Question

If A and B are sets, prove that $A - B$, $A \cap B$ and $B - A$ are pairwise disjoint.

Answer 1

Here we need to verify whether $A - B$, $A \cap B$ and $B - A$ are pairwise disjoint i.e.., intersection of two sets is a null set.

Complete step-by-step answer:

A & B are two sets. Pairwise disjoint means no two elements of any two sets are equal. So we have to proof:

$(A - B) \cap (B - A) = (A - B) \cap (A \cap B) = (B - A) \cap (A \cap B) = \emptyset$

Where $\emptyset$ is null set $\Rightarrow \emptyset = \\{ \\}$

Let,

$x \in (A - B) \cap (B - A)$

\therefore x \in (A - B){\text{ & }}x \in (B - A) \\\

\therefore x \in A{\text{ }}or{\text{ }}x \notin B\ & {\text{ }}x \in B\or x \notin A \\\

From this it is clear that we are at contradiction $(\because x \in A{\text{ or }}x \notin A)$

$\Rightarrow (A - B) \cap (B - A) = \emptyset \to (1)$

Now let $x \in (A - B) \cap (A \cap B)$

\therefore x \in (A - B) x \in (A \cap B) \\\

\therefore x \in A{\text{ }}or{\text{ }}x \notin B {\text{ }}x \in A\ x \in B \\\

From this it is clear that we are at contradiction $(\because x \in B{\text{ or }}x \notin B)$

$\Rightarrow (A - B) \cap (A \cap B) = \emptyset \to (2)$

Similarly, $(B - A) \cap (A \cap B) = \emptyset \to (3)$

From equation 1, 2 and 3 it is clear that it is pairwise disjoint.

Note: Pairwise disjoint means the intersection of any two sets is a null set. A venn diagram can also be drawn to solve this problem with a visual representation.