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

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The discussion revolves around proving that if x is the least upper bound (supremum) of a set S, then for every ε > 0, there exists an element a in S such that x - ε < a ≤ x. Participants explore two cases: when x is an element of S and when it is not. They emphasize that the interval (x - ε, x) must contain elements of S, leading to the conclusion that such an a exists. The importance of correctly applying the definition of least upper bound and understanding upper bounds is highlighted throughout the conversation. The discussion concludes with a focus on refining the proof to ensure clarity and correctness.
  • #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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