【Basic Abstract Algebra】Exercises of Section 1.1

1. Let \(A,B\) and \(C\) be sets. Show that

   • \((A\cap B)\text{\\} B=\empty\).
     Proof: Let \(x\in(A\cap B)\text{\\} B\), that means \(x\in A\cap B\) and \(x\not\in B\). Thus, we have \(x\in B\) and \(x\not\in B\). Therefore \((A\cap B)\text{\\} B\) must be the \(\empty\). #
   • \((A\cup B)\text{\\}B=A\text{\\}B\).
     Proof: On the one hand, let \(x\in (A\cup B)\text{\\} B\), that means \(x\in A\cup B\) and \(x\not\in B\). This implies that \(x\in A\) and \(x\not\in B\), Thus, \(x\in A\text{\\}B\), therefore \((A\cup B)\text{\\} B\subseteq A\text{\\}B\). On the other hand, let \(x\in A\text{\\}B\), we have \(x\in A\) and \(x\not\in B\). Since \(x\in A\), it is also in \(A\cup B\). Thus, \(x\in A\cup B\) and \(x\not\in B\). That is \(x\in (A\cup B)\text{\\}B\). Therefore \(A\text{\\}B\subseteq (A\cup B)\text{\\}B\). In a Word, \((A\cup B)\text{\\}B=A\text{\\}B\). #
   • \((B\cup C)\text{\\}A=(B\text{\\}A)\cup(C\text{\\}A)\).
     Proof: On the one hand, let \(x\in (B\cup C)\text{\\} A\), that means \(x\in B\cup C\) and \(x\not\in A\). This implies that \(x\in B\) or \(x\in C\) and \(x\not\in A\), i.e., \(x\in(B\text{\\}A)\) or \(x\in (C\text{\\}A)\). Thus, we have \(x\in(B\text{\\}A)\cup(C\text{\\}A)\), Therefore \((B\cup C)\text{\\}A\subseteq(B\text{\\}A)\cup(C\text{\\}A)\). On the other hand, let \(x\in(B\text{\\}A)\cup(C\text{\\}A)\), we have \(x\in(B\text{\\}A)\) or \(x\in (C\text{\\}A)\). That is \(x\in B\) and \(x\notin A\) or \(x\in C\) and \(x\notin A\). We have \(x\in B\cup C\) and \(x\notin A\), i.e. \(x\in(B\cup C)\text{\\}A\). Therefore \((B\text{\\}A)\cup(C\text{\\}A)\subseteq(B\cup C)\text{\\}A\). In a Word, \((B\cup C)\text{\\}A=(B\text{\\}A)\cup(C\text{\\}A)\). #

2. Let \(A,B,C\) be sets. Prove that \(A\text{\\}(B\text{\\}C)=(A\text{\\}B)\cup(A\cap C)\).
   Proof: Let \(x\in A\text{\\}(B\text{\\}C)\), then \(x\in A\) and \(x\notin B\text{\\}C\). That means \(x\in A\) and (\(x\notin B\) or \(x\in C\)). Thus, \(x\in (A\text{\\}B)\cup(A\cap C)\)， therefore \(A\text{\\}(B\text{\\}C)\subseteq(A\text{\\}B)\cup(A\cap C)\).
   On the other hand, let \(x\in (A\text{\\}B)\cup(A\cap C)\), then we have \(x\in A\text{\\}B\) or \(x\in A\cap C\). That means \(x\in A\) and \(x\notin B\) or \(x\in A\) and \(x\in C\). Thus, \(x\in A\) and (\(x\notin B\) or \(x\in C\)), which means \(x\in A\text{\\}(B\text{\\}C)\), therefore \((A\text{\\}B)\cup(A\cap C)\subseteq A\text{\\}(B\text{\\}C)\).
   In a word, \(A\text{\\}(B\text{\\}C)=(A\text{\\}B)\cup(A\cap C)\).

3. Let \(X,Y\) be finite sets. Prove that \(|X\cup Y|+|X\cap Y|=|X|+|Y|\).
   Proof: Note that

   \[\begin{aligned} |X\cup Y|&=|[X\text{\\}(X\cap Y)]\cup[Y\text{\\}(X\cap Y)]\cup(X\cap Y)|\\ &=|X\text{\\}(X\cap Y)|+|Y\text{\\}(X\cap Y)|+|X\cap Y|\\ &=|X|-|X\cap Y|+|Y|-|X\cap Y|+|X\cap Y|\\ &=|X|+|Y|-|X\cap Y|. \end{aligned} \]

   Therefore \(|X\cup Y|+|X\cap Y|=|X|+|Y|\). #