Question:

In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false,
give an example.
(i) If x \(\in\) A and A \(\in\) B, then x \(\in\) B
(ii) If A \(⊂\) B and B \(\in\) C, then A \(\in\) C
(iii) If A \(⊂\) B and B \(⊂\) C, then A \(⊂\) C
(iv) If A \(⊄\) B and B \(⊄\) C, then A \(⊄\) C
(v) If x\( ∈\) A and A \(⊄\) B, then x \(∈\) B
(vi) If A\( ⊂\) B and x \(∉\) B, then x \(∉\) A

Updated On: Oct 24, 2023
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Solution and Explanation

(i) False
Let A = {1, 2} and B = {1, {1, 2}, {3}}
Now,
\(∈\) {1,2} and {1,2} \(∈\) {{3},1,{1,2}}
\(∴ A ∈ B\)
However,
2\( ∉\) {{3},1,{1,2}}


(ii) False
Let 
A = {2}, B= {0,2}, and C = {1{0,2},3}
As \( A ⊂ B\)
\(B ∈ C\) 
However, \(A ∉ C\)


(iii) True
Let \(A ⊂ B\) and \(B ⊂ C\)
Let \(x ∈ A\) 
\(⇒ x ∈ B [∴ A ⊂ B]\)
\(⇒ x ∈ C [∴ B ⊂ C]\)
\(∴ A ⊂ C\)


(iv) False 
A = {1,2}, B = {0,6,8}, and C = {0,1,2,6,9}
\(A ⊄ B\) and \(B ⊄ C\)
Let Accordingly, 
However, \(A ⊂ C\)


(v) False 
Let A = {3, 5, 7} and B = {3, 4, 6} 
Now, \(5 ∈ A\) and \(A ⊄ B\) 
However, \(5 ∉ B\)


(vi) True 
Let \(A ⊂ B\) and \(x ∉ B\)
To show: \(x ∉ A\) If possible, 
suppose \(x ∈ A. \)
Then, \(x ∈ B\), which is a contradiction as \(x ∉ B \)
\(∴ x ∉ A\)

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Concepts Used:

Complement of a Set

The complement of a set is described as A’ = {x: x ∈ U and x ∉ A}

where,

A’ stands for the complement.

Complement of Sets Properties:

1. Complement Laws: The union of a set A and its complement A’ allows the universal set U of which, A and A’ are a subset.

A ∪ A’ = U

Also, the intersection of a set A and its complement A’ cause the empty set ∅.

A ∩ A’ = ∅

For Example: If U = {11, 12 , 13 , 14 , 15 } and A = {11 , 12 , 13 } then A’ = {14 , 15}. From this it can be seen that

A ∪ A’ = U = { 11 , 12 , 13 , 14 , 15}

Also, A ∩ A’ = ∅

2. Law of Double Complementation: According to the law, if we take the complement of the complemented set A’ then, we get the set A itself.

(A’)’ = A

In the previous example we can see that, if U = {11 , 12 , 13 , 14 , 15} and A = {11 , 12, 13} then A’ ={14 , 15}. Now if we consider the complement of set ‘A’ we get,

(A’)’ = {11 , 12 , 13} = A

This gives back the set A itself.

3. Law of empty set and universal set: According to this law the complement of the universal set gives us the empty set and vice-versa i.e.,

∅’ = U And U’ = ∅

This law is accessible.