Let A be a nonempty set of real numbers which is bounded below. Let -A be the set of all numbers -x, where x in A. Prove that inf A=- sup (-A)

genestesya

genestesya

Answered question

2022-09-04

Let A be a nonempty set of real numbers which is bounded below. Let -A be the set of all numbers -x, where x A . Prove that
inf A = sup ( A )

Answer & Explanation

soyafh

soyafh

Beginner2022-09-05Added 17 answers

Step 1
inf A = sup ( A )
Rudin's book gives definitions of the concepts involved, and I would stick close to what those definitions say.
A is bounded below, i.e. it has a lower bound x. That means a A ,   x a . Consequently a A ,   x a .
b A   a A   b = a , hence b A ,   x b . Thus −x is an upper bound of −A.
Thus we have proved that for every lower bound x of A, -x is an upper bound of -A. In particular -inf A is an upper bound of -A. In order to show that -inf A is the smallest upper bound of -A, one must show that no number less than -inf A is an upper bound of -A. Suppose c < inf A . Then c > inf A . Since -c is greater than the largest lower bound of A, -c is not a lower bound of A. Hence for some a A , a < c , and so a > c . Since a A, we have a member of -A that is greater than c, so c is not an upper bound of -A.
Dana Chung

Dana Chung

Beginner2022-09-06Added 14 answers

Step 1
Let A be a non-empty set of real numbers that is bounded below. By definition of bounded below, we may choose α R such that α x for every x A . This implies that x α for every x A so that α is an upper bound for -A. Thus, -A is a non-empty set of real numbers that is bounded above and, therefore, has a supremum, say β , by the axiom of completeness.
We must show that β is the infimum of A. First, note β is an upper bound for -A (by definition of supremum) or β x for every x A . Thus, β x for every x A and β is a lower bound for A. Next, we must show that β is the greatest lower bound of A. Thus, assume that β < γ. Then, γ < β so (since β is the supremum of -A), there is some x A with γ < x < β . Therefore, β < x < γ with x A so that γ cannot be a lower bound of -A.
To understand why these particular details are written out in grotesque detail, I would consider the material that you likely just learned. If you are trying to show that an infimum can be defined in terms of a supremum, then you have likely just learned these concepts, as well as concepts like upper and lower bounds. So I think you've really got to refer quite explicitly to those definitions. By contrast, I used the order properties, like x < y y < x without specific reference since that's probably at least a little bit in the past.

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