## 2018年10月24日水曜日

### 数学 - 集合と位相 - 位相空間 - R^nの距離と位相(内部と閉包)

1. $A=\left\{x\in {\text{ℝ}}^{2}|x=\left({x}_{1},{x}_{2}\right)\wedge {x}_{1}^{2}-4{x}_{2}\ge 0\right\}$

よって、 内部は

${A}^{0}=\left\{x\in {\text{ℝ}}^{2}|x=\left({x}_{1},{x}_{2}\right)\wedge {x}_{1}^{2}-4{x}_{2}>0\right\}$

閉包は

$\stackrel{-}{A}=\left\{x\in {\text{ℝ}}^{2}|x=\left({x}_{1},{x}_{2}\right)\wedge {x}_{1}^{2}-4{x}_{2}\ge 0\right\}$

2. $\begin{array}{}B=\left\{x\in {\text{ℝ}}^{2}|{x}_{1}^{2}-4{x}_{2}<0\right\}\\ {B}^{0}=\left\{x\in {\text{ℝ}}^{2}|{x}_{1}^{2}-4{x}_{2}<0\right\}\\ \stackrel{-}{B}=\left\{x\in {\text{ℝ}}^{2}|{x}_{1}^{2}-4{x}_{2}\le 0\right\}\end{array}$

3. $\begin{array}{}{C}^{0}=\varphi \\ \stackrel{-}{C}={\text{ℝ}}^{2}\end{array}$

4. $\begin{array}{}{D}^{0}=\varphi \\ \stackrel{-}{D}=\left\{x|0\le {x}_{1}={x}_{2}\le 1\right\}\end{array}$

5. $\begin{array}{}{x}_{1}>0,b={x}_{1}-a,0

よって、集合 E とその内部、閉包はそれぞれ

$\begin{array}{}E=\left\{x\in {\text{ℝ}}^{2}|{x}_{1}>0\wedge \frac{1}{2}{x}_{1}^{2}\le {x}_{2}<{x}_{1}^{2}\right\}\\ {E}^{0}=\left\{x\in {\text{ℝ}}^{2}|{x}_{1}>0\wedge \frac{1}{2}{x}_{1}^{2}<{x}_{2}<{x}_{1}^{2}\right\}\\ \stackrel{-}{E}=\left|x\in {\text{ℝ}}^{2}\right|{x}_{1}\ge 0\wedge \frac{1}{2}{x}_{1}^{2}\le {x}_{2}\le {x}_{1}^{2}\right\}\end{array}$

となる。