Proof of lower bound of a nonempty set of real numbers

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SUMMARY

The discussion centers on proving that for a nonempty set of real numbers A, which is bounded below, the infimum of A is equal to the negative of the supremum of the set -A, defined as -x for each x in A. The proof involves demonstrating that -sup(-A) serves as a lower bound for A and establishing it as the least lower bound through contradiction. This relationship is visually intuitive on a number line, reinforcing the connection between infimum and supremum in real analysis.

PREREQUISITES
  • Understanding of real number sets and their properties
  • Familiarity with the concepts of infimum and supremum
  • Basic knowledge of proof techniques, including proof by contradiction
  • Experience with number line representations in mathematics
NEXT STEPS
  • Study the properties of infimum and supremum in real analysis
  • Learn about proof techniques, particularly proof by contradiction
  • Explore the implications of bounded sets in real number theory
  • Investigate visual representations of mathematical concepts on number lines
USEFUL FOR

Mathematicians, students of real analysis, and anyone interested in understanding the foundational properties of sets of real numbers and their bounds.

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

Intuitively this makes sense if you draw it on a number line. But I am not sure how to formally prove it.
 
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tronter said:
1. Let A be a nonempty set of real numbers which is bounded below. Let -A be the set of numbers -x, where x \in A. Prove that \inf(A) = -\sup(-A).

Intuitively this makes sense if you draw it on a number line. But I am not sure how to formally prove it.

If -x is in -A then -x<sup(-A)
this implies:
x>-sup(-A)

Therefor -sup(-A) is a lower bound of A.

Now establish it is the least lower bound (probably via contradiction).
 

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