Exercise 3.1.8 (Absorption laws)

Question

Let $A$, $B$ be sets. Prove the absorption laws $A \cap (A \cup B) = A$
and $A \cup (A \cap B) = A$.

Khagan Gulusoy · Accepted Answer

\begin{proof}
\begin{enumerate}
    \item By Exercise 3.4, it suffices to show that $A\cap (A\cup B) \subseteq A$ and vice versa.
    \begin{description}
        \item[$A\cap (A\cup B) \subseteq A$] Let $x\in A\cap (A\cup B)$ be arbitrary. By Definition 3.1.23 we have $x\in A$ and $x\in A\cup B$, and thus, $x\in A$.
        \item[$A\subseteq A\cap (A\cup B)$] Let $x\in A$ be arbitrary. We have $x\in A\cup B$ by Axiom 3.4. Therefore, we get that $x\in A\cap (A\cup B)$ since $x\in A$ and $x\in A\cup B$.
    \end{description}

\item By Exercise 3.4, it suffices to show that $A\cup (A\cap B) \subseteq A$ and vice versa.
    \begin{description}
        \item[$A\cup (A\cap B) \subseteq A$] Let $x\in A\cup (A\cap B)$ be arbitrary. By Axiom 3.4 we have $x\in A$ or $x\in A\cap B$. In the first case, we in particular have $x\in A$. In latter case, we have $x\in A$ and $x\in B$, and thus, $x\in A$. Thus, in both cases we have $x\in A.$
        \item[$A\subseteq A\cup (A\cap B)$] This is a direct consequence of \href{./exercise_3-1-7}{Exercise 3.1.7}
        %Let $x\in A$ be arbitrary. We have $x\in A$ or $x\in A\cap B$ by Axiom 3.4. in the first case, we have $x\in A\cup (A\cap B)$ since $x\in A$ by Axiom 3.4. In the latter case, since $x\in (A\cap B)$ we have $x\in A\cup (A\cap B)$ by Axiom 3.4. Thus, in both cases we have $A\subseteq A\cup (A\cap B)$.
    \end{description}
\end{enumerate}
\end{proof}

Exercise 3.1.8 (Absorption laws)

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Exercise 3.1.8 (Absorption laws)

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