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If we have an ordered field $ \mathbb{F} $, can we consider a natural metric involved with this space? What should be this metric?

thanks

Giiovanna
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1 Answers1

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If $\mathbb{F}$ is Archimedean, then it has a unique embedding into $\mathbb{R}$, and $d(x,y) = |x-y|$ is our distance function.

Depending upon what you're doing, modifying the definition to the metric to take values in $\mathbb{F}$, or the real closure of $\mathbb{F}$, is a reasonable thing to do, and you can still use $d(x,y) = |x-y|$.

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    I don't have any hypotesis about my Field (except it is ordered) so I can't use it is or not archimedean. What I need to do is to show that if x =< y are non negative numbers in my ordered field, then d(x,0) <= d(y,0), which is easy to show if we use modulus. So, use |a-b| is a reasonable, in this case? – Giiovanna Sep 29 '13 at 19:34
  • @user2768645: Since it takes values in $\Bbb F$, not $\Bbb R$, it won’t actually be a metric, and if $\Bbb F$ happens to be non-Archimedean, it won’t be equivalent to a metric. It will, however, behave much like a metric: it’s a symmetric function that separates points and satisfies a triangle inequality. – Brian M. Scott Sep 29 '13 at 20:34
  • @BrianM.Scott : I understand. So, if I have just ordered field as my hypotesis, so the inequality is not true for all the metrics? Someone gave me a counterexample but i thought if we directly use the hypotesis that F is ordered, we could prove it for all metrics. – Giiovanna Sep 29 '13 at 21:15
  • @user2768645: The triangle inequality? It’s true in any ordered field. – Brian M. Scott Sep 29 '13 at 21:19
  • @BrianM.Scott : I know, but I couldn't figure out how to prove $ d(x,0) \leq d(y,0) $ in any ordered field only having the triangle inequality (except if my distance is defined using modulus). Can you give me a help with it? – Giiovanna Sep 29 '13 at 21:57
  • By the way, I don't even know if it is possible – Giiovanna Sep 29 '13 at 22:03