Exercise 3.1.10 (Disjoint partition of a union)

Question

Let $A$ and $B$ be sets. Show that the three sets $A \setminus B$, $A \cap B$, and $B \setminus A$ are disjoint, and that their union is $A \cup B$.

Khagan Gulusoy · Accepted Answer

Our proof consists from several parts.
\begin{enumerate}
    \item We should prove that $A\setminus B$ and $A\cap B$ are disjoint. This means that we should prove that $(A\setminus B)\cap (A\cap B) = \emptyset.$
    \item Similarly, we should prove that $A\cap B$ and $B\setminus A$ are disjoint. This means that we should prove that $(A\cap B)\cap (B\setminus A) = \emptyset.$
    \item Similarly, we prove that $A\setminus B$ and $B\setminus A$ are disjoint. This means that we should prove that $(A\setminus B)\cap (B\setminus A) = \emptyset.$
    \item Finally, we prove that all these sets form $A\cup B.$ I.e., we have to prove that $(A\setminus B)\cup (A\cap B)\cup (B\setminus A) = A\cup B.$
\end{enumerate}
\begin{proof}
\begin{itemize}
    \item ($(A\setminus B)\cap (A\cap B) = \emptyset$) Suppose for the sake of contradiction that $(A\setminus B)\cap (A\cap B) 
eq \emptyset$. By Lemma 3.1.6 (single choice), we can thus find $x$ such that $x\in A\setminus B$ and $x\in A\cap B$. $x\in A\setminus B$ means that $x\in A$ and $x
otin B.$ $x\in A\cap B$ means that $x\in A$ and $x\in B.$ But this causes a contradiction because $x\in B$ and $x
otin B$ cannot hold at once. 
    
    \item ($(A\cap B)\cap (B\setminus A) = \emptyset$) Follows similarly.
    
    \item ($(A\setminus B)\cap (B\setminus A) = \emptyset$) Follows similarly.
    %By \href{./exercise_3-1-4}{Exercise 3.1.4} it suffices to show that $(A\setminus B)\cap (B\cap A)\subseteq \emptyset$ and vice versa.
    %\begin{description}
     %   \item[$(A\setminus B)\cap (B\cap A)\subseteq \emptyset$] Let $x\in (A\setminus B)\cap (B\setminus A)$ be arbitrary. We have $x\in A\setminus B$ and $x\in B\setminus A.$ By Definition 3.1.23 and Definition 3.1.27 we have $[(x\in A \ 	ext{and} \ x
otin B)\ 	ext{and} \ (x\in B \ 	ext{and} \ x
otin A)].$ But $x\in B$ and $x
otin B$ cannot hold at once, and therefore, $x\in \emptyset.$
     %   \item[$\emptyset\subseteq (A\setminus B)\cap (B\setminus A)$] This is the direct consequence of Example 3.1.17.
    %\end{description}
    
    \item ($(A\setminus B)\cup (A\cap B)\cup (B\setminus A) = A\cup B$) 
    By \href{./exercise_3-1-4}{Exercise 3.1.4}, it suffices to show that $(A\setminus B)\cup (A\cap B)\cup (B\setminus A)\subseteq A\cup B$ and vice versa.

\begin{description}
    
        \item[$(A\setminus B)\cup (A\cap B)\cup (B\setminus A)\subseteq A\cup B$] Let $x\in (A\setminus B)\cup (A\cap B)\cup (B\setminus A)$ be arbitrary. By Axiom 3.4, this means that $x\in A\setminus B$ or $x\in A\cap B$ or $x\in B\setminus A.$ By Definition 3.1.23 and Definition 3.1.27 we have $(x\in A \ 	ext{and} \ x
otin B) \ 	ext{or} \ (x\in A\ 	ext{and} \ x\in B) \ 	ext{or} \ (x\in B \ 	ext{and} \ x
otin A).$ If $(x\in A \ 	ext{and} \ x
otin B)$, we have $x\in A,$ and therefore, $x\in A\cup B.$ In the case when $(x\in A\ 	ext{and} \ x\in B)$, we have $x\in A$, and thus, $x\in A\cup B.$ In the latter case, when $x\in B\setminus A,$ we have $x\in B,$ and thus, $x\in A\cup B.$ Since $x$ is an element of $A\cup B$ in all the cases, we can conclude that $(A\setminus B)\cup (A\cap B)\cup (B\setminus A)\subseteq A\cup B$.
        %No two of them can hold at once, because we'll get a contradiction such that $x\in A \ 	ext{and} x\in A$ or like $x\in B\ 	ext{and} x
otin B$. If $x\in A$ and $x
otin B$, we have $x\in A$ or $x\in B$, and thus, $x\in A\cup B.$ In the other case, when $x\in A$ and $x\in B$, we similarly have $x\in A\cap B$. In the last case, we have $x\in B$ and $x
otin A.$ Since $x\in B,$ we have $x\in A$ or $x\in B,$ and thus, $x\in A\cup B.$ Since in all cases we have $x\in A\cup B,$ we have $(A\setminus B)\cup (A\cap B)\cup (B\setminus A)\subseteq A\cup B$.
        %We have $x\in A\setminus B$ and $x\in A\cap B.$ By Definition 3.1.23 and Definition 3.1.27 we have $[(x\in A \ 	ext{and} \ x
otin B)\ 	ext{and} \ (x\in A \ 	ext{and} \ x\in B)].$ But $x\in B$ and $x
otin B$ cannot hold at once, and therefore, $x\in \emptyset.$
        
        \item[$A\cup B\subseteq (A\setminus B)\cup (A\cap B)\cup (B\setminus A)$] Let $x\in A\cup B$ be arbitrary. Then, we have $x\in A$ or $x\in B.$ 
        \begin{enumerate}
            \item ($x\in A$) In case when $x\in A,$ we have $x\in A\setminus B$ or $x\in A\cap B$. 
            \begin{enumerate}[label=(\ast)]
                \item ($x
otin B$) Since $x\in A\setminus B,$ we have $x\in (A\setminus B)\cup (A\cap B)\cup (B\setminus A)$.
                \item ($x\in B$) Since $x$ is contained in $A\cap B,$ we have $x\in (A\setminus B)\cup (A\cap B)\cup (B\setminus A)$ by Axiom 3.4.
            \end{enumerate}
            \item ($x\in B$) Follows similarly.
        \end{enumerate}
        
        %therefore, $x\in (A\setminus B)\cup (A\cap B)\cup (B\setminus A).$ In the latter case when $x\in B,$ we have $x\in A\cup B$, and therefore, $x\in (A\setminus B)\cup (A\cap B)\cup (B\setminus A).$ 
        
    \end{description}  
\end{itemize}
\end{proof}

Exercise 3.1.10 (Disjoint partition of a union)

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Exercise 3.1.10 (Disjoint partition of a union)

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