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Prove that the Poset has a unique least element
Prove that if (A, <) has a least element, then (A,≤) has a unique least element.
Ans: Let (A, ≤) be a poset. Suppose the poset A has two least elements x and y. Since x is the least element, it implies that x ≤ y. Using the same argument, we can say that y ≤ x, since y is supposed to be another least element of the same poset. ≤ is an anti-symmetric relation, so x ≤ y and y ≤ x ⇒ x = y. Thus, there is at most one least element in any poset.
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compare: 643,251: 633,512: 633,893. The answer is 633,512.
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pendaraban dan
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