Question

E = {2,{4,5}, 4}. Which statements are incorrect and why?
1) {4,5}⊂E
2) {4,51} ∈ E
3) {{4,5}}⊂E​

Answer 1

Step-by-step explanation:

(i) The statement {3, 4} ⊂ A is incorrect because 3 ∈ {3, 4}; however, 3∉A.

(ii) The statement {3, 4} ∈A is correct because {3, 4} is an element of A.

(iii) The statement {{3, 4}} ⊂ A is correct because {3, 4} ∈ {{3, 4}} and {3, 4} ∈ A.

(iv) The statement 1∈A is correct because 1 is an element of A.

(v) The statement 1⊂ A is incorrect because an element of a set can never be a subset of itself.

(vi) The statement {1, 2, 5} ⊂ A is correct because each element of {1, 2, 5} is also an element of A.

(vii)The statement {1, 2, 5} ∈ A is incorrect because {1, 2, 5} is not an element of A.

(viii) The statement {1, 2, 3} ⊂ A is incorrect because 3 ∈ {1, 2, 3}; however, 3 ∉ A.

(ix) The statement Φ ∈ A is incorrect because Φ is not an element of A

.x) The statement Φ ⊂ A is correct because Φ is a subset of every set.

(xi) The statement {Φ} ⊂ A is incorrect because Φ∈ {Φ}; however, Φ ∈ A.

Answer 2

Step-by-step explanation:

A = {1, 2, {3, 4}, 5} (i) The statement {3, 4} ⊂ A is incorrect because 3 ∈ {3, 4}; however, 3∉A. (ii) The statement {3, 4} ∈A is correct because {3, 4} is an element of A. (iii) The statement {{3, 4}} ⊂ A is correct because {3, 4} ∈ {{3, 4}} and {3, 4} ∈ A. (iv) The statement 1∈A is correct because 1 is an element of A. (v) The statement 1⊂ A is incorrect because an element of a set can never be a subset of itself. (vi) The statement {1, 2, 5} ⊂ A is correct because each element of {1, 2, 5} is also an element of A. (vii)The statement {1, 2, 5} ∈ A is incorrect because {1, 2, 5} is not an element of A. (viii) The statement {1, 2, 3} ⊂ A is incorrect because 3 ∈ {1, 2, 3}; however, 3 ∉ A. (ix) The statement Φ ∈ A is incorrect because Φ is not an element of A. (x) The statement Φ ⊂ A is correct because Φ is a subset of every set. (xi) The statement {Φ} ⊂ A is incorrect because Φ∈ {Φ}; however, Φ ∈ A.

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