Q.Show that for any sets A and B, A = ( A ∩ B ) ∪ ( A – B ) and A ∪ ( B – A ) = ( A ∪ B )

Concept: Set Difference — The set $A - B$ (also written $A \setminus B$) contains all elements of $A$ that are not in $B$.

Proof of $A = (A \cap B) \cup (A - B)$

1. Any element $x \in A$ is either in $B$ or not in $B$.
2. If $x \in B$, then $x \in A \cap B$. If $x \notin B$, then $x \in A - B$.
3. Hence every $x \in A$ belongs to $(A \cap B) \cup (A - B)$, and conversely any element of this union is clearly in $A$.

Proof of $A \cup (B - A) = (A \cup B)$

1. $B - A$ contains elements of $B$ that are not in $A$. …

Why This Works: The Logic of Set Difference

Set difference, $A - B$ (also written $A \setminus B$), picks out everything in $A$ that is not in $B$. So when you look at any element $x$ in $A$, there are exactly two possibilities: either $x$ is also in $B$, or $x$ is not in $B$. There is no third option. This is a fundamental partition of $A$ into two disjoint pieces: the part that overlaps with $B$ ($A \cap B$) and the part that is exclusive to $A$ ($A - B$). The first identity simply says that putting these two pieces back together gives you the original set $A$.

The second identity is about what happens when you take $A$ and add the part of $B$ that is not already in $A$. That is exactly the definition of the union $A \cup B$ — everything in $A$ plus everything in $B$ that isn't already covered.

Let's prove both formally.

1. Proving $A = (A \cap B) \cup (A - B)$

We show two-way inclusion: every element of the left side is in the right side, and vice versa.

Step 1a: Show $A \subseteq (A \cap B) \cup (A - B)$

Take any $x \in A$. There are two cases:

• Case 1: $x \in B$. Then $x \in A \cap B$, so $x \in (A \cap B) \cup (A - B)$.
• Case 2: $x \notin B$. Then $x \in A - B$, so $x \in (A \cap B) \cup (A - B)$.

Since every $x \in A$ falls into one of these two cases, we have $A \subseteq (A \cap B) \cup (A - B)$.

Step 1b: Show $(A \cap B) \cup (A - B) \subseteq A$

Take any $x \in (A \cap B) \cup (A - B)$. Then either:

• $x \in A \cap B$, which implies $x \in A$ (and $x \in B$), or
• $x \in A - B$, which implies $x \in A$ (and $x \notin B$).

In either case, $x \in A$. So $(A \cap B) \cup (A - B) \subseteq A$.

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

2. Proving $A \cup (B - A) = A \cup B$

Again, we prove two-way inclusion.

Step 2a: Show $A \cup (B - A) \subseteq A \cup B$ …

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. …