Q.Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C.

Concept: Set equality through double inclusion.

We prove $B = C$ by showing every element of $B$ belongs to $C$ and vice versa.

Step 1. Take any $x \in B$. Then $x \in A \cup B = A \cup C$, so either $x \in A$ or $x \in C$.

Step 2. If $x \in A$, then since $x \in B$ as well, we have $x \in A \cap B = A \cap C$, which gives $x \in C$.

Step 3. If $x \notin A$, then from $x \in A \cup C$ we must have $x \in C$.

In both cases $x \in C$, proving $B \subseteq C$. By symmetry (swapping $B$ and $C$ in the argument), we get $C \subseteq B$.

The heart of this problem is understanding what set equality means: two sets are equal if and only if they contain exactly the same elements. We'll prove $B = C$ by showing $B \subseteq C$ and $C \subseteq B$ simultaneously.

The strategy hinges on a simple observation. Any element $x$ in $B$ must either belong to $A$ or not belong to $A$. These two cases exhaust all possibilities, and in each case the given conditions force $x$ into $C$. The same reasoning works in reverse.

Proof

1. Show that $B \subseteq C$

Take an arbitrary element $x \in B$. We consider two cases:

Case 1: $x \in A$

Since $x \in A$ and $x \in B$, we have $x \in A \cap B$. But we're given that $A \cap B = A \cap C$, so $x \in A \cap C$. This immediately tells us $x \in C$.

Case 2: $x \notin A$

Since $x \in B$, we know $x \in A \cup B$ (an element in $B$ is certainly in the union $A \cup B$). We're given that $A \cup B = A \cup C$, so $x \in A \cup C$.

Now $x \in A \cup C$ means either $x \in A$ or $x \in C$. But we assumed $x \notin A$ in this case, so we must have $x \in C$.

In both cases, $x \in C$. Since $x$ was arbitrary, every element of $B$ belongs to $C$, giving us $B \subseteq C$.

2. Show that $C \subseteq B$

The argument is completely symmetric. Take any $x \in C$.

Case 1: $x \in A$

Then $x \in A \cap C = A \cap B$, so $x \in B$.

Case 2: $x \notin A$

Then $x \in A \cup C = A \cup B$. Since $x \notin A$, we must have $x \in B$.

Therefore $C \subseteq B$.

3. Conclude

Since $B \subseteq C$ and $C \subseteq B$, we have $B = C$.

Method: Element-Chasing Proof (Set Equality via Double Inclusion)

This is the standard method to prove two sets are equal — show that every element of $B$ is in $C$, and every element of $C$ is in $B$.

Step 1: Show $B \subseteq C$

Take any element $x \in B$. We must prove $x \in C$.

• Case 1: $x \in A$

  Since $x \in B$ and $x \in A$, we have $x \in A \cap B$.

  Given $A \cap B = A \cap C$, it follows that $x \in A \cap C$.

  Therefore $x \in C$.

• Case 2: $x \notin A$

  Since $x \in B$, we have $x \in A \cup B$.

  Given $A \cup B = A \cup C$, it follows that $x \in A \cup C$.

  But $x \notin A$, so $x$ must be in $C$.

In both cases, $x \in C$. Hence $B \subseteq C$.

Step 2: Show $C \subseteq B$

Take any element $x \in C$. We must prove $x \in B$.

• Case 1: $x \in A$

  Then $x \in A \cap C$.

  Since $A \cap C = A \cap B$, we get $x \in A \cap B$, so $x \in B$.

• Case 2: $x \notin A$

  Then $x \in A \cup C$.

  Since $A \cup C = A \cup B$, we have $x \in A \cup B$.

  But $x \notin A$, so $x \in B$.

In both cases, $x \in B$. Hence $C \subseteq B$.

Step 3: Conclude

Since $B \subseteq C$ and $C \subseteq B$, we have $B = C$.

Key insight: The two given conditions split the proof into two natural cases — whether the element is inside $A$ or outside $A$. This covers all possibilities without any gaps.

Here are the most common mistakes students make when proving B = C from the given set conditions, along with how to avoid each.

Mistake 1: Assuming the result without proof

What students do:

They write something like:

“Since $A \cup B = A \cup C$ and $A \cap B = A \cap C$, it is obvious that $B = C$.”

Why it’s wrong:

This is circular reasoning — you are trying to prove that $B = C$, so you cannot assume it. The problem is designed to test your ability to use set algebra or element chasing.

How to avoid:

Always start with an element approach or algebraic manipulation.

• Element method: Take an arbitrary $x \in B$ and show $x \in C$ (and vice versa).
• Algebraic method: Use identities like $B = (B \cap A) \cup (B \cap A^c)$ and substitute given equalities.

Mistake 2: Only proving one direction of inclusion

What students do:

They show that if $x \in B$ then $x \in C$, and stop there.

Why it’s wrong:

To prove $B = C$, you need two-way inclusion:

• $B \subseteq C$
• $C \subseteq B$

How to avoid:

Explicitly write both parts. A clean structure is:

1. Part 1: Let $x \in B$. Show $x \in C$.
2. Part 2: Let $x \in C$. Show $x \in B$.

Mistake 3: Forgetting to consider the case when $x \notin A$

What students do:

When proving $B \subseteq C$, they take $x \in B$ and immediately use $A \cup B = A \cup C$ to say $x \in A \cup C$. Then they conclude $x \in C$ — but this is only valid if $x \notin A$.

Why it’s wrong:

If $x \in A$, then $x \in A \cup C$ is true, but $x$ might not be in $C$. You need to use the intersection condition to handle the $x \in A$ case.

How to avoid:

Split into two cases for $x \in B$:

• Case 1: $x \in A$ → then $x \in A \cap B$. Since $A \cap B = A \cap C$, we get $x \in C$.
• Case 2: $x \notin A$ → then $x \in A \cup B$ implies $x \in A \cup C$. Since $x \notin A$, we must have $x \in C$.

This case analysis is the core of the proof.

Mistake 4: Misusing set algebra identities

What students do:

They try to manipulate $B = (A \cup B) \cap (A \cap B)$ or similar incorrect identities.

Why it’s wrong:

Set algebra has specific laws (distributive, absorption, De Morgan’s). Using a wrong identity leads to an invalid proof.

How to avoid:

Stick to proven identities only:

• $B = (B \cap A) \cup (B \cap A^c)$ (partition)
• $B = (A \cup B) \cap (A^c \cup B)$
• Absorption: $A \cup (A \cap B) = A$

If you are unsure, use the element method — it is safer and always works.

Mistake 5: Not handling the complement properly

What students do:

They write $B \cap A^c$ without defining the universal set or assuming $A^c$ exists.

Why it’s wrong:

In set theory, complement is relative to a universal set. If not specified, you cannot assume $A^c$ is well-defined.

How to avoid:

Either:

• Use the element method (no complement needed), or
• Explicitly define the universal set $U$ (e.g., $U = A \cup B \cup C$) and use $A^c = U \setminus A$.

Mistake 6: Skipping the “why” in the final step

What students do:

They show $x \in B \implies x \in C$ and $x \in C \implies x \in B$, then stop.

Why it’s wrong:

You must explicitly state:

“Therefore, $B \subseteq C$ and $C \subseteq B$, so $B = C$.”

How to avoid:

End every set equality proof with the conclusion sentence. It’s a small step but examiners look for it.

Quick Summary Checklist

Final tip: The cleanest proof for this exact problem is the element method with case analysis. Practice writing it step-by-step — it will serve you well in exams.