# Real Analysis - Supremum / Infimum

Let sup A $<$ inf B. Prove that there exist $ε > 0$ and $c ∈ R$ so that $c + ε$ is a lower bound for B and $c−ε$ is an upper bound for A.

Solution Removed

• Sorry, please use MathJax to type out your attempts. Pictures are not recommended, otherwise some user would not see them due to various reasons. – xbh Sep 14 '18 at 4:22

Your proof is correct. Alternately, you may choose $c = \frac{\sup A + \inf B}{2}$ and $\epsilon = \frac{\inf B - \sup A}{4}$ : by drawing a diagram, you can check that these work , since I am just dividing the number line between $\sup A$ and $\inf B$ into four equal parts, and taking $c$ as the midpoint and $c-\epsilon$ and $c + \epsilon$ as the other two points.
However, your proof shows that for any arbitrary $\sup A < c < \inf B$ such an $\epsilon$ can be chosen.
Also note that if $\sup A = \inf B$ then such a choice cannot be made, so the sign in the question is very important.