Given a non-empty set X, define the relation R in P(X) as follows: For A, B ∈ š‘ƒ(š‘‹), (š“, šµ) ∈ š‘… iff š“ āŠ‚ šµ. Prove that R is reflexive, transitive and not symmetric.

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[Sample paper] Given a non-empty set X, define the relation R in P(X) - CBSE Class 12 Sample Paper for 2023 Boards

part 2 - Question 33 (Choice 2) - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 3 - Question 33 (Choice 2) - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12 part 4 - Question 33 (Choice 2) - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12 part 5 - Question 33 (Choice 2) - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

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Question 33 (Choice 2) - Introduction Given a non-empty set X, define the relation R in P(X) as follows: For A, B ∈ š‘ƒ(š‘‹), (š“, šµ) ∈ š‘… iff š“ āŠ‚ šµ. Prove that R is reflexive, transitive and not symmetric.Taking an example Let X = {1, 2, 3} P(X) = Power set of X = Set of all subsets of X = { šœ™, {1} , {2} , {3}, {1, 2} , {2, 3} , {1, 3}, {1, 2, 3} } Since {1} āŠ‚ {1, 2} ∓ {1} R {1, 2} Question 33 (Choice 2) Given a non-empty set X, define the relation R in P(X) as follows: For A, B ∈ š‘ƒ(š‘‹), (š“, šµ) ∈ š‘… iff š“ āŠ‚ šµ. Prove that R is reflexive, transitive and not symmetric.ARB means A āŠ‚ B Here, relation is R = {(A, B): A & B are sets, A āŠ‚ B} Check reflexive Since every set is a subset of itself, A āŠ‚ A ∓ (A, A) ∈ R. ∓R is reflexive. Check symmetric To check whether symmetric or not, If (A, B) ∈ R, then (B, A) ∈ R If (A, B) ∈ R, A āŠ‚ B. But, B āŠ‚ A is not true Example: Let A = {1} and B = {1, 2}, As all elements of A are in B, A āŠ‚ B But all elements of B are not in A (as 2 is not in A), So B āŠ‚ A is not true ∓ R is not symmetric. Checking transitive Since (A, B) ∈ R & (B, C) ∈ R If, A āŠ‚ B and B āŠ‚ C. then A āŠ‚ C ⇒ (A, C) ∈ R So, If (A, B) ∈ R & (B, C) ∈ R , then (A, C) ∈ R ∓ R is transitive. Hence, R is reflexive and transitive but not symmetric.

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