Question

Show that the following four conditions are equivalent:

Accepted Answer

To show that the four conditions are equivalent, we establish a cyclic implication: (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i). Step 1: (i) ⇒ (ii) First, assume (i) A ⊂ B is true. We need to show (ii) A – B = φ. By definition, A – B contains elements in A that are not in B. Since A is a subset of B, every element of A is in B. Thus, there are no elements in A that are not in B, meaning A – B contains no elements. ∴ A – B = φ. Step 2: (ii) ⇒ (iii) Next, assume (ii) A – B = φ. We need to show (iii) A ∪ B = B. Since A – B is empty, there are no elements in A that are outside B. This implies that all elements of A are contained within B. Consequently, the union of A and B is simply the set B itself. ∴ A ∪ B = B. Step 3: (iii) ⇒ (iv) Then, assume (iii) A ∪ B = B. We need to show (iv) A ∩ B = A. The equality A ∪ B = B implies that A is a subset of B. When we take the intersection of a set A with its superset B, the result is the set A itself. ∴ A ∩ B = A. Step 4: (iv) ⇒ (i) Finally, assume (iv) A ∩ B = A. We need to show (i) A ⊂ B. The condition A ∩ B = A means that all elements of A are common to both A and B. This implies every element of A is an element of B. ∴ A ⊂ B. This completes the proof of equivalence. — This solution follows the standard cyclic proof method recommended for establishing equivalence between multiple set conditions in NCERT Class 11 Mathematics. It uses the fundamental definitions of subset, difference, union, and intersection operations provided in the chapter…

Explanation

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