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$\mathbb{R}$ 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?

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    $\begingroup$ Isn’t any topological field a dense subset of itself? $\endgroup$ – Lubin Sep 25 '15 at 3:28
  • $\begingroup$ @Lubin: the question has been edited to ask for a proper subfield. $\endgroup$ – Ross Millikan Sep 25 '15 at 5:01
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As $\Bbb Q$ is the smallest ordered field, it does not contain a proper subfield, dense or not.

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  • $\begingroup$ 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! $\endgroup$ – user263630 Sep 25 '15 at 5:12
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    $\begingroup$ No, you can have $\Bbb Q[\sqrt 2]$, for example. $\endgroup$ – Ross Millikan Sep 25 '15 at 5:28
  • $\begingroup$ 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. $\endgroup$ – Ross Millikan Sep 25 '15 at 13:40

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