Question

give a direct proof ,as Well as a proof by contradiction,of the following statement A∩B⊆A⋃B for any two sets A and B

Answer 1

Answer:

Direct proof 1

x ∈ A∩B  =>  x ∈ A  =>  x ∈ A∪B

Therefore A∩B ⊆ A∪B.

[ In other words, A∩B ⊆ A  and  A ⊆ A∪B,  therefore  A∩B ⊆ A∪B. ]

Direct proof 2

(A∩B) ∩(A∪B)

= (A∩B∩A) ∪ (A∩B∩B)         [ ∩ distributes over ∪ ]

= (A∩B) ∪ (A∩B)                   [ X∩X=X ]

= A∩B                                 [ X∪X=X ]

Since X ⊆ Y  <=>  X∩Y = X, it follows that A∩B ⊆ A∪B.

Proof by contradiction

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

Then there exists an element x ∈ A∩B such that x ∉ A∪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∩B ⊆ A∪B.

Answer 2

note it down........