# 凸优化-对偶问题（Convex Optimization-Duality）

2022-05-19 04:31:42  阅读：189  来源： 互联网

# 凸优化

## Duality

$\begin{array}{*{20}{c}} {}&{\min {f_0}\left( x \right)}&{}\\ {s.t.}&{{f_i}\left( x \right) \le 0,i = 1, \ldots ,m}&{}\\ {}&{{h_i}\left( x \right) \le 0,i = 1, \ldots ,p}&{} \end{array} \tag{1}$

### Lagrange function（拉格朗日函数）

#### 定义

$L\left(x,\lambda,\nu \right)=f_{0}\left(x\right)+\sum_{i=1}^{m}\lambda_{i}f_{i}\left(x\right)+\sum_{i=1}^{p}\nu_{i}h_{i}\left(x\right)$

### Lagrange duality function（拉格朗日对偶函数）

#### 定义

$g\left(\lambda,\nu\right)=\mathop {\inf }\limits_{x \in D}L\left(x,\lambda,\nu \right)=\mathop {\inf }\limits_{x \in D}\left(f_{0}\left(x\right)+\sum_{i=1}^{m}\lambda_{i}f_{i}\left(x\right)+\sum_{i=1}^{p}\nu_{i}h_{i}\left(x\right)\right)$

#### 性质

• Lagrange duality function 是一族关于 $$\left(\lambda,\nu\right)$$ 的一族仿射函数的逐点下确界，对偶函数一定是凹函数。

Reference：

$g\left(\theta \lambda_{1}+\left(1- \theta \right)\lambda_{2},\theta \nu_{1}+\left(1- \theta \right)\nu_{2}\right) \geq \theta g\left(\lambda_{1},\nu_{1}\right)+\left(1- \theta \right)g\left(\lambda_{2},\nu_{2}\right) \\ g\left(\lambda,\nu\right)=min\left\{L\left(x_{1},\lambda,\nu \right),L\left(x_{2},\lambda,\nu \right),...,L\left(x_{n},\lambda,\nu \right) \right\}$

\begin{align} & g\left( \theta {{\lambda }_{1}}+\left( 1-\theta \right){{\lambda }_{2}},\theta {{\nu }_{1}}+\left( 1-\theta \right){{\nu }_{2}} \right) \\&=min\left\{ L\left( {{x}_{1}},\theta {{\lambda }_{1}}+\left( 1-\theta \right){{\lambda }_{2}},\theta {{\nu }_{1}}+\left( 1-\theta \right){{\nu }_{2}} \right),L\left( {{x}_{2}},\theta {{\lambda }_{1}}+\left( 1-\theta \right){{\lambda }_{2}},\theta {{\nu }_{1}}+\left( 1-\theta \right){{\nu }_{2}} \right),...,L\left( {{x}_{n}},\theta {{\lambda }_{1}}+\left( 1-\theta \right){{\lambda }_{2}},\theta {{\nu }_{1}}+\left( 1-\theta \right){{\nu }_{2}} \right) \right\} \\ & \ge min\left\{ \theta L\left( {{x}_{1}},{{\lambda }_{1}},{{\nu }_{1}} \right)+\left( 1-\theta \right)L\left( {{x}_{1}},{{\lambda }_{2}},{{\nu }_{2}} \right),\theta L\left( {{x}_{2}},{{\lambda }_{1}},{{\nu }_{1}} \right)+\left( 1-\theta \right)L\left( {{x}_{2}},{{\lambda }_{2}},{{\nu }_{2}} \right),...,\theta L\left( {{x}_{n}},{{\lambda }_{1}},{{\nu }_{1}} \right)+\left( 1-\theta \right)L\left( {{x}_{n}},{{\lambda }_{2}},{{\nu }_{2}} \right) \right\} \\ & \ge \theta min\left\{ L\left( {{x}_{1}},{{\lambda }_{1}},{{\nu }_{1}} \right),L\left( {{x}_{2}},{{\lambda }_{1}},{{\nu }_{1}} \right),...,L\left( {{x}_{n}},{{\lambda }_{1}},{{\nu }_{1}} \right) \right\}+\left( 1-\theta \right)min\left\{ L\left( {{x}_{1}},{{\lambda }_{2}},{{\nu }_{2}} \right),L\left( {{x}_{2}},{{\lambda }_{2}},{{\nu }_{2}} \right),...,L\left( {{x}_{n}},{{\lambda }_{2}},{{\nu }_{2}} \right) \right\} \\ & \ge \theta g\left( {{\lambda }_{1}},{{\nu }_{1}} \right)+\left( 1-\theta \right)g\left( {{\lambda }_{2}},{{\nu }_{2}} \right) \end{align}

• Lagrange duality function 提供了原问题 $$\left(1\right)$$ 的最优值 $$p^{*}$$ 的下界，对任意 $$\lambda \geq 0$$ 和 $$\nu$$ 下式成立

$g\left(\lambda,\nu\right)\leq p^{*}$

$$$\begin{split} g\left(\lambda,\nu\right)&=\mathop {\inf }\limits_{x \in D}L\left(x,\lambda,\nu \right) \\&=\mathop {\inf }\limits_{x \in D}\left(f_{0}\left(x\right)+\sum_{i=1}^{m}\lambda_{i}f_{i}\left(x\right)+\sum_{i=1}^{p}\nu_{i}h_{i}\left(x\right)\right)\\ & \leq f_{0}\left(\overset{\sim }{\mathop{x}}\right)+\sum_{i=1}^{m}\lambda_{i}f_{i}\left(\overset{\sim }{\mathop{x}}\right)+\sum_{i=1}^{p}\nu_{i}h_{i}\left(\overset{\sim }{\mathop{x}}\right)\\ {\color{Blue}\because \lambda_{i} \geq 0,f_{i}\left(x\right) \leq 0, \therefore} &\leq f_{0}\left(\overset{\sim }{\mathop{x}}\right)\\ &=p^{*} \end{split}$$$

#### Lagrange duality function 和 Conjugate function（共轭函数）

##### Conjugate function 定义

$$f:R^{n}\rightarrow R$$，则其共轭函数 $$f^{*}$$ 为

$f^{*}\left(y\right)=\mathop {\sup }\limits_{x \in D}\left\{ y^{T}x-f\left(x\right)\right\}$

##### Lagrange duality function 与 Conjugate function 的关系

$\begin{array}{*{20}{c}} {}&{\min {f_0}\left( x \right)}&{}\\ {s.t.}&{Ax\leq b}&{}\\ {}&{Cx=d}&{} \end{array} \tag{2}$

$\begin{array}{*{20}{c}} {}&{\min {f_0}\left( x \right)}&{}\\ {s.t.}&{Ax-b\leq 0}&{}\\ {}&{Cx-d=0}&{} \end{array} \tag{3}$

\begin{align*} & g\left( \lambda ,\nu \right)=\underset{x\in D}{\mathop{\inf }}\,L\left( x,\lambda ,\nu \right) \\ & =\underset{x\in D}{\mathop{\inf }}\,\left( {{f}_{0}}\left( x \right)+{{\lambda }^{T}}\left( Ax-b \right)+{{\nu }^{T}}\left( Cx-d \right) \right) \\ & =\underset{x\in D}{\mathop{\inf }}\,\left( {{f}_{0}}\left( x \right)+{{\lambda }^{T}}Ax+{{\nu }^{T}}Cx \right)-{{\lambda }^{T}}b-{{\nu }^{T}}d \\ & =-\underset{x\in D}{\mathop{\sup }}\,-\left( {{f}_{0}}\left( x \right)+{{\lambda }^{T}}Ax+{{\nu }^{T}}Cx \right)-{{\lambda }^{T}}b-{{\nu }^{T}}d \\ & =-\underset{x\in D}{\mathop{\sup }}\,\left( {{\left( -{{A}^{T}}\lambda -{{C}^{T}}\nu \right)}^{T}}x-{{f}_{0}}\left( x \right) \right)-{{\lambda }^{T}}b-{{\nu }^{T}}d \\ & =-f_{0}^{*}\left( -{{A}^{T}}\lambda -{{C}^{T}}\nu \right)-{{\lambda }^{T}}b-{{\nu }^{T}}d \end{align*}

$\underset{x\in D}{\mathop{\inf }}\,S\left( x \right)=-\underset{x\in D}{\mathop{\sup }}\,-S\left( x \right)$

### Lagrange 对偶问题

• 原问题：

$\begin{array}{*{20}{c}} {}&{\min {f_0}\left( x \right)}&{}\\ {s.t.}&{{f_i}\left( x \right) \le 0,i = 1, \ldots ,m}&{}\\ {}&{{h_i}\left( x \right) \le 0,i = 1, \ldots ,p}&{} \end{array} \tag{1}$

• 对任意 $$\lambda \geq 0$$ 以及 $$\nu$$ ，Lagrange duality function 提供了原问题最优值的一个下界

$g\left(\lambda,\nu\right)\leq p^{*}$

#### 定义

$\begin{array}{*{20}{c}} {}&{\max {g\left(\lambda,\nu\right)}}&{}\\ {s.t.}&{\lambda \ge 0}&{}\tag{4} \end{array}$

#### Weak Duality (弱对偶性)

$$$\begin{split} g\left(\lambda,\nu\right) \leq p^{*}\\ max \text { }g\left(\lambda,\nu\right) \leq p^{*}\\ d^{*} \leq p^{*}\\ \end{split}$$$

！！！注意，到现在为止，我们讨论的优化问题并不局限于凸问题，例如即使原问题不是凸问题，弱对偶性仍然成立

#### Strong Duality (强对偶性)

$d^{*} = p^{*}$

#### 强对偶性的充分条件：Slater 条件 + convex function

##### 仿射集合

$V = C-x_{0} =\left\{x-x_{0}|x\in C\right\}$

$$$\begin{split} \alpha v_{1}+\beta v_{1}+x_{0}=\alpha \left(v_{1}+x_{0}\right)+\beta \left(v_{2}+x_{0}\right)+\left(1-\alpha-\beta\right)x_{0}&\Rightarrow \alpha v_{1}+\beta v_{1}+x_{0}\in C\\ &\Rightarrow \alpha v_{1}+\beta v_{1}\in C-x_{0}\\ &\Rightarrow \alpha v_{1}+\beta v_{1}\in V\\ \end{split}$$$

$C = V + x_{0} =\left\{v+x_{0}|v\in V\right\}$

##### 仿射包

$C = \left\{\theta_{1}x_{1}+\theta_{2}x_{2}+...+\theta_{k}x_{k}|\forall x_{1},x_{2},...,x_{k} \in C,\theta_{1}+\theta_{2}+...+\theta_{k}=1 \right\}$

$$aff \text{ }C$$ 是包含 $$C$$ 的最小的仿射集合。也就是说：如果 $$S$$ 是满足 $$C \subseteq S$$ 的仿射集合，那么 $$aff \text{ }C\subseteq S$$

##### 相对内点集

$relint \text{ }D =\left\{x \in D|\exists r >0,s.t.B\left(x,r\right)\cap aff\text{ }D \subseteq D\right\}$

##### Slater 条件

$\begin{split} \exist x \in relint \text{ }D,s.t. f_{i}(x)&<0,i=1,...,m\\ h_{i}(x)&=0,i=1,...,p \end{split}$

##### 修正 Slater 条件

$$$\begin{split} \exist x \in relint \text{ }D,s.t. f_{i}(x)&\leq 0,i=1,...,k,f_{i}\left(x\right)\text{为仿射函数} \\f_{i}(x)&<0,i=k+1,...,m,f_{i}\left(x\right)\text{为非线性函数} \\h_{i}(x)&=0,i=1,...,p \end{split}$$$

#### 证明Slater 条件 + convex function $\Rightarrow$ 强对偶性

##### 支撑超平面

$x_{0} \in bd \text{ }C=cl\text{ }C\setminus int \text{ }C$

##### 超平面分离定理

$g=\left\{ \underbrace{\left( {{f}_{1}}\left( x \right),...,{{f}_{m}}\left( x \right) \right)}_{u},\underbrace{\left( {{h}_{1}}\left( x \right),...,{{h}_{p}}\left( x \right) \right),}_{u}\underbrace{{{f}_{0}}\left( x \right)}_{t}|x\in D \right\}$

$$$\begin{split} \Alpha =\{\left( u,v,t \right)|\exists x\in D,{{f}_{0}}\left( x \right)&\le {{u}_{i}},i=1,...,m\\ {{h}_{i}} \left( x \right)&\le {{v}_{i}},i=1,...,p,\\ {{f}_{0}} \left ( x \right)&\le t\} \in g +\left(R_{+}^{m} \times \left\{0\right\} \times R_{+}^{p}\right) \end{split}$$$

$$$\begin{split} \left( u,v,t \right) \in B &\Rightarrow u=0, v=0, \color{Blue}{t<p^{*}}\\ \left( u,v,t \right) \in A &\Rightarrow \exists x\in D,{{f}_{i}}\left( x \right)\le {{u}_{i}}=0,{{h}_{i}}\left( x \right)=0,\color{Blue}{{{f}_{0}}\left( x \right)=t\ge {{p}^{*}}} \end{split}$$$

$$$\begin{split} \exists \left( \widetilde{\lambda },\widetilde{\nu },\mu \right)\ne 0,\alpha \in R\\ {\color{red}\forall\left( u,v,t \right) \in A } &{\color{red}\Rightarrow \left( \widetilde{\lambda },\widetilde{\nu },\mu\right) ^{T} \left( u,v,t \right) \geq \alpha}\\ &{\color{red}\Rightarrow \sum_{i=1}^{m}\widetilde{\lambda }_{i}f_{i}\left(x\right)+\sum_{i=1}^{p}\widetilde{\nu }_{i}h_{i}\left(x\right)+\mu f_{0}\left(x\right) \geq \alpha}\\ {\color{green}\forall\left( u,v,t \right) \in B} &{\color{green}\Rightarrow \left( \widetilde{\lambda },\widetilde{\nu },\mu \right) ^{T} \left( u,v,t \right) \leq \alpha}\\ &{\color{green}\Rightarrow \mu t\leq \alpha} \end{split}$$$

$\mu t\leq \mu p^{*}\leq\alpha$

$$$\begin{split} {\color{red}\forall\left( u,v,t \right) \in A } &{\color{red}\Rightarrow \left( \widetilde{\lambda },\widetilde{\nu },\mu\right) ^{T} \left( u,v,t \right) \geq \alpha}\\ &{\color{red}\Rightarrow \sum_{i=1}^{m}\widetilde{\lambda }_{i}f_{i}\left(x\right)+\sum_{i=1}^{p}\widetilde{\nu }_{i}h_{i}\left(x\right)+\mu f_{0}\left(x\right) \geq \alpha}\\ &\Rightarrow \sum_{i=1}^{m}\widetilde{\lambda }_{i}f_{i}\left(x\right)+\sum_{i=1}^{p}\widetilde{\nu }_{i}h_{i}\left(x\right)+\mu f_{0}\left(x\right) \geq \mu p^{*}\\ &\Rightarrow \sum_{i=1}^{m}\frac{\widetilde{\lambda }_{i}}{\mu}f_{i}\left(x\right)+\sum_{i=1}^{p}\frac{\widetilde{\nu }_{i}}{\mu}h_{i}\left(x\right)+f_{0}\left(x\right) \geq p^{*}\\ 令L\left(x,\frac{\widetilde{\lambda }_{i}}{\mu},\frac{\widetilde{\nu }_{i}}{\mu}\right)=L\left(x,\lambda,\nu\right)&\Rightarrow L\left(x,\lambda,\nu\right) \geq p^{*}\\ \end{split}$$$

$g\left(\lambda,\nu\right) \leq p^{*}$

$g\left(\lambda,\nu\right) = p^{*}$

$$$\begin{split} {\color{red}\sum_{i=1}^{m}\widetilde{\lambda }_{i}f_{i}\left(\widetilde{x}_{0}\right)+\sum_{i=1}^{p}\widetilde{\nu }_{i}h_{i}\left(\widetilde{x}_{0}\right)+\mu f_{0}\left(\widetilde{x}_{0}\right) }&\geq 0\\ (\mu=0) \Rightarrow \sum_{i=1}^{m}\widetilde{\lambda }_{i}f_{i}\left(\widetilde{x}_{0}\right)+\sum_{i=1}^{p}\widetilde{\nu }_{i}h_{i}\left(\widetilde{x}_{0}\right)&\geq 0\\ (\widetilde{x}_{0}\in D 为满足 slater 条件的一点,h_{i}\left(\widetilde{x}_{0}\right)=0) \Rightarrow \sum_{i=1}^{m}\widetilde{\lambda }_{i}f_{i}\left(\widetilde{x}_{0}\right)&\geq 0\\ (\widetilde{x}_{0}\in D 为满足 slater 条件的一点,f_{i}\left(\widetilde{x}_{0}\right)<0,\widetilde{\lambda }_{i}\geq 0) \Rightarrow \widetilde{\lambda }_{i}&= 0 \end{split}$$$

$$$\begin{split} {\color{red}\sum_{i=1}^{m}\widetilde{\lambda }_{i}f_{i}\left(x\right)+\sum_{i=1}^{p}\widetilde{\nu }_{i}h_{i}\left(x\right)+\mu f_{0}\left(x\right) }&\geq 0\\ \sum_{i=1}^{p}\widetilde{\nu }_{i}h_{i}\left(x\right)&\geq 0\\ \widetilde{\nu }^{T}\left(Ax-b\right)&\geq 0 \end{split}$$$

$\exists \varepsilon ,x=\widetilde{x}_{0}-\varepsilon A^T\widetilde{\nu }\in D,s.t.\widetilde{\nu }^T(A(\widetilde{x}_{0}-\varepsilon A^T\widetilde{\nu })-b)＝-\varepsilon \widetilde{\nu }^TAA^T\widetilde{\nu }\leq0$

Reference：

### 最优性条件

#### 次优解认证和终止准则

$$$\begin{split} p^{*}&\geq L\left(x,\lambda,\nu \right)\\ &\geq \mathop {\inf }\limits_{x \in D}L\left(x,\lambda,\nu \right)\\ &= g\left(\lambda,\nu\right)\\ \\ \Rightarrow f_{0}\left(\overset{\sim }{\mathop{x}}\right)-p^{*}&\leq f_{0}\left(\overset{\sim }{\mathop{x}}\right)- g\left(\lambda,\nu\right) \end{split}$$$

$p^{*}\in\left [ g\left(\lambda,\nu\right), f_{0}\left(x\right)\right],d^{*}\in\left [ g\left(\lambda,\nu\right), f_{0}\left(x\right)\right]$

$f_{0}\left(x_{k}\right)- g\left(\lambda_{k},\nu_{k}\right)\leq \varepsilon_{abs}$

$$if \text{ }g\left(\lambda_{k},\nu_{k}\right)>0:$$

$\frac{f_{0}\left(x_{k}\right)- p^{*}}{\left |p^{*} \right |}\leq \frac{f_{0}\left(x_{k}\right)- g\left(\lambda_{k},\nu_{k}\right)}{g\left(\lambda_{k},\nu_{k}\right)}\leq \varepsilon_{abs}\\$

$$if \text{ } f_{0}\left(x_{k}\right)<0:$$

$\frac{f_{0}\left(x_{k}\right)- p^{*}}{\left |p^{*} \right |}\leq \frac{f_{0}\left(x_{k}\right)- g\left(\lambda_{k},\nu_{k}\right)}{-f_{0}\left(x_{k}\right)}\leq \varepsilon_{abs}\\$

#### 互补松驰性

$\lambda_{i}^{*}f_{i}^{*}\left(x\right)=0,i=1,...,m$

\begin{align*} & g\left( {{\lambda }^{*}},{{\nu }^{*}} \right)=f_{0}^{*}\left( {{x}^{*}} \right) \\ & \ge \underset{\lambda ,\nu }{\mathop{\max }}\,\left( {{f}_{0}}\left( {{x}^{*}} \right)+\sum\limits_{i=1}^{m}{{{\lambda }_{i}}}{{f}_{i}}\left( {{x}^{*}} \right)\left( \le 0 \right)+\sum\limits_{i=1}^{p}{{{\nu }_{i}}}{{h}_{i}}\left( {{x}^{*}} \right) \right) \\ & =\underset{\lambda ,\nu }{\mathop{\max }}\,L\left( {{x}^{*}},\lambda ,\nu \right) \\ & \ge {{f}_{0}}\left( {{x}^{*}} \right)+\sum\limits_{i=1}^{m}{\lambda _{i}^{*}}{{f}_{i}}\left( {{x}^{*}} \right)+\sum\limits_{i=1}^{p}{\nu _{i}^{*}}{{h}_{i}}\left( {{x}^{*}} \right) \\ & \ge inf\text{ }L\left( x,{{\lambda }^{*}},{{\nu }^{*}} \right) \\ & \ge g\left( {{\lambda }^{*}},{{\nu }^{*}} \right) \end{align*}

$$$\begin{split} \color{red}{g\left(\lambda^{*},\nu^{*}\right)} &= f_{0}\left(x^{*}\right)\\ &= \mathop {\max }\limits_{\lambda,\nu}\left(f_{0}\left(x^{*}\right)+\sum_{i=1}^{m}\lambda_{i}f_{i}\left(x^{*}\right)+\sum_{i=1}^{p}\nu_{i}h_{i}\left(x^{*}\right)\right)\\ &= \mathop {\max }\limits_{\lambda,\nu}L\left(x^{*},\lambda,\nu \right)\\ &= \color{red}{f_{0}\left(x^{*}\right)+\sum_{i=1}^{m}\lambda_{i}^{*}f_{i}\left(x^{*}\right)+\sum_{i=1}^{p}\nu_{i}^{*}h_{i}\left(x^{*}\right)}\\ &= inf \text{ }L\left(x,\lambda^{*},\nu^{*}\right)\\ &= g\left(\lambda^{*},\nu^{*}\right) \end{split}$$$

$f_{0}\left(x^{*}\right)=g\left(\lambda^{*},\nu^{*}\right)=L\left(x^{*},\lambda^{*},\nu^{*}\right)=f_{0}\left(x^{*}\right)+{\color{red}\sum_{i=1}^{m}\lambda_{i}^{*}f_{i}\left(x^{*}\right)}+\sum_{i=1}^{p}\nu_{i}^{*}h_{i}\left(x^{*}\right)$

$\sum_{i}^{m}\lambda_{i}^{*}f_{i}^{*}\left(x\right)=0\\ \Rightarrow\lambda_{i}^{*}f_{i}^{*}\left(x\right)=0,i=1,...,m$

#### KKT 最优性条件

KKT 最优性条件如下

$$$\begin{split} f_{i}^{*}\left(x\right)&\leq0,i=1,...,m\\ h_{i}^{*}\left(x\right)&\leq0,i=1,...,p\\ \lambda_{i}^{*}&\geq0,i=1,...,m\\ \lambda_{i}^{*}f_{i}^{*}\left(x\right)&=0,i=1,...,m\\ \bigtriangledown f_{0}\left ( x^{*} \right )+\sum_{i}^{m}\lambda_{i}^{*}\bigtriangledown f_{i}\left(x^{*}\right)+\sum_{i}^{p}\nu_{i}^{*}\bigtriangledown h_{i}\left(x^{*}\right)&=0\\ \end{split}$$$

• 强对偶条件 $$p^{*}=d^{*}$$，$$f_{i}$$，$$h_{i}$$ 可微 $$\Rightarrow$$ KKT 条件

$$$\begin{split} f_{i}^{*}\left(x\right)&\leq0,i=1,...,m\\ h_{i}^{*}\left(x\right)&\leq0,i=1,...,p\\ \lambda_{i}^{*}&\geq0,i=1,...,m\\ \lambda_{i}^{*}f_{i}^{*}\left(x\right)&=0,i=1,...,m \end{split}$$$

$$$\begin{split} g\left(\lambda^{*},\nu^{*}\right)&=L\left(x^{*},\lambda^{*},\nu^{*}\right)\\ &=inf \text{ }L\left(x,\lambda^{*},\nu^{*}\right) \end{split}$$$

$\bigtriangledown _{x^{*}} L = \bigtriangledown f_{0}\left ( x^{*} \right )+\sum_{i}^{m}\lambda_{i}^{*}\bigtriangledown f_{i}\left(x^{*}\right)+\sum_{i}^{p}\nu_{i}^{*}\bigtriangledown h_{i}\left(x^{*}\right)=0$

• KKT 条件， $$f_{i}$$，$$h_{i}$$ 可微 + 凸问题 $$\Rightarrow$$ 强对偶条件 $$p^{*}=d^{*}$$

$$$\begin{split} g\left(\lambda^{*},\nu^{*}\right)&=L\left(x^{*},\lambda^{*},\nu^{*}\right)\\ &=f_{0}\left(x^{*}\right)+\sum_{i=1}^{m}\lambda_{i}^{*}f_{i}\left(x^{*}\right)+\sum_{i=1}^{p}\nu_{i}^{*}h_{i}\left(x^{*}\right)\\ &=f_{0}\left(x^{*}\right) \end{split}$$$

## 度量拓扑

• 任意个开集的并集是开集
• 有限个开集的交集是开集

### 闭集

#### 性质

• $$\varnothing$$ 和 $$X$$ 是闭集
• 任意个闭集的交集是闭集
• 有限个闭集的并集是闭集

（1）$$\left(\cup_{\alpha\in\Lambda}A_{\alpha}\right)^{c}=\cap_{\alpha\in\Lambda}A_{\alpha}^{c}$$

（2）$$\left(\cap_{\alpha\in\Lambda}A_{\alpha}\right)^{c}=\cup_{\alpha\in\Lambda}A_{\alpha}^{c}$$

### 极限点

#### 性质

• $$x_{0}$$ 为 $$F$$ 的极限点
• 包含 $$x_{0}$$ 任何一个开集都含有 $$F$$ 异于 $$x_{0}$$ 的无穷多个点（若有限，则在有限个点中取半径 $$min \text{ }d$$ 的开球，不满足极限点定义）
• 在 $$F$$ 中存在序列 $$x_{n}$$， $$x_{n}\neq x_{0}$$，且 $$\lim_{n\rightarrow \infty }x_{n} = x_{0}$$（注意，一定是 $$x_{n}\neq x_{0}$$，因为极限点要求任意开集都含有不同于 $$x_{0}$$ 的点）

### 导集与闭包

#### 性质

1. 下列条件等价：
• $$F$$ 是闭集
• $$F^{’} \subset F$$
• $$\bar{F} = F$$
1. 设 $$\left(X,d\right)$$ 是度量空间，$$F$$ 是 $$X$$ 的子集，$$x \in X$$ ，则下列条件等价：
• $$x \in \bar{F}$$

• $$x$$ 的每个开球都包含有 $$F$$ 的点

• 存在序列 $$\left\{x_{n}\right\}$$，使得 $$\lim_{n\rightarrow \infty }x_{n} = x_{0}$$（注意，这时不一定是 $$x_{n}\neq x_{0}$$，因为闭包包括$$F$$ 本身的点）

### 内部

#### 性质

• $$G$$ 是开集当且仅当 $$G^{0}=G$$
• $$G^{0}\subseteq G \subseteq \bar G$$（第一个关系 $$G^{0}\subseteq G$$，是因为 $$G$$ 可能包含孤立点；第一个关系 $$G \subseteq \bar G$$，是因为 $$\bar G$$ 包含了极限点（更形象去理解为边界点，因为内点也是极限点）
• 当 $$G \subset F$$ 时，一定有 $$G^{0} \subset F^{0}$$，$$\bar G\ \subset \bar F$$

Reference

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