2
$\begingroup$

If F is Archimedean ordered field, the subfield of F isomorphic to $\mathbb{Q}$ is dense in F ?

How can i approach to this question i have no idea and the field is new to me.

$\endgroup$
  • $\begingroup$ What does density mean here? Elements arbitrarily close? $\endgroup$ – Jacob Wakem Mar 31 '17 at 7:12
  • $\begingroup$ Assume Q is not dense around f. That means there is some distance between f and the nearest interval of rational numbers. $\endgroup$ – Jacob Wakem Mar 31 '17 at 7:32
  • $\begingroup$ @Alephnull. Perhap referring to order-denseness,i.e. a member of Q lying between any 2 members of F. $\endgroup$ – DanielWainfleet Mar 31 '17 at 7:39
  • $\begingroup$ @user254665 this is probably the same as topological density. $\endgroup$ – Jacob Wakem Mar 31 '17 at 7:51
  • $\begingroup$ @Alephnull. Yes. If a subset A is order-dense it is dense in the order topology unless A is empty and the space has just 1 point. $\endgroup$ – DanielWainfleet Mar 31 '17 at 20:04
2
$\begingroup$

Let $x \in F$ and $\epsilon >0$ in $F$. Because $F$ is archimedian, we may find $b \in \mathbb{N}^*$ such that $b. 1_F>\frac{1}{\epsilon}$ (so that $\frac{1}{b.1_F} < \epsilon$). Then let $a \in \mathbb{Z}$ the smallest integer such that $a. 1_F > b.x$ (once again, this integer exists because $F$ is archimedian). Then from the minimality of $a$, the distance between $a.1_F$ and $b.x$ is smaller than $1_F$, so dividing by $b. 1_F >0$, we get that the distance between $\frac{a.1_F}{b.1_F}$ and $x$ is smaller than $\frac{1}{b.1_F} < \epsilon$.

$\endgroup$
1
$\begingroup$

Wouldn't $\mathbb{R}$ work? Or $\mathbb{Q}[\sqrt{2}]$ or any subfield of the real numbers.

$\endgroup$
  • $\begingroup$ You mean R is dense in blah-blah, and Q is dense in R? $\endgroup$ – Jacob Wakem Mar 31 '17 at 7:13

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.