Q.Using properties of sets, show that

Concept: Absorption Laws of sets – a set "absorbs" any intersection or union with a subset.

(i) $A \cup (A \cap B) = A$

Every element of $A \cap B$ is already in $A$ (by definition of intersection). Taking the union of $A$ with a subset of itself adds nothing new.

Proof:

• If $x \in A$, then $x \in A \cup (A \cap B)$ automatically.
• If $x \in A \cup (A \cap B)$, then either $x \in A$ or $x \in A \cap B$. But $A \cap B \subseteq A$, so in both cases $x \in A$.

Hence $A \cup (A \cap B) = A$.

(ii) $A \cap (A \cup B) = A$

Every element of $A$ lies in $A \cup B$ (since $A \subseteq A \cup B$). Intersecting $A \cup B$ with $A$ retains precisely the elements of $A$.

Proof: …

These two results capture a fundamental property of sets: when you combine a set with something "derived from itself," you simply get the original set back. The first says that adding elements that are already in $A$ (namely $A \cap B$) doesn't change $A$. The second says that keeping only elements of $A$ from a larger collection ($A \cup B$) leaves you with $A$ itself.

The standard technique is to prove set equality by showing each side is a subset of the other: $X = Y$ if and only if $X \subseteq Y$ and $Y \subseteq X$.

(i) Proving $A \cup (A \cap B) = A$

1. Show $A \cup (A \cap B) \subseteq A$

Take any element $x \in A \cup (A \cap B)$. By definition of union, either $x \in A$ or $x \in A \cap B$.

If $x \in A$, we're done.

If $x \in A \cap B$, then by definition of intersection, $x \in A$ and $x \in B$. In particular, $x \in A$.

Either way, $x \in A$. Therefore $A \cup (A \cap B) \subseteq A$.

2. Show $A \subseteq A \cup (A \cap B)$

Take any element $x \in A$. By definition of union, $x \in A$ immediately implies $x \in A \cup (A \cap B)$ (since the union contains everything in $A$).

Therefore $A \subseteq A \cup (A \cap B)$.

3. Conclude equality

Since both inclusions hold, $A \cup (A \cap B) = A$.

(ii) Proving $A \cap (A \cup B) = A$

1. Show $A \cap (A \cup B) \subseteq A$

Take any element $x \in A \cap (A \cup B)$. By definition of intersection, $x \in A$ and $x \in A \cup B$.

The first condition alone tells us $x \in A$.

Therefore $A \cap (A \cup B) \subseteq A$.

2. Show $A \subseteq A \cap (A \cup B)$ …

Method: Roster Form Intersection

This method works when sets are given in roster form (listing elements inside curly braces). The intersection of two sets contains only the elements that appear in both sets.

Steps

1. Write both sets clearly in roster form.
2. Identify common elements — scan each element of the first set and check if it is also present in the second set.
3. List only the common elements inside curly braces $\{ \}$.
4. If no element is common, the intersection is the empty set, written as $\emptyset$ or $\{ \}$.

Example (from typical Question 1 type)

Let:

• $A = \{1, 2, 3, 4\}$
• $B = \{3, 4, 5, 6\}$

Step 1: Sets are already in roster form.

Step 2: Common elements:

• $1$ is not in $B$ …

The Correct Answer

If $A \subset B$, then $A \cup B = B$.

Why?

Because every element of $A$ is already inside $B$. So when you take the union (all elements in $A$ or in $B$), you don’t add anything new beyond what $B$ already has. The union just gives back $B$.

Common Mistakes & How to Avoid Them

Mistake 1: Writing $A \cup B = A$

• What students think: “Since $A$ is inside $B$, the union is just the smaller set $A$.”
• Why it’s wrong: The union must include everything from both sets. $B$ has extra elements that $A$ doesn’t have — those must be included.
• How to avoid: Draw a Venn diagram. Shade $A$ and $B$ separately, then shade the union. You’ll see the larger set $B$ is fully covered.

Mistake 2: Writing $A \cup B = A \cap B$

• What students think: “If one is inside the other, union and intersection are the same.”
• Why it’s wrong:
  • $A \cup B$ = all elements in either set = $B$ (the bigger one).
  • $A \cap B$ = only elements in both sets = $A$ (the smaller one). They are equal only if $A = B$.
• How to avoid: Memorise the difference:
  • Union → bigger set (or equal).
  • Intersection → smaller set (or equal).

Mistake 3: Forgetting the special case $A = B$

• What students think: “If $A \subset B$, then $A$ is strictly smaller.”
• Why it’s wrong: In many textbooks, $A \subset B$ allows $A = B$ (some use $\subseteq$ for that). If $A = B$, then $A \cup B = A = B$ — still correct, but students sometimes panic. …