Decide, among the following sets, which sets are subsets of one and another: A = x : x ∈ R a n d x s a t i s f y x 2 - 8 x + 12 = 0 , B = 2 , 4 , 6 , C = 2 , 4 , 6 , 8 … , D = 6 .

According to the question we have, A = x : x ∈ R a n d x s a t i s f y x 2 - 8 x + 12 = 0 Let us write set A in roster form, x 2 - 8 x + 12 = 0 x 2 - 6 x - 2 x + 12 = 0 x x - 6 - 2 x - 6 = 0 x - 6 x - 2 = 0 x = 2 , x = 6 ∴ A = 2 , 6 We are given, B = 2 , 4 , 6 , C = 2 , 4 , 6 , 8 , . . . , D = 6 If the elements of set A are present in the set B , then A ⊂ B else it will be A ⊄ B . Therefore, we can write it as D ⊂ A ⊂ B ⊂ C Thus, A ⊂ B , A ⊂ C , B ⊂ C , D ⊂ A , D ⊂ B and D ⊂ C .

Show that the following four conditions are equivalent: A ⊂ B A - B = ϕ A ∪ B = B A ∩ B = A

Let us first show that A ⊂ B is equivalent to A - B = ϕ Let, A ⊂ B To prove: A - B = ϕ Proof: Let us suppose that A - B ≠ ϕ This means that there exists an element x such that x ∈ A but x ∉ B which is not possible as A ⊂ B . This means our assumption is wrong. ∴ A - B = ϕ Thus A ⊂ B ⇒ A - B = ϕ Conversely, Let, A - B = ϕ To show: A ⊂ B proof: Let x ∈ A Clearly, x ∈ B because if x ∉ B , then A - B ≠ ϕ ∴ A - B = ϕ ⇒ A ⊂ B ∴ A ⊂ B ⇔ A - B = ϕ Let A ⊂ B To show: A ∪ B = B proof: Clearly, B ⊂ A ∪ B ..... 1 Now let us prove that A ∪ B ⊂ B For that, let x ∈ A ∪ B ..... 2 ⇒ x ∈ A or x ∈ B Case I: x ∈ A ⇒ x ∈ B ∵ A ⊂ B ...... 3 From 2 and 3 we have, A ∪ B ⊂ B Case II: x ∈ B Since if x ∈ A ∪ B then, x ∈ B ∴ A ∪ B ⊂ B and it is clear that B ⊂ A ∪ B Therefore, A ∪ B = B Conversely, Let A ∪ B = B Let x ∈ A ⇒ x ∈ A ∪ B ∵ A ⊂ A ∪ B ⇒ x ∈ B ∵ A ∪ B = B ∴ A ⊂ B Therefore, A ⊂ B ⇔ A ∪ B = B Now, we have to show that A ⊂ B ⇔ A ∩ B = A Given, A ⊂ B To show, A ∩ B = A Proof: 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 Since if any element x ∈ A then x must belongs to A ∩ B ∴ A ⊂ A ∩ B Therefore, A = A ∩ B Conversely, suppose A ∩ B = A To prove: A ⊂ B Proof: Let x ∈ A ⇒ x ∈ A ∩ B ⇒ x ∈ A and x ∈ B ⇒ x ∈ B ∴ A ⊂ B Therefore, A ⊂ B ⇔ A ∩ B = A .