Prove That A − B, A∩B and B − A Are Pairwise Disjoint

Solution

To prove that the sets are pairwise disjoint, we show that the intersection of every pair is empty.

1. Show that \( (A-B)\cap(A\cap B)=\phi \)

Let \( x\in(A-B)\cap(A\cap B) \).

Then,

\[ x\in A-B \quad \text{and} \quad x\in A\cap B \]

From \( x\in A-B \),

\[ x\in A \quad \text{and} \quad x\notin B \]

From \( x\in A\cap B \),

\[ x\in A \quad \text{and} \quad x\in B \]

This is impossible because \( x \) cannot belong and not belong to \( B \) simultaneously.

Therefore,

\[ (A-B)\cap(A\cap B)=\phi \]

2. Show that \( (A\cap B)\cap(B-A)=\phi \)

Let \( x\in(A\cap B)\cap(B-A) \).

Then,

\[ x\in A\cap B \quad \text{and} \quad x\in B-A \]

From \( x\in A\cap B \),

\[ x\in A \quad \text{and} \quad x\in B \]

From \( x\in B-A \),

\[ x\in B \quad \text{and} \quad x\notin A \]

This is impossible because \( x \) cannot belong and not belong to \( A \) simultaneously.

Therefore,

\[ (A\cap B)\cap(B-A)=\phi \]

3. Show that \( (A-B)\cap(B-A)=\phi \)

Let \( x\in(A-B)\cap(B-A) \).

Then,

\[ x\in A-B \quad \text{and} \quad x\in B-A \]

From \( x\in A-B \),

\[ x\in A \quad \text{and} \quad x\notin B \]

From \( x\in B-A \),

\[ x\in B \quad \text{and} \quad x\notin A \]

This is impossible.

Therefore,

\[ (A-B)\cap(B-A)=\phi \]

Hence,

\[ A-B,\quad A\cap B,\quad B-A \]

are pairwise disjoint sets.