1. **Problem Statement:** Given a set $A$, let $A'$ be the set of all subsets of $A$. For example, if $A = \{1, 2, 3\}$, then $$A' = \{\emptyset, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, A\}.$$ Given $X, Y \in A'$, define $X \cup Y$ as the union of $X$ and $Y$, and $X \cap Y$ as the intersection of $X$ and $Y$. We need to answer several questions about these operations on $A'$. 2. **(a) Why are $\cup$ and $\cap$ binary operations on $A'$?** A binary operation on a set $S$ is a function $*: S \times S \to S$ that takes any two elements of $S$ and returns another element of $S$. - Since $X, Y \in A'$, both are subsets of $A$. - The union $X \cup Y$ is also a subset of $A$, so $X \cup Y \in A'$. - Similarly, the intersection $X \cap Y$ is a subset of $A$, so $X \cap Y \in A'$. Therefore, both $\cup$ and $\cap$ take two elements from $A'$ and return an element in $A'$, so they are binary operations on $A'$. 3. **(b) Prove that $\cup$ and $\cap$ are associative on $A'$.** For any $X, Y, Z \in A'$, we need to show: $$ (X \cup Y) \cup Z = X \cup (Y \cup Z) $$ $$ (X \cap Y) \cap Z = X \cap (Y \cap Z) $$ - By the properties of set union and intersection, both union and intersection are associative operations. - This means the grouping of sets does not affect the result. Hence, $\cup$ and $\cap$ are associative on $A'$. 4. **(c) Does $A'$ have an identity element with respect to $\cup$? If yes, find it.** - The identity element $e$ for $\cup$ satisfies $X \cup e = X$ for all $X \in A'$. - The empty set $\emptyset$ satisfies this because $X \cup \emptyset = X$. Therefore, the identity element for $\cup$ on $A'$ is $\emptyset$. 5. **(d) Does $A'$ have an identity element with respect to $\cap$? If yes, find it.** - The identity element $e$ for $\cap$ satisfies $X \cap e = X$ for all $X \in A'$. - The set $A$ itself satisfies this because $X \cap A = X$. Therefore, the identity element for $\cap$ on $A'$ is $A$. 6. **(e) Why is $(A', \cup)$ not a group?** A group requires four properties: closure, associativity, identity, and inverses. - We have closure, associativity, and identity ($\emptyset$). - However, inverses do not exist for all elements under $\cup$. - For example, there is no subset $Y$ such that $X \cup Y = \emptyset$ unless $X = \emptyset$. Thus, $(A', \cup)$ is not a group because not every element has an inverse. 7. **(f) Is $(A', \cap)$ a group? Explain.** - Closure, associativity, and identity ($A$) hold for $\cap$. - But inverses do not exist for all elements under $\cap$. - For example, no $Y$ satisfies $X \cap Y = A$ unless $X = A$. Therefore, $(A', \cap)$ is not a group. **Final answers:** - (a) Both $\cup$ and $\cap$ are binary operations on $A'$. - (b) Both are associative. - (c) Identity for $\cup$ is $\emptyset$. - (d) Identity for $\cap$ is $A$. - (e) $(A', \cup)$ is not a group due to lack of inverses. - (f) $(A', \cap)$ is not a group due to lack of inverses.