# Let A = {1, 2, {3

The parts of the question are solved below:

(i) Here, we know that {3, 4} is a member of set A.

= {3, 4} A

Thus, the given statement is incorrect.

(ii) Here, we know that {3, 4} is a member of set A.

Thus, the given statement is incorrect.

(iii) Here, we know that {3, 4} is a member of set A.

={{3, 4}} is a set.

Thus, the given statement is correct.

(iv) Here, we can observe that 1 is a member of set A.

Thus, the given statement is correct.

(v) Here, we can see that 1 is a member of set A but is not any set itself.

Thus, the given statement is incorrect.

(vi) Here, we can see that 1, 2, 5 is a member of set A

Thus, the given statement is correct.

(vii) Here, we can see that 1, 2, 5 is a member of set A

=, {1, 2, 5} is a subset of A

Thus, the given statement is incorrect.

(viii) Here, we can see that 3 is not a member of set A

={1, 2, 3} is not a subset of A

Thus, the given statement is incorrect.

(ix) Here, we can see that ϕ is not a member of set A

Thus, the given statement is correct.

(x) Here, as we know the null set is a subset of every set.

Thus, the given statement is correct.

(xi) {ϕ } is the set containing the null set.

{ϕ } ⊂ A is only possible if ϕ is in set A. But it is not there.
So, the statement is incorrect.

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