Proving the Uniqueness of inf(A) in a Subset of ℝ | Epsilon Proof Technique

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Homework Help Overview

The discussion revolves around proving the uniqueness of the infimum of a subset A of ℝ, under the assumption that the infimum exists. Participants are exploring the epsilon proof technique in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are discussing the use of an epsilon proof and questioning the appropriate definitions of "infimum" that may affect the proof approach. There is mention of starting with two potential infimums and showing their equality.

Discussion Status

Some participants are providing guidance on how to structure the proof, suggesting starting with two elements assumed to be infimums and exploring the implications of their relationship. There is an acknowledgment of the need for clarity in definitions and methods.

Contextual Notes

Participants are considering different definitions of infimum, which may lead to varying proof strategies. There is also a focus on ensuring logical consistency in the statements made about the infimums.

jaqueh
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Homework Statement


Let A be a subset of ℝ, now prove if inf(A) exists, then inf(A) is unique

The Attempt at a Solution


I am using an epsilon proof but i don't think i am going about it the right way, can someone nudge me in the right direction
 
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jaqueh said:

Homework Statement


Let A be a subset of ℝ, now prove if inf(A) exists, then inf(A) is unique

The Attempt at a Solution


I am using an epsilon proof but i don't think i am going about it the right way, can someone nudge me in the right direction

How you do it depends very much on what *definition* you are using for "inf". Different (but equivalent) definitions would need different types of proofs.

RGV
 
Usually, for something like this, it is best to start of saying "Let x and y be inf(S)" and then show that x and y are, in fact, the same. Now, showing that |x - y| < epsilon might be a good strategy, but there are probably more efficient methods. But, as RGV said, it depends on how inf is defined for you. If you have defined it as a greatest lower bound, the method I mentioned works very well.
 
yes i think I am going to revise it as x and y are infimums for A then x<y and y<x so x=y
 
jaqueh said:
yes i think I am going to revise it as x and y are infimums for A then x<y and y<x so x=y

That's a contradiction the way you wrote it. Did you mean [itex]x \leq y, \ \text{and} \ y \leq x,\ \text{so} \ x=y[/itex]?
 
yes that is what i meant and i did write it that way x≥y y≥x
 

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