# How can I prove that $\inf B=\sup A$? [duplicate]

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?

• What did you try?
– Mark
Mar 21 '19 at 0:45
• Consider the definition of the supremum. How does it relate to all of the other upper bounds of a set? Mar 21 '19 at 0:54
• This has been asked many times before. Please search the site before asking a question. Mar 21 '19 at 1:03

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?