Is it true that if $A \subset C$ and $B \subset D$, then $(A \cup B) \subset (C \cup D)$ and $(A \cap B) \subset (C \cap D)$. It seems that this should be the case, but I'm having trouble recalling.
if $A \subset C$ and $B \subset D$ is $(A \cup B) \subset (C \cup D)$ and $(A \cap B) \subset (C \cap D)$
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elementary-set-theory
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0Dont recall, think. If a is in A U B then a is in A or B so a is in C or D so a in C U D. If a is in A cap B, the a is in A and a is in B. So a is in C and a is in D. So a is in C cap D. – 2017-02-12
2 Answers
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If $x\in A\cup B$ then $x\in A$ or $x\in B$. But $A\subset C, B\subset D$ so $x\in C$ or $x\in D$, namely $x\in C\cup D$.
If $x\in A\cap B$ then $x\in A$ and $x\in B$. But $A\subset C, B\subset D$ so $x\in C$ and $x\in D$, namely $x\in C\cap D$.
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For the first part. Let $x \in A \cup B \implies x \in A \vee x \in B$. If $x \in A \implies x \in C \implies x \in C \cup D$. Otherwise, if $x \in B \implies x \in D \implies x \in C \cup D$. Therefore, $(A\cup B) \subseteq (C\cup D)$
For the next one. Let $x \in A \cap B \implies x \in A \wedge x \in B$. Since $x \in A \implies x \in C$ and since $x \in B \implies x \in D$. Therefore, $x \in C \cap D \implies (A\cap B) \subseteq (C\cap D)$