- #1

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prove:

let A be a set of real numbers bounders above.

then, inf (-A)=-sup(A) [-A= {-a : a∈A}]

really...thanks......i have no idear..how to prove it....?

- Thread starter ShengyaoLiang
- Start date

- #1

- 23

- 0

prove:

let A be a set of real numbers bounders above.

then, inf (-A)=-sup(A) [-A= {-a : a∈A}]

really...thanks......i have no idear..how to prove it....?

- #2

- 23

- 0

so do i need to prove A is a non-empty set?

Or it said is real numbers, then it means is non empty set?

- #3

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A is a set of real numbers already means there exists an element x which is a real number and belongs to A. Since there exists such an element, thus A is non-empty.

- #4

HallsofIvy

Science Advisor

Homework Helper

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To prove your theorem- keep multiplying by -1!

For example, sup(A) is, by definition, an upper bound on A.

Let b be any member of -A. Then b=-a for some a in A. We must have [itex]a\le sup(A)[/itex] so, multiplying by -1, [itex]-a= b\ge -sup(A)[/itex]. That tells you that -sup(A) is a lower bound on -A. I'll leave the rest to you.

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