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# System prepended metadata

title: Optimization Assignment Week 3
tags: [最佳化作業]

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# Optimization Assignment Week 3
###### tags: `最佳化作業`
### 2.Let $A, B \subseteq \mathbb{R}^n$. Prove that $cl(A ∩ B) \subseteq cl(A) ∩ cl(B)$. Given an example in which the inclusion is proper.

We have $A ∩ B \subseteq A \subseteq cl(A)$, and $A ∩ B \subseteq B \subseteq cl(B)$
$\implies A ∩ B \subseteq cl(A) ∩ cl(B)$
The latter is an intersection of closed sets, hence is closed
$\implies A ∩ B$ is closed $\iff (A ∩ B)'\equiv$ {accumulation points of A ∩ B} $\subseteq A ∩ B$
$\iff A ∩ B \supseteq (A ∩ B) ∪ (A ∩ B)'$
$\iff A ∩ B = cl(A ∩ B)$
Thus we have $cl(A ∩ B) \subseteq cl(A) ∩ cl(B)$

Example: Let A=(0,1) and B=(1,2).
Then $A ∩ B=\emptyset$, which is closed, so $cl(A ∩ B)=\emptyset$.
On the other hand, $cl(A)=[0,1]$, and $cl(B)=[1,2]$, so $cl(A) ∩ cl(B)$={1}.
Then we have $cl(A ∩ B) \subseteq cl(A) ∩ cl(B)$

### 3.Let $A, B ⊆ \mathbb{R}^n$. Prove that $int(A∩B) = int(A)∩int(B)$ and $int(A)∪int(B) ⊆ int(A∪B)$. Show an example in which the latter inclusion is proper.

* $int(A)=$ the largest open subset of A
Proof. Let $A_1=∪${$S:S\subseteq A$ and $S$ is open}. Then $A_1$ is open.
For any $x \in A_1,\exists \delta >0$ s.t.$D(x,\delta)\subseteq A_1 \subseteq A \implies A_1 \subseteq int(A)$
On the other hand, $int(A)$ is open and $int(A)\subseteq A \implies int(A)\subseteq A_1$

$int(A) ∩ int(B)= int(A∩B)$
Proof. Since $int(A) ∩ int(B) \subseteq A ∩ B$ and $int(A) ∩ int(B)$ is open,
●$\implies int(A) ∩ int(B) \subseteq int(A∩B)$.
Since $A∩B \subseteq A$ and $A∩B \subseteq B$,
$\implies int(A∩B) \subseteq int(A)$ and $int(A∩B) \subseteq int(B)$.
$\implies int(A∩B) \subseteq int(A) ∩ int(B)$
Thus, $int(A) ∩ int(B)= int(A∩B)$.

$int(A)∪int(B) ⊆ int(A∪B)$
Proof. Since $int(A)∪int(B)\subseteq A ∪ B$ and $int(A)∪int(B)\subseteq A ∪ B$ is open, then by ● we have that $int(A)∪int(B) ⊆ int(A∪B)$.

Example: Let $A=(0,1]$ and $B=[1,2)$.
Then $int(A)=(0,1)$ and $int(B)=(1,2)$,
so $int(A)∪int(B)=(0,1)∪(1,2)$.
On the other hand, $A∪B=(0,2)$ is open, so $int(A∪B)=(0,2)$.
Then we have $int(A)∪int(B) ⊆ int(A∪B)$.




