Q.Show that A ∪ B = A ∩ B implies A = B

Concept: Set equality through mutual subset inclusion and De Morgan-style reasoning.

We need to prove that if $A \cup B = A \cap B$, then $A = B$.

Take any element $x \in A$. Then $x \in A \cup B$ (by definition of union), so $x \in A \cap B$ (by the given equality). This forces $x \in B$. Hence $A \subseteq B$. …

The union $A \cup B$ collects everything in either set, while the intersection $A \cap B$ keeps only what both share. Normally $A \cap B \subseteq A \cup B$, and equality holds only when nothing exists in one set without being in the other. This constraint is so tight that it forces the two sets to coincide.

The key insight: if $A \cup B = A \cap B$, then an element cannot belong to exactly one of the sets. Any element in $A$ must also be in $B$, and vice versa.

Proof

We need to show $A \subseteq B$ and $B \subseteq A$.

1. Show $A \subseteq B$

   Take any $x \in A$. Since $x$ is in at least one of the sets, $x \in A \cup B$. But we are given that $A \cup B = A \cap B$, so $x \in A \cap B$. By definition of intersection, $x \in A$ and $x \in B$. Therefore every element of $A$ is in $B$, giving us $A \subseteq B$.

2. Show $B \subseteq A$

   Take any $y \in B$. Then $y \in A \cup B$ (since $y$ is in at least one set). Using $A \cup B = A \cap B$ again, we have $y \in A \cap B$. This means $y \in A$ and $y \in B$. So every element of $B$ is in $A$, giving us $B \subseteq A$.

3. Conclude $A = B$ …

Method: Set Equality via Distributive Law and Given Conditions

We will prove $A = B$ by showing $A \subseteq B$ and $B \subseteq A$ simultaneously using the Distributive Law of set intersection over union.

Step 1: Express $A$ using the hint

Given:

$A = A \cap (A \cup X)$

But we are told $A \cup X = B \cup X$. So substitute:

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

Step 2: Apply Distributive Law

Using $A \cap (B \cup X) = (A \cap B) \cup (A \cap X)$:

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

Step 3: Use the given intersection condition

We know $A \cap X = \phi$. So:

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

$A = A \cap B$

Step 4: Interpret the result

$A = A \cap B$ means every element of $A$ is also in $B$.

Thus:

$\boxed{A \subseteq B}$

Step 5: Repeat symmetrically for $B$

Start with $B = B \cap (B \cup X) = B \cap (A \cup X)$

Apply Distributive Law:

$B = (B \cap A) \cup (B \cap X)$ …

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. …