Consider the following relation on set B = {a, b, {a}, {b}, {a, b}}: P = {(a, b), (b, {a, b}), ({a, b}, a), ({b}, a), (a, {a})}. The relation P does not satisfy trichotomy.Which ordered pairs should be included in P so that an extended relation P1 (say) would satisfy trichotomy?Choose the alternative that provides all the missing ordered pairs that should be included as elements of P1 in order for P1 to be a relation that satisfies trichotomy.(For trichotomy, each element of B must be paired with each other different element in B to form elements of P1. For example, we see that b ≠ {a} but neither (b, {a}) nor ({a}, b) are elements of P, but at least one of these elements should be an element of P1.We include (b, {a}) in P1: P1 = {(a, b), (b, {a, b}), ({a, b}, a), ({b}, a), (a, {a}), (b, {a}), ...}.Now check all the different elements of B and if two different elements of B are not already grouped in an ordered pair of P, they should be paired and be included as elements of P1. You can ask the question: Which ordered pairs should be included in P1 so that it will be true that forall x, y ∈ B with x ≠ y, we have (x, y) ∈ P1 or (y, x)∈ P1? Refer to study guide, p 78.) Choose the alternative that provides all the missing ordered pairs that should be included as elements of P1 in order for P1 to be a relation that satisfies trichotomy.a.(b, {a}), (b, {b}), (b, a), ({a}, {a, b}) and ({a, b}, {b})b.({a}, b), (b, {b}), ({b}, {a, b}) and ({a, b}, {a})c.(b, {a}), ({b}, b), ({a}, {b}), ({b}, {a, b}) and ({a, b}, {a})d.(b, {a}), (b, {b}), ({a}, {b}) and ({a, b}, {a, b})
Question
Consider the following relation on set B = {a, b, {a}, {b}, {a, b}}: P = {(a, b), (b, {a, b}), ({a, b}, a), ({b}, a), (a, {a})}. The relation P does not satisfy trichotomy.Which ordered pairs should be included in P so that an extended relation P1 (say) would satisfy trichotomy?Choose the alternative that provides all the missing ordered pairs that should be included as elements of P1 in order for P1 to be a relation that satisfies trichotomy.(For trichotomy, each element of B must be paired with each other different element in B to form elements of P1. For example, we see that b ≠ {a} but neither (b, {a}) nor ({a}, b) are elements of P, but at least one of these elements should be an element of P1.We include (b, {a}) in P1: P1 = {(a, b), (b, {a, b}), ({a, b}, a), ({b}, a), (a, {a}), (b, {a}), ...}.Now check all the different elements of B and if two different elements of B are not already grouped in an ordered pair of P, they should be paired and be included as elements of P1. You can ask the question: Which ordered pairs should be included in P1 so that it will be true that forall x, y ∈ B with x ≠ y, we have (x, y) ∈ P1 or (y, x)∈ P1? Refer to study guide, p 78.) Choose the alternative that provides all the missing ordered pairs that should be included as elements of P1 in order for P1 to be a relation that satisfies trichotomy.a.(b, {a}), (b, {b}), (b, a), ({a}, {a, b}) and ({a, b}, {b})b.({a}, b), (b, {b}), ({b}, {a, b}) and ({a, b}, {a})c.(b, {a}), ({b}, b), ({a}, {b}), ({b}, {a, b}) and ({a, b}, {a})d.(b, {a}), (b, {b}), ({a}, {b}) and ({a, b}, {a, b})
Solution
To satisfy trichotomy, every pair of distinct elements in B must be ordered in one way or the other. Let's go through each pair of distinct elements in B and see which pairs are missing in P.
- (a, b) is in P.
- (a, {a}) is in P.
- (a, {b}) is not in P.
- (a, {a, b}) is not in P.
- (b, {a}) is not in P.
- (b, {b}) is not in P.
- (b, {a, b}) is in P.
- ({a}, {b}) is not in P.
- ({a}, {a, b}) is not in P.
- ({b}, {a, b}) is not in P.
- ({a, b}, {a}) is in P.
- ({a, b}, {b}) is not in P.
So, the missing ordered pairs are: (a, {b}), (a, {a, b}), (b, {a}), (b, {b}), ({a}, {b}), ({a}, {a, b}), ({b}, {a, b}), and ({a, b}, {b}).
Therefore, the correct answer is not listed in the options provided. The correct set of missing ordered pairs should be: (a, {b}), (a, {a, b}), (b, {a}), (b, {b}), ({a}, {b}), ({a}, {a, b}), ({b}, {a, b}), and ({a, b}, {b}).
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