###### tags: `108上助教` `數學`
# 數學導論HW1
## ch2-2
### 33. **Theorem 2.6**
Let $A$ and $B$ be subsets of some universal set $U$. Then
$\mbox{c) } (A \cap B)^{'} = A^{'} \cup B^{'}$
$\mbox{d) }A - B = A \cap B^{'}$

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### **proof**
#### c）
#### method 1（左包右包）

$(\mbox{Case 1: }\subseteq )$
$\mbox{Let } x \in (A \cap B)^{'}$
$\therefore x \notin A \cap B$
$\Rightarrow x \notin A \mbox{ or } x \notin B$ (by Proposition 1.1 (d))
$\Rightarrow x \in A^{'} \mbox{ or } x \in B^{'}$
$\Rightarrow x \in A^{'} \cup B^{'}$
$\Rightarrow (A \cap B)^{'} \subseteq A^{'} \cup B^{'}$

$(\mbox{Case 2: }\supseteq )$
$\mbox{Let } x \in A^{'} \cup B^{'}$
$\therefore x \in A^{'} \mbox{ or } x \in B^{'}$
$\Rightarrow x \notin A \mbox{ or } x \notin B$
$\Rightarrow x \notin A \cap B$(by Proposition 1.1 (d))
$\Rightarrow x \in (A \cap B)^{'}$
$\Rightarrow A^{'} \cup B^{'} \subseteq (A \cap B)^{'}$

$\mbox{Hence , } (A \cap B)^{'} = A^{'} \cup B^{'}\ \Box$

#### method 2

(by Theorem 2.6 (a) and (b))
$\mbox{We have } (A^{'} \cup B^{'})^{'} = A \cap B$
$\Rightarrow A^{'} \cup B^{'} = ((A^{'} \cup B^{'})^{'})^{'} = (A \cap B)^{'}\ \Box$

#### d）
$(\mbox{Case 1: }\subseteq )$
$\mbox{Let } x \in A-B$
$i.e. x \in A \mbox{ and } x \notin B$
$\Rightarrow x \in A \mbox{ and } x \in B^{'}$
$\Rightarrow x \in A \cap B^{'}$
$\Rightarrow A - B \subseteq A \cap B^{'}$

$(\mbox{Case 2: }\supseteq )$
$\mbox{Let } x \in A \cap B^{'}$
$\Rightarrow x \in A \mbox{ and } x \in B^{'}$
$\Rightarrow x \in A \mbox{ and } x \notin B$
$\Rightarrow x \in A-B$
$\Rightarrow A \cap B^{'} \subseteq A - B$

$\mbox{Hence } A - B = A \cap B^{'}\ \Box$

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### 46. Prove or find a counterexample to each of the followling.
$\mbox{a) }(A-B)\cup C = A - (B \cup C)$
$\mbox{b) }(A^{'} \cup B) \cap (B^{'} \cup C) \subseteq A^{'} \cup C$

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### **solution**
#### a）
$\mathrm{We\ give\ a\ counterexample.}\\
\mbox{Let } A = \{1, 2, 3, 4, 5\} , B = \{2, 3\} , C = \{4\}\\
\mbox{then } A-B = \{1, 4, 5\} , \mbox{and }(A-B)\cup C = \{1, 4,5\}\\
\mbox{but } B \cup C = \{2,3,4\}, \mbox{and } A - (B \cup C) = \{1, 5\}\\
\{1, 4, 5\} \neq \{1, 5\}\\
\mathrm{This\ is\ a\ counterexample.}\Box$

#### b）
$\mbox{We give a proof.}$
$\mbox{Let } x \in (A^{'} \cup B) \cap (B^{'} \cup C)$
$\Rightarrow x \in A^{'} \cup B \mbox{ and } x \in B^{'} \cup C$

$\mbox{case 1 : } x \in A^{'} \mbox{ and } x \in B^{'} , \mbox{ implies } x \in A^{'}$
$\mbox{case 2 : }x \in A^{'} \mbox{ and } x \in C , \mbox{ implies } x \in A^{'} \cup C$
$\mbox{case 3 : }x \in B \mbox{ and } x \in B^{'} , \mbox{ which is impossible.}$
$\mbox{case 4 : }x \in B \mbox{ and } x \in C , \mbox{ implies } x \in C$

$\mbox{By case 1 ~ case 4 , we have } x \in A^{'} \cup C\\
\mbox{Hence }(A^{'} \cup B) \cap (B^{'} \cup C) \subseteq A^{'} \cup C\ \Box$