Help, analsysis proof - inf (-A)=-sup(A) [-A= {-a : a∈A}]

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Discussion Overview

The discussion revolves around proving the statement that the infimum of the set of negative elements of a set A of real numbers is equal to the negative of the supremum of A. The scope includes mathematical reasoning and proof techniques related to real analysis.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant seeks help in proving that inf(-A) = -sup(A), where -A is defined as {-a : a ∈ A}.
  • Another participant questions whether A needs to be proven as a non-empty set, given that it is described as a set of real numbers.
  • A different participant asserts that A being a set of real numbers implies it contains at least one element, thus it is non-empty.
  • Another participant clarifies the distinction between a "set of real numbers" and a "subset of the set of real numbers," noting that the latter could be empty.
  • This participant suggests a method for proving the theorem by multiplying by -1 and provides an example involving the definition of supremum and lower bounds.

Areas of Agreement / Disagreement

Participants generally agree that A is non-empty based on its definition as a set of real numbers, but there is some discussion about the implications of this definition and the need for clarity regarding the nature of A.

Contextual Notes

There is a lack of consensus on whether the proof requires A to be explicitly stated as non-empty, and the discussion touches on the definitions and implications of sets in real analysis.

ShengyaoLiang
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help...

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...?
 
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the question didn't mention if A is a non-empty set...
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?
 
You don't have to prove A is non-empty.

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.
 
Notice that a "set of real numbers" is not a "subset of the set of real numbers" since the latter might be empty! It is more common for a theorem like this to start "Let A be a non-empty subset of R."

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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