Ultrametric completion does not introduce new norms 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?
 A: 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 $K$.
A: 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.
