(i) False
Let A = {1, 2} and B = {1, {1, 2}, {3}}
Now,
2 \(∈\) {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 \(x ∉ B \)
\(∴ x ∉ A\)
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
The complement of a set is described as A’ = {x: x ∈ U and x ∉ A}
where,
A’ stands for the complement.
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.