Q.If $A$ and $B$ are two sets, then $A \cap (A \cup B)$ equals
(A) $A$
(B) $B$
(C) $\phi$
(D) $A \cap B$

The key idea is applying Set Identities, specifically the Absorption Law.

1. The set $A \cup B$ contains all elements that are in $A$ or in $B$ (or both).
2. By definition of union, $A$ is a subset of $A \cup B$. This means every element belonging to $A$ also belongs to $A \cup B$.
3. We are looking for the intersection $A \cap (A \cup B)$, which represents the elements common to both set $A$ and set $(A \cup B)$. …

When we work with sets, understanding what each operation means for an individual element is key. The expression $A \cap (A \cup B)$ asks us to find the elements that are common to set $A$ AND the set $(A \cup B)$. Let's break this down by considering an arbitrary element and where it must belong. This method, often called the element-wise definition or subset listing, builds understanding from first principles.

1. Understand the Goal: We want to determine which of the given options ($A$, $B$, $\phi$, or $A \cap B$) is equivalent to the set $A \cap (A \cup B)$. Two sets are equal if and only if they contain exactly the same elements.

2. Define the Operations:

   • The union $A \cup B$ consists of all elements that are in $A$ OR in $B$ (or both).
   • The intersection $X \cap Y$ consists of all elements that are in $X$ AND in $Y$.

3. Consider an Arbitrary Element: Let $x$ be an arbitrary element. We will show that $x$ is in $A \cap (A \cup B)$ if and only if $x$ is in $A$. This proves the two sets are equal.

   • Part 1: If $x \in A \cap (A \cup B)$, then $x \in A$.

     If $x \in A \cap (A \cup B)$, then by the definition of intersection, $x$ must be in both sets:

     $x \in A$ AND $x \in (A \cup B)$.

     The condition $x \in (A \cup B)$ means $x \in A$ OR $x \in B$.

     So, we have the logical statement: $x \in A$ AND ($x \in A$ OR $x \in B$).

     Let $P$ be the statement "$x \in A$" and $Q$ be the statement "$x \in B$". Our logical statement is $P \land (P \lor Q)$.

     This logical expression simplifies to $P$. If you are in $A$, and you are also in ($A$ or $B$), then you must be in $A$. The "($A$ or $B$)" part doesn't add any new restriction beyond being in $A$.

     Therefore, if $x \in A \cap (A \cup B)$, it implies $x \in A$. This means $A \cap (A \cup B) \subseteq A$.

   • Part 2: If $x \in A$, then $x \in A \cap (A \cup B)$.

     If $x \in A$, then it is certainly true that $x \in A$ OR $x \in B$. (If $P$ is true, then $P \lor Q$ is true, regardless of $Q$).

     So, if $x \in A$, then $x \in (A \cup B)$.

     Now we have two facts: $x \in A$ (our initial assumption) AND $x \in (A \cup B)$.

     By the definition of intersection, if $x$ is in $A$ and $x$ is in $(A \cup B)$, then $x \in A \cap (A \cup B)$. …