Let K be an ordered field. Then K contains the smallest ordered filed $\mathbb{Q}$.

  1. If $\mathbb{Q}$ is dense and proper in K, is K a complete ordered field?
  2. If $\mathbb{Q}$ is dense and proper in K, and K has the Archimedean property, is K a complete ordered field?

(i.e. satisfying any one of the equivalent definitions of a complete ordered field.) (If you know the answer you may give hints first.)

  • 3
    $\begingroup$ Have you thought about the case $K=\Bbb Q$? $\endgroup$ – David C. Ullrich Oct 3 '15 at 0:14
  • $\begingroup$ @DavidC.Ullrich: You are right. I have to edit or delete this post... $\endgroup$ – TCHuang Oct 3 '15 at 0:16
  • 2
    $\begingroup$ There are many such ordered fields, such as the reals of the form $a+b\sqrt{3}$, where $a$ and $b$ are rational, or the real algebraics. $\endgroup$ – André Nicolas Oct 3 '15 at 0:20
  • 1
    $\begingroup$ You really might have given the matter a little thought before "fixing" the question by just adding the word "proper"... $\endgroup$ – David C. Ullrich Oct 3 '15 at 0:24
  • $\begingroup$ @DavidC.Ullrich: I apologize for such possibly low quality question... $\endgroup$ – TCHuang Oct 3 '15 at 0:29

No. For example, consider the smallest field that includes every rational number and also $\sqrt 2$. That is an ordered field that is not complete. You can show it is not complete by using the fact that it is countable, as in this page.

  • $\begingroup$ Thank you! :) So if an ordered filed is complete then it contains a dense sub ordered filed. But the converse is not true. $\endgroup$ – TCHuang Oct 3 '15 at 0:25
  • $\begingroup$ Correct. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 3 '15 at 0:26

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.