Show that the following four conditions are equivalent: (i) A⊆B (ii) A−B=∅ (iii) A∪B=B (iv) A∩B=A

First, we have to show that (i)↔(ii) (i)⇒(ii).Let A⊆B To show: A−B=∅ If possible, suppose A−B≠∅ This means that there exists x∈A, x∉B, which is not possible as A⊆B. ∴A−B=∅ ∴A⊆B ⇒A−B=∅ (ii)⇒(i).Let A−B=∅ To show: A⊆B Let x∈A Clearly, x∈B because if x∉B, then A−B≠∅ ∴A−B=∅ ⇒A⊆B Hence,(ii)↔(i) (i)⇒(iii).Let A⊆B To show: A∪B=B Clearly, B⊆A∪B Let x∈A∪B ⇒x∈A or x∈B Case I: x∈A ⇒x∈B [∵A⊆B] ∴A∪B⊆B Case II: x∈B Then A∪B⊆B So, A∪B=B (iii)⇒(i).Conversely, let A∪B=B To show : A⊆B Let x∈A ⇒x∈A∪B [∵A⊆A∪B] ⇒x∈B [∵A∪B=B] ∴A⊆B Hence,(iii)↔(i) Now, we have to show that (i)↔(iv).Let A⊆B Clearly A∩B⊆A Let x∈A We have to show that x∈A∩B As A⊆B, x∈B ∴x∈A∩B ∴A⊆A∩B Hence, A=A∩B Conversely, suppose A∩B=A Let x∈A ⇒x∈A∩B ⇒x∈A and x∈B ⇒x∈B ∴A⊆B Hence,(i)↔(iv).