方企勤 第五章 多元函数微分学 第2题
📝 题目
例 2 设 $A \subset {\mathbf{R}}^{m},B \subset {\mathbf{R}}^{m}$ ,证明:
(1) ${\left( A \cap B\right) }^{ \circ } = {A}^{ \circ } \cap {B}^{ \circ }$ ； (2) $\overline{A \cup B} = \overline{A} \cup \overline{B}$ .
💡 答案解析
证明 (1) 证法 $1\forall x \in {\left( A \cap B\right) }^{ \circ }$ ,由内点定义, $\exists$ 邻域 $U(x$ ; $\delta ) \subset A \cap B$ ,因而 $U\left( {x;\delta }\right) \subset A,U\left( {x;\delta }\right) \subset B$ ,故 $x \in {A}^{ \circ },x \in {B}^{ \circ }$ ,所以 $\mathbf{x} \in {A}^{ \circ } \cap {B}^{ \circ }$ ,即得
$$ {\left( A \cap B\right) }^{ \circ } \subset {A}^{ \circ } \cap {B}^{ \circ }\text{ . } \tag{1.1} $$
$\forall x \in {A}^{ \circ } \cap {B}^{ \circ }$ ,这等价于 $x \in {A}^{ \circ },x \in {B}^{ \circ }$ ,由内点定义, $\exists U\left( {x;{\delta }_{1}}\right)$ $\subset A,U\left( {x;{\delta }_{2}}\right) \subset B$ ,取 $= \min \left( {{\delta }_{1},{\delta }_{2}}\right) > 0$ ,有 $U\left( {x;\delta }\right) \subset A \cap B$ ,所以 $x \in$ ${\left( A \cap B\right) }^{ \circ }$ ,即得
$$ {A}^{ \circ } \cap {B}^{ \circ } \subset {\left( A \cap B\right) }^{ \circ }\text{ . } \tag{1.2} $$
由 (1.1) 与 (1.2) 式便得 ${\left( A \cap B\right) }^{ \circ } = {A}^{ \circ } \cap {B}^{ \circ }$ .
证法 2 因 ${A}^{ \circ } \subset A,{B}^{ \circ } \subset B$ ,所以 ${A}^{ \circ } \cap {B}^{ \circ } \subset A \cap B$ ,由于 ${A}^{ \circ } \cap {B}^{ \circ }$ 为开集,推出 ${A}^{ \circ } \cap {B}^{ \circ } = {\left( {A}^{ \circ } \cap {B}^{ \circ }\right) }^{ \circ } \subset {\left( A \cap B\right) }^{ \circ }$ . 反之,由 $A \cap B \subset A,A \cap$ $B \subset B$ ,可得 ${\left( A \cap B\right) }^{ \circ } \subset {A}^{ \circ },{\left( A \cap B\right) }^{ \circ } \subset {B}^{ \circ }$ ,从而推出 ${\left( A \cap B\right) }^{ \circ } \subset {A}^{ \circ } \cap$ ${B}^{ \circ }$ . 最后即得 ${\left( A \cap B\right) }^{ \circ } = {A}^{ \circ } \cap {B}^{ \circ }$ .
(2)类似于上面用聚点定义和闭包性质两种证法外,还可用开闭集的关系来证.
因为 $\bar{E} = {\left( {\left( {E}^{c}\right) }^{ \circ }\right) }^{c}$ ,所以
$$ \bar{A} \cup \bar{B} = {\left( {\left( {A}^{c}\right) }^{ \circ }\right) }^{c} \cup {\left( {\left( {B}^{c}\right) }^{ \circ }\right) }^{c} = {\left( {\left( {A}^{c}\right) }^{ \circ } \cap {\left( {B}^{c}\right) }^{ \circ }\right) }^{c} $$
$$ = {\left( {\left( {A}^{c} \cap {B}^{c}\right) }^{ \circ }\right) }^{c} = {\left( {\left( {\left( A \cup B\right) }^{c}\right) }^{ \circ }\right) }^{c} = \overline{A \cup B}\text{ . } $$