Question

To give a direct proof, as well as a proof by contradiction, of the following statement:
'A∩B is contained in A∪B for any two sets A and B.'

Answer 1

Direct Proof:

Let $x \in A \bigcap B$ It means

$x \in A$ and $x \in B$

$\implies$ $x \in A \bigcup B$

It means A∩B ⊆ A∪B

So $A\bigcap B$ is contained in $A \bigcup B$ for any two sets A and B

Proof by Contradiction:

Suppose to the contrary that A∩B ⊄ A∪B.

Then there exists an element $x \in A \bigcap B$ such that $x \notin A\bigcap B$ . That is, there is an element $x$ that belongs to both set A and set B and at the same time belongs to neither. This is a contradiction, so the original assumption is false. 

It means A∩B ⊆ A∪B.

So $A\bigcap B$ is contained in $A \bigcup B$ for any two sets A and B.