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On $\mathbb R$, every archimedean absolute value $|\cdot|$, such that $(\mathbb R, |\cdot|)$ is complete, is equivalent to the usual absolute value defined for $x\in \mathbb R$ by: $$\mathbb |x|_{\mathbb R}= \begin{cases} x & \text{if } x\ge 0 \\ -x & \text{if } x\le 0 \end{cases}$$

Now, I am trying to show that $\forall x, |x|=\mathbb |x|^a_{\mathbb R}$, for some $a>0$. Using the same steps as in the proof of the Ostrowski Theorem for $\mathbb Q$ in the archimedean case, one knows that there exists an $a>0$ such that $|r|=\mathbb |r|^a_{\mathbb R}, \forall r\in \mathbb Q$. Now I want to show that this "$a$" does work for all reals, not just the rationals. So, for an arbitrary $x\in \mathbb R$, I was thinking to start with a sequence of rationals $(r_i)$, such that $r_i\rightarrow x$ with respect to "usual" absolute value. Now, since $|r|=\mathbb |r|^a_{\mathbb R}$, it means that $(r_i)$ is Cauchy w.r.t. $|\cdot|$, hence converges to some $t\in \mathbb R$, because of completeness asumption. Now I want to show that $t=x$, but I am not able to do that.

Note: I use the definition of absolute value from Neukirch, chapter 2.

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  • $\begingroup$ might be a good idea to use the archimedean property $\endgroup$ – Mikhail Katz Nov 6 '16 at 14:59
  • $\begingroup$ Can you suggest me how exactly to do that? $\endgroup$ – user223794 Nov 6 '16 at 16:17
  • $\begingroup$ What I was pointing out is that the procedure you outlined works for all valuations, whether p-adic or archimedean, so it couldn't yield a solution. There is only one Euclidean valuation on Q so it must be the one yielding the real numbers as completion. $\endgroup$ – Mikhail Katz Nov 6 '16 at 16:20
  • $\begingroup$ @MikhailKatz: It is fair to say that the archimedean property is already used for showing that the restriction to $\mathbb{Q}$ is equivalent to the usual value. However, I do not see how to prove the statement without also using the order structure on $\mathbb{R}$. $\endgroup$ – Torsten Schoeneberg Jan 3 '18 at 1:03
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It seems that we really have to make use of more extra structure on $\mathbb{R}$, namely, the usual order $<$.

So let $t \in \mathbb{R}$, wlog $t \ge 0$, and $r_i$ a sequence of rationals that converge to $t$ with respect to $|\cdot|$. By replacing $r_i$ with $2t-r_i$ if necessary, we can make sure that

(*) for all indices $i\in \Bbb{N}$, we have $r_i \le t$ for odd $i$ and $r_i\ge t$ for even $i$; in particular, $t$ lies between $r_i$ and $r_{i+1}$.

I think you already got the following, and it does not need (*): $r_i$ is a Cauchy sequence w.r.t. $|\cdot|$. Then the sequence of values $|r_i| = |r_i|_\Bbb{R}^a$ is a Cauchy sequence in $\mathbb{R}$ w.r.t. the usual value (! -- this is how the metric is defined), hence converges to some $x_0 \in \mathbb{R}_{\ge 0}$. So $|t| = x_0$ by continuity of the value. On the other hand, if $|r_i| = |r_i|_{\mathbb{R}}^a \to x_0$, then by continuity of $a$-exponentiation, $|r_i|_{\mathbb{R}} \to x_0^{1/a}$. In other words, with respect to the usual value, $r_i$ converges to $x := x_0^{1/a}$. So $|t| = x^a = |x|_\mathbb{R}^a$.

So what is left to show is $x=t$. But by (*), we have $t\in \bigcap_{n\in \Bbb{N}} [r_{2n-1}, r_{2n}]$, and because the sequence is Cauchy w.r.t the usual value, that intersection is actually a singleton, hence $\{t\}$. But one also sees (going to monotonous subsequences in the odd resp. even indices) that $x\in \bigcap_{n\in \Bbb{N}} [r_{2n-1}, r_{2n}]$. Hence $x=t$ and $|x|=|x|_\mathbb{R}^a$.

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  • $\begingroup$ The only problem with this solution is the second sentence. How do we know that every real number $r$ can be written as a limit of rational numbers with respect to the (unknown) absolute value $|\cdot|$ (I don't mean the usual one $|\cdot|_{\mathbb R}$)? The rest of the solution would of course be fine. $\endgroup$ – user223794 Feb 5 '18 at 23:13
  • $\begingroup$ Good point. I admit I'm not sure how to circumvent this. Maybe a) one can start with rationals converging to $x$ wrt the usual value, as in your attempt, and then possibly reorder them similar to (*); or b) (approach very different from my answer) one could show that the completion of $\Bbb Q$ wrt the new value is isomorphic to $\Bbb R$, and by the universal property of completion we get an injection of that $\Bbb R$ into the usual $\Bbb R$ which then has to be the identity. $\endgroup$ – Torsten Schoeneberg Feb 7 '18 at 22:41

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