Question

Show that AUB(A union B) = A Intersection B implies A=B.

Answer 1

Let A = B

AUB = A 

A intersection B = A implies A U B  = A intersection B

A = B implies AU B = intersection B ---- (1)

Now let A U B  = A intersection B. Then we have to prove that A = B. For this let x belongs to A implies x belongs to A U B

= x belongs to A intersection B

= x belongs to A and x Belongs to B

= x belongs to B

so A subset of B --- (2)

Now we will let y belong to B which implies y belongs to A U B 

= y belongs to A intersection B
= y belongs to A and y belongs to B
= y belongs to A

Therefore, B subset of A --- (3)

from (2) and (3) we get A=B

so  A union B will be equal to A intersection B implies A = B ---(4)

From (1) and (4) we will get  

A union B = A INTERSECTION B Implies A = B

Answer 2

Prove: (A∪B=A∩B)⟹(A=B)

Suppose x∈A, then x∈A∪B. So then x∈A∩B. Thus x∈B and thus A⊆B

Suppose x∈B, then x∈A∪B. So then x∈A∩B. Thus x∈A and thus B⊆A

Therefore, A=B

Is this correct? Another way to prove this? tips for a better proof writing?

Step-by-step explanation:

An other way

A⊂A∪B=A∩B⊂A

and

B⊂A∪B=A∩B⊂B.

Therefore

A=A∪B=A∩B=B.A simple proof by contradiction, A≠B, then either:

∃x:x∈A∧x∉B

∃x:x∈B∧x∉A

Let's look at 1 (2 is symmetrical). From 1 ⇒x∉A∩B which means (from A∩B=A∪B assumption) x∉A∪B which is a contradiction (with A∪B={x:x∈A∨x∈B}).