Topo Homework - Week 3

T1

T1. (ER) 确定下面每种情况中的 $\text{Int}(A), \text{Cl}(A)$ 和 $\partial A$ :

(1) 在下极限拓扑空间 $\mathbb{R}_l$ 中, $A = [0, 1]$ ;
(2) $X = \{a, b, c\}, \mathcal{T} = \{X, \emptyset, \{a\}, \{a, b\}\}, A = \{a, c\}$ ;
(3) 在欧氏直线 $\mathbb{R}$ 上, $A = (-1, 1) \cup \{2\}$ ;
(4) 在下极限拓扑空间 $\mathbb{R}_l$ 中, $A = (-1, 1) \cup \{2\}$ ;
(5) 在欧氏平面 $\mathbb{R}^2$ 上, $A = \{(\sin \theta, \cos \theta) \in \mathbb{R}^2 \mid 0 < \theta < \pi\}$ .

(1): $\operatorname{Int}A=[0,1),\operatorname{Cl}A=[0,1],\partial A=\{ 1 \}$

(2): $\operatorname{Int}A=\{ a \} ,\operatorname{Cl}A=X,\partial A=\{ b,c \}$

(3): $\operatorname{Int}A=(-1,1),\operatorname{Cl}A=[-1,1]\cup \{ 2 \} ,\partial A=\{ -1,1,2 \}$

(4): $\operatorname{Int}=(-1,1),\operatorname{Cl}A=[-1,1)\cup \{ 2 \} ,\partial A=\{ -1,2 \}$

(5): $\operatorname{Int}A=\varnothing,\operatorname{Cl}A=A\cup \{ (0,1),(0,-1) \},\partial A=A\cup \{ (0,1),(0,-1) \}$

T2

T5. (MRH) 在 $\mathbb{R}$ 上采用标准拓扑. 是否存在集合 $A \subset \mathbb{R}$ , 分别使得

(1) $A, \text{Cl}(A), \text{Int}(A)$ 和 $\text{Cl}(\text{Int}(A))$ 两两不同?

(2) $A, \text{Cl}(A), \text{Int}(A)$ 和 $\text{Int}(\text{Cl}(A))$ 两两不同?

(3) $A, \text{Cl}(A), \text{Int}(A), \text{Int}(\text{Cl}(A))$ 和 $\text{Cl}(\text{Int}(A))$ 两两不同?

(1):取 $A=Q\cup (0,1)$ ,则 $A=Q\cup (0,1),\operatorname{Cl}A=R,\operatorname{Int}A=(0,1),\operatorname{Cl}\operatorname{Int}A=[0,1]$

(2):取 $A=Q\cap (0,1)$ ,则 $\operatorname{Cl}A=[0,1],\operatorname{Int}A=\varnothing,\operatorname{Int}\operatorname{Cl}A=(0,1)$

(3):取 $A=((Q\cup (0,1))\cap (-1,2)) \cup (Q\cap (3,4))$ ,则

\begin{gathered} \operatorname{Int}A=(0,1) \\ \operatorname{Cl}A=[-1,2]\cup [3,4] \\ \operatorname{Int}\operatorname{Cl}A=(-1,2)\cup (3,4) \\ \operatorname{Cl} \operatorname{Int}A=[0,1] \end{gathered}

T3

T12. (MR) 设 $A$ 和 $B$ 都是拓扑空间 $X$ 上的稠密子集, 且 $A$ 是开集. 证明 $A \cap B$ 也是稠密子集.

\begin{gathered} A,B \text{ is dense} \\ \implies \forall x\in U_x\subset X, \\ \exists a\in U_a\subset A\cap U_x, \\ \exists b\in U_a\cap B \\ b\in U_a\subset A \implies b\in A\cap B \\ b\in U_x \implies A\cap B \text{ is dense} \end{gathered}

T4

T20. (ERH) 设 $X$ 为一个拓扑空间, $A \subset X$ . 证明:

(1) $\partial A$ 是闭集;
(2) $\partial A \cap \text{Int}(A) = \emptyset$ ;
(3) $\partial A \cup \text{Int}(A) = \text{Cl}(A)$ ;
(4) $\partial A \subset A$ 当且仅当 $A$ 是闭集;
(5) $\partial A \cap A = \emptyset$ 当且仅当 $A$ 是开集;
(6) $\partial A = \emptyset$ 当且仅当 $A$ 既是开集又是闭集.

(1):

\begin{gathered} \partial A=\operatorname{Cl}A\cap (X-\operatorname{Int}A) \\ \operatorname{Cl}A \text{ is close} \\ X-\operatorname{Int}A \text{ is close} \\ \implies \partial A \text{ is close} \end{gathered}

(2):

\begin{gathered} \partial A=(\operatorname{Cl}A-\operatorname{Int}A)\cap \operatorname{Int}A=\varnothing \end{gathered}

(3):

\begin{gathered} \partial A\cup \operatorname{Int}A=(\operatorname{Cl}A-\operatorname{Int}A)\cup \operatorname{Int}A=\operatorname{Cl}A \end{gathered}

(4):

\begin{gathered} \operatorname{Int}A\subset A \\ \implies \operatorname{Cl}A-\operatorname{Int}A\subset A \iff \operatorname{Cl}A\subset A \iff A \text{ is close} \end{gathered}

(5):

\begin{gathered} \partial A\cap A=\operatorname{Cl}A\cap (X-\operatorname{Int}A)\cap A \\ =A-A\cap \operatorname{Int}A \\ \implies \partial A\cap A=\varnothing \iff A\cap \operatorname{Int}A=A \iff A \text{ is open} \end{gathered}

(6):

By (4),(5):

\begin{gathered} A \text{ is open and close} \\ \iff \partial A\subset A \land \partial A\cap A=\varnothing \\ \iff \partial A = \varnothing \end{gathered}

T5

T2. (ER) 设 $f: X \to Y$ . 证明下列陈述等价:

(4) 对 $X$ 的任意子集 $A \subset X$ , $f(\text{Cl}(A)) \subset \text{Cl}(f(A))$ ;
(5) 对 $Y$ 的任意子集 $B \subset Y$ , $\text{Cl}(f^{-1}(B)) \subset f^{-1}(\text{Cl}(B))$ .

\begin{gathered} (4) \implies (5): \\ \text{let } A=f^{-1}(B) \\ \implies f(\operatorname{Cl} f^{-1} (B))\subset \operatorname{Cl}f(f^{-1}(B))\subset \operatorname{Cl}B \\ \implies \operatorname{Cl}f^{-1}(B) \subset f^{-1}(f(\operatorname{Cl}f^{-1} (B)))\subset f^{-1}(\operatorname{Cl}B) \\ (5) \implies (4): \\ \text{let } B=f(A) \\ \implies \operatorname{Cl}A\subset \operatorname{Cl}f^{-1}(f(A))\subset f^{-1}(\operatorname{Cl}f(A)) \\ \implies f(\operatorname{Cl}A)\subset f(f^{-1}(\operatorname{Cl}f(A)))\subset \operatorname{Cl}f(A) \end{gathered}

T6

\begin{gathered} X,Y \text{are topo spaces} ,A\subset X,B\subset Y \\ \implies \operatorname{Int}(A\times B)=\operatorname{Int}A\times \operatorname{Int}B \end{gathered}

设 $X,Y$ 的拓扑基分别是 $C,D$ ,则让

\begin{gathered} \operatorname{Int}A=\bigcup_{i\in S} C_i,\operatorname{Int} B=\bigcup_{i\in T} D_i \end{gathered}

左边由定义是被 $A\times B$ 包含的开集的并,等价于被 $A\times B$ 包含的拓扑基的并,而某个拓扑基有:

\begin{gathered} C_i\times D_j\subset A\times B \\ \iff C_i\subset A\land D_j\subset D \\ \iff C_i\subset \operatorname{Int}A\land D_j\subset \operatorname{Int}B \end{gathered}

所以 $A\times B$ 内的拓扑基恰好是所有 $C_i\times D_j$ .

而右侧是

\begin{gathered} \operatorname{Int}A\times \operatorname{Int}B \\ =\bigcup_{i\in S} C_i \times \bigcup_{i\in T} D_i \\ =\bigcup_{i\in S,j\in T} C_i\times D_j \end{gathered}

所以两侧相等.

T7

\begin{gathered} \overline{A\cap B}\subset \overline{ A } \cap \overline{ B } \\ \overline{ A\cup B } =\overline{ A } \cup \overline{ B } \end{gathered}

(1):

\begin{gathered} \overline{ A\cap B } \subset \overline{ A } ,\overline{ A\cap B } \subset \overline{ B } \implies \overline{ A\cap B } \subset \overline{ A } \cap \overline{ B } \end{gathered}

(2):

\begin{gathered} \overline{ A } \subset \overline{ A\cap B } ,\overline{ B } \subset \overline{ A\cap B } \\ \implies \overline{ A } \cup \overline{ B } \subset \overline{ A\cup B } \end{gathered}

对另一侧, $\forall x\in \overline{ A\cup B }$ ,假设 $X$ 不在 $A$ 和 $B$ 的闭包中,那么 $\exists x\in U_1,x\in U_2,U_1,U_2 \text{ is open},U_1\cap A=U_2\cap B=\varnothing$ ,则 $(U_1\cap U_2)\cap (A\cup B)=\varnothing$ ,推出 $x\notin \overline{ A\cup B }$ ,矛盾,得证.