Question

Give a direct proof, as well as a proof by contradiction, of the following statement:
‘ B A ∩ B ⊆ A ∪ for any two sets A and B .’

Answer 1

Direct proof:

$$\text{If}\ x\in A\cap B\implies x\in A\implies x\in A \ \text{or} \ x\in B \implies$$

$$\implies x\in A\cup B$$

Therefore

$$A\cap B\sube A\cup B$$

Proof by contradiction

Suppose to the contrary that $A\cap B \subsetneq A\cup B$

Then an element $x\in A\cap B$ exists such that $x\notin A\cup B.$

That is, there is an element x that belongs to both A and B and ($x\in A$ and $x\in B$ ) at the same time belongs to neither A nor B ($x\notin A$ and $x\notin B$ ). This is a contradiction, so the original assumption is false. It follows that

$$A\cap B\sube A\cup B$$