【拓扑空间】示例及详解3

时间：2024-04-07 14:22:32

目录

例1 

例2

例3

例4

例5

例1 

列出X=\left \{ a,b \right \}的所有拓扑

solution

\tau_1=\left \{\varnothing, X \right \}=\tau_{tri}

\tau_2=\left \{\varnothing, X,\left \{ a \right \}\right \}

\tau_3=\left \{\varnothing, X,\left \{b \right \}\right \}

\tau_4=\left \{\varnothing, X,\left \{ a \right \},\left \{ b \right \}\right \}=\tau_{dis}

例2

设三元素集X=\left \{ a,b,c \right \}，在X上有多种可能的拓扑，可以有代表性地找到X上的9类拓扑（这里的代表性实际上是同胚的体现up to homeomorphism）

solution：

\tau_1=\left \{\varnothing, X \right \}=\tau_{tri}

\tau_2=\left \{\varnothing, X,\left \{ a \right \} \right \}

\tau_3=\left \{\varnothing, X,\left \{ a \right \},\left \{ a,b \right \} \right \}

\tau_4=\left \{\varnothing, X,\left \{ a \right \},\left \{ b,c \right \} \right \}

\tau_5=\left \{\varnothing, X,\left \{ a \right \},\left \{ b \right \},\left \{ a,b \right \} \right \}

\tau_6=\left \{\varnothing, X,\left \{ a \right \},\left \{ b \right \},\left \{ a,c \right \} \right \}

\tau_7=\left \{\varnothing, X,\left \{ a \right \},\left \{ a,b \right \},\left \{ a,c \right \} \right \}

\tau_8=\left \{\varnothing, X,\left \{ a \right \},\left \{ a,b \right \},\left \{ a,c \right \} \right \}

\tau_9=\left \{\varnothing, X,\left \{ a \right \},\left \{ b \right \},\left \{ c \right \},\left \{ a,b \right \},\left \{ a,c \right \} ,\left \{ b,c \right \}\right \}=\tau_{dis}

离散拓扑\tau_{dis}：X的所有子集的族是X的一个拓扑

平凡拓扑\tau_{tri}：仅由\varnothing和X组成的族是X的一个拓扑

拓扑空间（开集公理）：=

1.\varnothing ,X

2.任意并封闭

3.有限交封闭

例3

证明余有限拓扑\tau_{cof}符合开集公理

\tau_{cof}=\left \{ u\subseteq X:X-u \ is \ finite \right \}\cup\left \{ \varnothing \right \}​​​​​​​

(complement finite topology)

Proof:

1.X-X=\varnothing \ is \ finite\ , so\ \varnothing,X \in \tau_{cof}

2.let \left \{ u_i:i\in I,u_i\in\tau_{cof},u_i \neq\varnothing \right \}

firstly,\ \varnothing \cup u_i=\cup u_i,then\ consider \ the \ \cup u_i

X-\cup u_i=\cap(X-u_i)\ is \ finite\ , \ from \ X-u_i \ is \ finite

so\ \cup u_i \in \tau_{cof}

3.let \ non empty \ u_1,u_2,...,u_n\in\tau_{cof}

firstly,\varnothing \cap u_i=\varnothing

X-\bigcap_{i=1}^{n}u_i=\bigcup_{i=1}^{n}(X-u_i),X-u_i\ is \ finite

finite\ cup\ of\ finite\ set\ is \ finite

so \ \bigcap_{i=1}^{n}u_i \in \tau_{cof}

In \ sum,\tau_{cof}\ is\ a\ topo.​​​​​​​

例4

证明余可数拓扑\tau_{coc}符合开集公理

\tau_{coc}=\left \{ u\subseteq X:X-u \ is \ countable \right \}\cup\left \{ \varnothing \right \}

Proof:

1.X-X=\varnothing \ is \ countable\ , so\ \varnothing,X \in \tau_{coc}

2.let \left \{ u_i:i\in I,u_i\in\tau_{coc},u_i \neq\varnothing \right \}c

firstly,\ \varnothing \cup u_i=\cup u_i,then\ consider \ the \ \cup u_i

X-\cup u_i=\cap(X-u_i)\ is \ countable\ , \ from \ X-u_i \ is \ countable

so\ \cup u_i \in \tau_{coc}

3.let \ non empty \ u_1,u_2,...,u_n\in\tau_{coc}

firstly,\varnothing \cap u_i=\varnothing

X-\bigcap_{i=1}^{n}u_i=\bigcup_{i=1}^{n}(X-u_i),X-u_i\ is \ countable

countable\ cup\ of\ countable\ set\ is \ countable

so \ \bigcap_{i=1}^{n}u_i \in \tau_{coc}

In \ sum,\tau_{coc}\ is\ a\ topo.​​​​​​​

例5

E^2的子集A=\left \{ (x,sin\frac{1}{x}) :x\in (0,1)\right \}，求\bar{A}

设A是拓扑空间X的子集，x\in X

聚点：x的每个领域都含有A\setminus \left \{ x \right \}中的点，则称x为A的聚点

导集：A的所有聚点的集合，记A{}'

闭包：\bar{A}=A\cup A{}'

solution:

\bar{A}=\left \{ (x,sin\frac{1}{x}) :x\in (0,1]\right \}\cup\left \{ (0,y):y\in[-1,1] \right \}

 proof\ of \ \left \{ (0,y):y\in[-1,1] \right \}\subseteq A{}'\ as \ follow:

\forall (0,y),y\in[-1,1],\exists x_0\in[0,2\pi],sinx_0=y

let\ a_n=\frac{1}{x_0+2n\pi},\forall u\in \mathcal{N}_{(0,y)},\exists n\in \mathbb{N},(a_n,sin\frac{1}{a_n})\in u

therefore, (0,y)\in A'

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