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Let $B$ be the set of all the upper bounds of the non-empty bounded subset $A\subseteq\Bbb R$. Prove that $\inf B=\sup A$.

I divided it into two areas ($\inf B>\sup A$, $\inf B<\sup A$) and tried to show a contradiction. Is it the right approach? How can I prove it?

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  • $\begingroup$ What did you try? $\endgroup$
    – Mark
    Mar 21 '19 at 0:45
  • $\begingroup$ Consider the definition of the supremum. How does it relate to all of the other upper bounds of a set? $\endgroup$
    – Brian
    Mar 21 '19 at 0:54
  • $\begingroup$ This has been asked many times before. Please search the site before asking a question. $\endgroup$
    – John Douma
    Mar 21 '19 at 1:03
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The following is what we can say:

  • $\sup A\leq b\;\forall\;b\in B\quad$ (Why?)

  • Also by definition, $\inf B\leq b\;\forall\;b\in B$. In particular $\sup A\in B$.

Can you complete?

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