Can anyone check my proof involving least-upper-bounds?

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Homework Help Overview

The discussion revolves around proving a property of the least upper bound (supremum) of a set S, specifically that if x is the supremum of S, then for any positive ε, there exists an element a in S such that x - ε < a ≤ x. Participants are exploring the implications of the definition of supremum and how it relates to the elements of the set S.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss different cases regarding whether the supremum is an element of S or not. They explore the implications of the definition of supremum and question the validity of certain assumptions made in the proof attempts. There are discussions about the existence of elements in the interval (x - ε, x) and whether x - ε can be considered an upper bound for S.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's reasoning and suggesting clarifications. Some participants have pointed out potential flaws in the reasoning related to the existence of elements in S and the implications of the supremum definition. There is no explicit consensus yet, but the dialogue is fostering deeper exploration of the concepts involved.

Contextual Notes

Participants note that ε can be any positive number, which raises questions about the relationship between ε and the elements of S. There is also a recognition that S might not contain intervals, complicating the proof further.

  • #31
So, ##x-\epsilon## is not an upper bound for ##S## because there exists an ##a∈S## such that ##a>x-\epsilon##.

...Oh, so because ##x-\epsilon## is not an upper bound for ##S##, then it is implied that there must exist an element ##a∈S## greater than ##x-\epsilon##?
 
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  • #32
Eclair_de_XII said:
So, ##x-\epsilon## is not an upper bound for ##S## because there exists an ##a∈S## such that ##a>x-\epsilon##.

...Oh, so because ##x-\epsilon## is not an upper bound for ##S##, then it is implied that there must exist an element ##a∈S## greater than ##x-\epsilon##?

Yes!
 
  • #33
Thanks for bearing with me and helping me finish my homework. I'm kind of apprehensive about how I'll handle the rest of the class, to be honest...
 
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  • #34
Eclair_de_XII said:
So, ##x-\epsilon## is not an upper bound for ##S## because there exists an ##a∈S## such that ##a>x-\epsilon##.

...Oh, so because ##x-\epsilon## is not an upper bound for ##S##, then it is implied that there must exist an element ##a∈S## greater than ##x-\epsilon##?
I like the fact that you italicized the word "must" above. That's the first inkling you have given that you actually get it. Good for you. That's what this forum is here for.
 

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