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
(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.