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I could not solve this problem:

Prove that for a non-Archimedian field $K$ with completion $L$, $$\left\{|x|\in\mathbb R \mid x\in K\right\} =\left\{|x|\in\mathbb R \mid x\in L\right\}$$

I considered a Cauchy sequence in $K$ with norms having limit $l$, but I could not construct an element of $K$ with norm $l$ from the sequence.

Will anyone please show how to prove it?

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it's right. Try to show that ultrametric inequality forces Cauchy sequence to be eventually constant. – Soarer May 16 '11 at 18:56
@Soarer, thanks but I could still not prove it. If the sequence is eventually constant then isn't the field equal to its own closure? – quanta May 16 '11 at 20:03
I think he means showing that if the sequence is Cauchy, then the sequence of real numbers $\{|x_n|\}$ will be eventually constant (rather than the sequence $\{x_n\}$ being eventually constant). – Arturo Magidin May 16 '11 at 20:50
Using $|x+y|=\max(|x|,|y|)$ when $|x|\not = |y|$? – quanta May 16 '11 at 20:51
If x_1,x_2,... is a Cauchy sequence in a non-archimedean valued field which does not tend to 0, then the real numbers |x_1|,|x_2|,... are eventually constant by the ultrametric inequality. – KCd May 16 '11 at 21:56
up vote 5 down vote accepted

Suppose that $\{x_n\}$ is a Cauchy sequence under the ultrametric $|\cdot|$, and consider the sequence $\{|x_n|\}$ of real numbers.

If the sequence is eventually constant, then there is nothing to do: the absolute value of the limit is equal to the absolute value of some element of $K$.

If the sequence is not eventually constant, then we can find a subsequence in which $|x_{n_k}|\neq|x_{n_{k+1}}|$ for all $k$. Since a sequence is Cauchy in an ultrametric if and only if the sequence of consecutive differences $|a_i-a_{i+1}|$ goes to $0$, and we have $$|x_{n_k} - x_{n_{k+1}}| = \max\{|x_{n_k}|,|x_{n_{k+1}}|\}$$ (since the norms of $x_{n_k}$ and of $x_{n_{k+1}}$ are different), then we must have $$\lim_{k\to\infty}|x_{n_k}| = 0,$$ and hence, since the original sequence is Cauchy, $$\lim_{n\to\infty}|x_n| = 0.$$ Therefore, the absolute value of the limit is $0$, which is the absolute value of some element of $K$ as well.

In either case, the absolute value of the limit of a Cauchy sequence of elements of $K$ is always equal to the absolute value of an element of $K$, so the set of absolute values of elements of the completion is equal to the set of absolute values of elements of $K$.

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Excellent! Thank you! – quanta May 16 '11 at 23:38

Here is another solution: Let $(K,| \ |)$ be any non-Archimdean normed field. If $\alpha \in \mathbb{R}^{> 0}$, then the corona $\{x \in K \ | \ |x| = \alpha \}$ is open (possibly empty). (The analogous thing holds in any ultrametric space: c.f. Exercise 2.12 in these notes.) This is a counterintuitive fact even for those who have developed some good intuition for ultrametric spaces: a small part of me still feels instinctively that $|K^{\times}|$ should be discrete in order for this to hold, but it is quite straightforward to prove it in the general case.

Now, let $(\hat{K},| \ |)$ be the completion. No additional calculations are necessary to observe that $K$ is dense in $\hat{K}$ -- that's the whole point of the completion of a metric space, after all. But by the above paragraph, if the group of nonzero norms of the completion were any larger than the group of nonzero norms of $K$, there would be a nonempty, open corona in $\hat{K}$ without any $K$-valued points: contradiction.

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