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I am confused on this:

Suppose A is a non-empty set of $\mathbb{R}$ with finite supremum sup A = L. Then how do I prove that $\forall \epsilon>0, \exists a \in A $ s.t $L-a < \epsilon$. Maybe I can assume that $L-a>\epsilon$ or $L-a=\epsilon$ for a contradiction, but I am not sure how to proceed. Intuitively it seems to say that the difference between the supremum and any element of the subset can be as small as we want. How should I prove this point?

Jayden Rice
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  • It follows directly from the definition of supremum, doesn't it? What is your definition of supremum? – TonyK Jan 30 '21 at 02:00
  • Supremum is the least upper bound. I understand that it is the "least" upper bound, but how can I argue that with contradiction? – Jayden Rice Jan 30 '21 at 02:04
  • If there existed $\epsilon>0$ for which that statement was not true, then $L-\epsilon$ would be an upper bound. So $L$ would not be the least upper bound. – TonyK Jan 30 '21 at 10:24

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