Answer to Question #86949 in Algebra for sonali mansingh

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 that

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

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

Therefore

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

Or other direct proof

We have that

$$X\sube Y \iff X\cap Y=X$$

Consider

$$(A\cap B)\cap (A\cup B)$$

$$(A\cap B)\cap (A\cup B)=(A\cap B\cap A)\cup (A\cap B\cap B)=$$

$$=(A\cap B)\cup (A\cap B)=A\cap B$$

Therefore

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

Proof by contradiction

Suppose to the contrary that

$$A\cap B\nsubseteq A\cup B$$

Then there exists an element

$$x \in A\cap B$$

such that

$$x \notin A\cup B$$

That is, there is an element x that belongs to both A and B and at the same time belongs to neither.  This is a contradiction, so the original assumption is false.  It follows that

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