1.

if A = {3, {4, 5}, 6} find which of the following statements are true.(i) {4, 5} ⊄A (ii) {4, 5} ϵA (iii) {{4, 5}} ⊆A (iv) 4ϵA (v) {3} ⊆A (vi) {ϕ} ⊆A (vii) ϕ⊆A (viii) {3, 4, 5} ⊆A (ix) {3, 6} ⊆A

Answer»

(i) True 

Explanation: we have, A = {3, {4, 5}, 6} 

Let {4,5} = x 

Now, A = {3, x, 6} 

4,5 is not in A, {4,5} is an element of A and element cannot be subset of set,thus {4, 5} ⊄A. 

(ii) True 

Explanation: we have, A = {3, {4, 5}, 6} 

Let {4,5} = x 

Now, A = {3, x, 6} 

Now, x is in A. 

So, x ∈ A. 

Thus, {4, 5} ϵ A 

(iii) True

Explanation: {4,5} is an element of set {{4,5}}. 

Let {4,5} = x 

{{4,5}} = {x} 

we have, A = {3, {4, 5}, 6} 

Now, A = {3, x, 6} 

So, x is in {x} and x is also in A. 

So , {x} is a subset of A. 

Hence, {{4, 5}} ⊆A 

(iv) False 

Explanation: 4 is not an element of A.

(v) True 

Explanation: 3 is in {3} and also 3 is in A. 

(vi) False 

Explanation: ϕ is an element in { ϕ} but not in A. 

Thus, {ϕ} ⊄ A 

(vii) True 

Explanation: ϕ is a subset of every set. 

(viii) False 

Explanation: we have, A = {3, {4, 5}, 6} 

Let {4,5} = x 

Now, A = {3, x, 6} 

4,5 is in {3,4,5} but not in A, thus {3,4, 5} ⊄A. 

(ix) True 

Explanation: 3,6 is in {3,6} and also in A, thus {3, 6} ⊆A.



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