Chapter 2 - Section 2.2 - Set Operations - Exercises - Page 136: 18

Prove each inclusion

a) We have to prove that $$(A\cup B)\subseteq (A\cup B\cup C).$$ We take an arbitrary element $x\in A\cup B$. This means that $$x\in A\text{ or }x\in B.$$ Therefore $x\in (A\cup B\cup C)$ because $x\in (A\cup B\cup C)$ is equivalent to $x\in A$ or $x\in B$ or $x\in C$. So we got that $$(A\cup B)\subseteq (A\cup B\cup C).$$ b) We have to prove that $$(A\cap B\cap C)\subseteq (A\cap B).$$ We take an arbitrary element $x\in A\cap B\cap C$. This means that $$x\in A\text{ and } x\in B\text{ and } x\in C.$$ Because $x\in A\text{ and } x\in B$, it means $x\in A\cap B$. We got: $$(A\cap B\cap C)\subseteq (A\cap B).$$ c) We have to prove that $$(A-B)-C\subseteq A-C.$$ We take an arbitrary element $x\in (A-B)-C$. This means that $$x\in A-B\text{ and } x\not \in C.$$ Because $x\in A-B$ it follows that $x\in A$ and $x\not\in B$. Because $x\in A\text{ and } x\not\in C$, it means $x\in A-C$. We got: $$(A-B)-C\subseteq A-C.$$ d) We have to prove that $$(A-C)\cap(C-B)=\emptyset.$$ We take an arbitrary element $x\in (A-C)\cap(C-B)$. This means that $$x\in A-C\text{ and } x\in C-B.$$ Because $x\in A-C$ it follows that $x\in A$ and $x\not\in C$. Because $x\in C-B$ it follows that $x\in C$ and $x\not\in B$. The intersection of the sets $(A-C)$ and $(C-B)$ contains the points which are in the same time in $C$, but not in $A$ and in $C$, but not in $B$. Since a point cannot be in the same time in $C$ and not in $C$, the intersection is the empty set. We got: $$(A-C)\cap(C-B)=\emptyset.$$ e) We have to prove that $$(B-A)\cup(C-A)=(B\cup C)-A.$$ We take an arbitrary element $x\in (B-A)\cup(C-A)$. This is true if and only if $$x\in B-A\text{ or } x\in C-A.$$ Because $x\in B-A$ if and only if $x\in B$ and $x\not\in A$ and $x\in C-A$ if and only if $x\in C$ and $x\not\in A$, it follows that $x\in B$ or $x\in C$ and $x\not\in A$, which can be written: $$x\in (B\cup C)-A.$$ We got: $$(B-A)\cup(C-A)=(B\cup C)-A.$$