrace$ has $0$ as its limit point.
E.g. The set of rational numbers has every real number as a limit point.

定義： $S \subseteq X$ , and $x \in X$ , $x$ is adherent to $S$ if
- $\forall r >0, N_r(x) \cap S \neq \phi$ .
- 任意 $x$ 的鄰域 $N_r(x)$ 均包含至少一個 $S$ 中的元素。
- adherent point與limit point的唯一差異為與 $S$ 交集中的點不可為 $x$ 本身。
If $x \in S$ , $x$ must adherent to $S$ (集合中所有點均為附著點)。

Theorem: if $x$ is a limit point of set $S$ , then $\forall r > 0$ , $N_r(x)$ contains infinitely many point of $S$ .
- proof: 假設 $N_r(x)$ 只包含了 $s_1,\cdots, s_m$ 個元素，取 $r=\min \lbrace \lVert x-s_1 \rVert, \cdots, \lVert x-s_m \rVert \rbrace$ ，則 $N_{r/2}(x)$ 不包含除了 $x$ 以外的任何元素(矛盾)。

定義： If $x \in S$ , but $x$ is not a limit point of $S$ , then $x$ is an isolated point of $S$ .
- $\exists r > 0 \ni N_r(x) \cap S - \lbrace x \rbrace = \phi$ .
Let $x \in S$ , if $x$ is not a limit point of $S$ , then $N_r(x) \cap S = \lbrace x \rbrace$ .
E.g. $S = \lbrace \frac{1}{n} \vert \ n\in \mathbb{N} \rbrace$ , then every point in $S$ is an isolated point of $S$ .

定義 1： $S \subseteq X$ is a closed set $\Leftrightarrow$ $X\backslash S$ is an open set.
定義 2： $S \subset X$ is a closed set if every limit point of $S$ is a point of $S$ .

有限個閉集合的聯集仍然是閉集合。
- $A_n$ is a closed set, then $\cup_{n=1}^{\infty} A_n$ is a closed set.
- 無限多個閉集合的聯集可能會變成開集合。E.g. $A_n = (-1/n, 1/n)$ .
任意(無限)閉集合的交集仍是閉集合
- $A_n$ is a closed set, then $\cap_{n=1}^{\infty} A_n$ is a closed set.

If $A$ is an open set, $B$ is an close set, then $A-B$ is an open set, and $B-A$ is a closed set.
- $\because A-B = A \cap (X\backslash B)$ ，而兩個open sets的有限個交集仍為open set。
- $\because B-A= B \cap (X \backslash A)$ ，為兩個closed set的交集，仍為closed set。
$S \subseteq X$ is a closed set $\Leftrightarrow$ $S$ contains all its adherent(limit) points.
- " $\Rightarrow$ " 假設 $S$ is closed set，且 $x$ 為 $S$ 的adherent point，to prove $x \in S$ 。
  - If $x \notin S$ , then $x \in X\backslash S$ 。
  - $\because X\backslash S$ is open set, $\therefore \exists r > 0 \ni N_r(x) \in X \backslash S$ , thus $N_r(x)$ contains no points of $S$ (矛盾)。
- " $\Leftarrow$ " Assume $x \in X \backslash S$ , then $x \notin S$ , $x$ is not adherent to $S$ , $\therefore \exists r > 0 \ni N_r(x) \cap S = \phi$ , then $N_r(x) \in X\backslash S$ , $\therefore X\backslash S$ is open set, $\therefore S$ is closed set.

定義 1： The set of all adherent points of a set $S$ is called the closure of $S$ , denoted by $\bar{S}$ .
定義 2: $S \subseteq X$ and $S^{'}$ is the set of all limit points of $S$ in $X$ , the closure of $S$ is the set $\bar{S} = S \cup S^{'}$ .
- 依定義 $\bar{S}$ 為包含 $S$ 的最小閉集合。
- $S \subseteq \bar{S}$ (Every point of $S$ is adherent to $S$ ).
- $\bar{S} \subseteq S$ $\Leftrightarrow$ $S$ is closed.
  - 令 $x$ is limit point of $S$ or $S^{'}$ .
  - 如果 $x$ is limit point of $S$ $\Rightarrow$ $x\in \bar{S}$ .
  - 若 $x$ is limit of $S^{'}$ ，因 $S^{'}$ is closed $\therefore x \in S^{'} \subseteq \bar{S}$ .
- $S$ is closed $\Leftrightarrow$ $S=\bar{S}$ .
- $S$ is closed set $\Leftrightarrow$ $S$ contains all its limit points.
- Theorem: $\phi \neq S subseteq X$ is bounded above. Let $y = \sup S$ , then $y \in \bar{S}$ .

定義： $S \subseteq X$ is perfect set if
- $S$ is closed set and
- every point of $S$ is a limit point of $S$ .
- i.e. $S = S^{'}$
E.g. $[0,1]$ , $[0, \infty)$ , $(-\infty, 0]$ are all perfect sets.
E.g. $N_r(x)$ , $r > 0$ is not perfect set.
E.g. $\mathbb{N}$ is not perfect set.
E.g. $\mathbb{R}^n$ , $\forall n \in \mathbb{N}$ are perfect sets，但是 $\mathbb{R}^n$ 中有限元素的子集合均不為perfect set。

定義： $S \subseteq X$ is dense set if every point of $X$ is a limit point of $S$ or a point of $E$ .
E.g. The set of all rational numbers $\mathbb{Q}$ is dense in $\mathbb{R}$ , and so is the set of all irrational numbers.

定義： $S \subseteq X$ is convex set if $\forall x,y \in S$ and $0 < \lambda < 1$ , $\lambda x + (1-\lambda)y \in S$ ,