Show that for any sets A and B, \(A = (A ∩ B) ∪ (A – B)\) and \(A ∪ (B – A) = (A ∪ B)\)
Solution and Explanation
To show: \(A = (A ∩ B) ∪ (A – B) \)
Let \(x ∈ A \)
We have to show that \(x ∈ (A ∩ B) ∪ (A – B) \)
Case I \(x ∈ A ∩ B \)
Then, \(x ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B) \)
Case II \(x ∉ A ∩ B \)
\(⇒ x ∉ A or x ∉ B \)
\(∴ x ∉ B [x ∉ A] \)
\(∴ x ∉ A – B ⊂ (A ∪ B) ∪ (A – B) \)
\(∴ A ⊂ (A ∩ B) ∪ (A – B) … (1) \)
It is clear that
\(A ∩ B ⊂ A\) and \((A – B) ⊂ A \)
\(∴ (A ∩ B) ∪ (A – B) ⊂ A … (2) \)
From (1) and (2), we obtain
\(A = (A ∩ B) ∪ (A – B) \)
To prove: \(A ∪ (B – A) ⊂ A ∪ B \)
Let \(x ∈ A ∪ (B – A)\)
\(⇒ x ∈ A \) or \(x ∈ (B – A) \)
\(⇒ x ∈ A\) or \((x ∈ B \space and \space x ∉ A) \)
\(⇒ (x ∈ A \space or x ∈ B)\) and \((x ∈ A \space or x ∉ A) \)
\(⇒ x ∈ (A ∪ B)\)
\(∴ A ∪ (B – A) ⊂ (A ∪ B) … (3) \)
Next, we show that \((A ∪ B) ⊂ A ∪ (B – A). \)
Let \(y ∈ A ∪ B \)
\(⇒ y ∈ A \space or\space y ∈ B \)
\(⇒ (y ∈ A\) or \(y ∈ B)\) and\( (y ∈ \)A or \(y ∉ A) \)
\(⇒ y ∈ A\) or \((y ∈ B\) and \(y ∉ A) \)
\(⇒ y ∈ A ∪ (B – A) \)
\(∴ A ∪ B ⊂ A ∪ (B – A) … (4) \)
Hence, from (3) and (4), we obtain \(A ∪ (B – A) = A ∪B.\)
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Concepts Used:
Operations on Sets
Some important operations on sets include union, intersection, difference, and the complement of a set, a brief explanation of operations on sets is as follows:
1. Union of Sets:
- The union of sets lists the elements in set A and set B or the elements in both set A and set B.
- For example, {3,4} ∪ {1, 4} = {1, 3, 4}
- It is denoted as “A U B”
2. Intersection of Sets:
- Intersection of sets lists the common elements in set A and B.
- For example, {3,4} ∪ {1, 4} = {4}
- It is denoted as “A ∩ B”
3.Set Difference:
- Set difference is the list of elements in set A which is not present in set B
- For example, {3,4} - {1, 4} = {3}
- It is denoted as “A - B”
4.Set Complement:
- The set complement is the list of all elements present in the Universal set except the elements present in set A
- It is denoted as “U-A”