# Does any Archimedean ordered field contain a proper dense subfield?

$\mathbb{R}$ is an Archimedean ordered field that contains a proper dense subfield $\mathbb{Q}$. And the proof of $\mathbb{Q}$ being dense in $\mathbb{R}$ uses only Archimedean property and ordered field properties. So one may want to ask:

Does any Archimedean ordered field contain a proper dense subfield? Does there exist an Archimedean ordered field that contains no proper dense subfield?

• Isn’t any topological field a dense subset of itself? – Lubin Sep 25 '15 at 3:28
• @Lubin: the question has been edited to ask for a proper subfield. – Ross Millikan Sep 25 '15 at 5:01

As $\Bbb Q$ is the smallest ordered field, it does not contain a proper subfield, dense or not.
• So is $\mathbb{R}$ the unique Archimedean ordered field containing a proper dense subfield up to isomorphism? Should I better ask this in a new post? Thanks for answering! – user263630 Sep 25 '15 at 5:12
• No, you can have $\Bbb Q[\sqrt 2]$, for example. – Ross Millikan Sep 25 '15 at 5:28
• Every infinite field includes $\Bbb Q$. The only reason $\Bbb Q$ fails as a response to your question is that you asked for a proper subset. Every subfield of $\Bbb R$, of which there are many, will contain $\Bbb Q$ as a dense subset. – Ross Millikan Sep 25 '15 at 13:40