If A⊆B, prove that A×A⊆(A×B)∩(B×A)
Let (a, b) be an arbitrary element of A×A
Then, (a,a)ϵA×A⇒aϵA and aϵA
⇒aϵA and aϵB [∵A⊆B]
⇒(a,b)ϵA×B and (a,b)ϵB×A
⇒(a,b)ϵ(A×B)∩(B×A)
∴A×A⊆(A×B)∩(B×A)
Hence, A⊆B⇒A×A⊆(A×B)∩(B×A).
If (A∪B)=(A∩B) then prove that A=B
If A∩B′=ϕ then prove that A=A∩B and hence show that A⊆B.