3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Sorry, please use MathJax to type out your attempts. Pictures are not recommended, otherwise some user would not see them due to various reasons. $\endgroup$ – xbh Sep 14 '18 at 4:22
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.