91Ó°ÊÓ

To prove this, we will show that \(M_{1} \cap\left(M_{1} \cup M_{2}\right) \subseteq M_{1}\) and \(M_{1} \subseteq M_{1} \cap\left(M_{1} \cup M_{2}\right)\). If both of these are true, we can conclude that both sets are equal. 1. Let \(x \in M_{1} \cap\left(M_{1} \cup M_{2}\right)\). By definition of intersection, we know that \(x \in M_{1}\) and \(x \in \left(M_{1} \cup M_{2}\right)\). However, we only need the first condition to claim that \(x \in M_{1}\). Therefore, \(M_{1} \cap\left(M_{1} \cup M_{2}\right) \subseteq M_{1}\). 2. Now let \(x \in M_{1}\). From this, we know that \(x\) is in both \(M_{1}\) and \(\left(M_{1} \cup M_{2}\right)\). Thus, by definition of the set intersection, we have \(x \in M_{1} \cap\left(M_{1} \cup M_{2}\right)\), and therefore \(M_{1} \subseteq M_{1} \cap\left(M_{1} \cup M_{2}\right)\). Since \(M_{1} \cap\left(M_{1} \cup M_{2}\right) \subseteq M_{1}\) and \(M_{1} \subseteq M_{1} \cap\left(M_{1} \cup M_{2}\right)\), we conclude that \(M_{1} \cap\left(M_{1} \cup M_{2}\right) = M_{1}\).

To prove this, we will show that \(M_{1} \cup\left(M_{1} \cap M_{2}\right) \subseteq M_{1}\) and \(M_{1} \subseteq M_{1} \cup\left(M_{1} \cap M_{2}\right)\). If both of these are true, we can conclude that both sets are equal. 1. Let \(x \in M_{1} \cup\left(M_{1} \cap M_{2}\right)\). By definition of the union, we know that \(x \in M_{1}\) or \(x \in \left(M_{1} \cap M_{2}\right)\). In either case, \(x \in M_{1}\). Therefore, \(M_{1} \cup\left(M_{1} \cap M_{2}\right) \subseteq M_{1}\). 2. Now let \(x \in M_{1}\). From this, we know that \(x\) is in either \(M_{1}\) or \(\left(M_{1} \cap M_{2}\right)\). Thus, by definition of the set union, we have \(x \in M_{1} \cup\left(M_{1} \cap M_{2}\right)\), and therefore \(M_{1} \subseteq M_{1} \cup\left(M_{1} \cap M_{2}\right)\). Since \(M_{1} \cup\left(M_{1} \cap M_{2}\right) \subseteq M_{1}\) and \(M_{1} \subseteq M_{1} \cup\left(M_{1} \cap M_{2}\right)\), we conclude that \(M_{1} \cup\left(M_{1} \cap M_{2}\right) = M_{1}\).