Prove Lemma 0.1 Please with details and not just copying any thing from other si

Question

Prove Lemma 0.1

Please with details and not just copying any thing from other sites. thank you

Let A be a set. Then, a number x is an upper bound of the set A if x ge; y for all y A. A set A is bounded above if there exists an upper bound x of the set A. For this case, we say that A is bounded above by x. A number x is said to be the supremum (the least upper bound) of the set A if it satisfies the following two conditions: the number x is an upper bound of A. and if y is an upper bound of A, then x le y. For this case, we use the symbol sup A = x. Let A be a nonempty bounded above set. Then there exists the sup A. Let A be a nonempty set. Suppose that there exists the sup A. Then, for any delta > 0, there exists a number x [sup A - delta , sup A] A.

Explanation / Answer

Let A be a non-empty set such that supA exists, and let >0 be some positive number. Assume that there are no numbers in the intersection of S = [supA-,supA] and A; i.e. S and A are disjoint. Then b = supA- and supA cannot be in A. There are two cases: b is a lower bound of A or b is an upper bound of A (because b can't be in A). If b is a lower bound of A, then because supA is an upper bound of the non-empty set A, S would have to contain elements of A, which is impossible because we just said S and A are disjoint. If b is an upper bound of A, then b = supA - for positive so b < supA, which contradicts the definition of the supremum of A. Either case leads to a contradiction, meaning our original assumption that S and A are disjoint is incorrect. Therefore S = [supA-,supA] intersect A must contain some numbers.

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