State whether the given statement is true false:(i) If A ⊂ B and x�

1.	State whether the given statement is true false:(i) If A ⊂ B and x∉B than x ∉A.(ii) If A ⊆ϕ then A = ϕ(iii) If A, B and C are three sets such than A ϵ B and B ⊂ C then A ⊂ C.(iv) If A, B and C are three sets such than A ⊂B and B ϵ C then A ϵC.(v) If A, B and C are three sets such that A ⊄B and B ⊄C then A ⊄C.(vi) If A and B are sets such that x A and A ϵ B then x ϵ B.
Answer» (i) True Explanation: We have A ⊂ B since A is a subset of B then all elements of A should be in B. Let A = {1,2} and B = {1,2,3} Let x=4∉B Also we observe that 4∉A. Hence, If A ⊂ B and x∉B than x ∉A. (ii) True Explanation: We have, A ⊆ ϕ Now, A is a subset of null set , this implies A is also an empty set. ⇒ A =ϕ (iii) False Explanation: Let A = {a}, B = {{a}, b} here , A ϵ B Now, let C = {{a}, b, c}. Since, {a},b is in B and also in C thus, B ⊂ C. But, A ={a} and {a} is an element of C, since the element of a set cannot be a subset of a set. Hence,A ⊄ C. (iv) False Explanation: Let A = {a},B = {a, b} and C = {{a, b}, c}. Then,A⊂ B and B ϵC. But, A ∉ C since {a} is not an element of C. (v) False. Explanation: Let A = {a}, B = {b, c} and C = {a, c}. Since a ∈ A and a ∉ B.Then, A ⊄ B Now, b ∈ B and b ∉ C ⇒ B ⊄C. But, A ⊂ C since, a ∈ A and a ∈ C. (vi) False. Explanation: Let A = {x}, B = {{x}, y} Now, x ϵ A and {x} is an element of B ⇒ A ϵ B But, x is not an element of B. Thus, x∉B.

State whether the given statement is true false:(i) If A ⊂ B and x∉B than x ∉A.(ii) If A ⊆ϕ then A = ϕ(iii) If A, B and C are three sets such than A ϵ B and B ⊂ C then A ⊂ C.(iv) If A, B and C are three sets such than A ⊂B and B ϵ C then A ϵC.(v) If A, B and C are three sets such that A ⊄B and B ⊄C then A ⊄C.(vi) If A and B are sets such that x A and A ϵ B then x ϵ B.

Answer»

(i) True

Explanation: We have A ⊂ B since A is a subset of B then all elements of A should be in B.

Let A = {1,2} and B = {1,2,3}

Let x=4∉B

Also we observe that 4∉A.

Hence, If A ⊂ B and x∉B than x ∉A.

(ii) True

Explanation: We have, A ⊆ ϕ

Now, A is a subset of null set , this implies A is also an empty set.

⇒ A =ϕ

(iii) False

Explanation: Let A = {a}, B = {{a}, b}

here , A ϵ B

Now, let C = {{a}, b, c}.

Since, {a},b is in B and also in C thus, B ⊂ C.

But, A ={a} and {a} is an element of C, since the element of a set cannot be a subset of a set.

Hence,A ⊄ C.

(iv) False

Explanation: Let A = {a},B = {a, b} and C = {{a, b}, c}.

Then,A⊂ B and B ϵC. But, A ∉ C since {a} is not an element of C.

(v) False.

Explanation: Let A = {a}, B = {b, c} and C = {a, c}.

Since a ∈ A and a ∉ B.Then, A ⊄ B

Now, b ∈ B and b ∉ C ⇒ B ⊄C.

But, A ⊂ C since, a ∈ A and a ∈ C.

(vi) False.

Explanation: Let A = {x}, B = {{x}, y}

Now, x ϵ A and {x} is an element of B ⇒ A ϵ B

But, x is not an element of B. Thus, x∉B.