Completion of $\mathbb{Q}$ under trivial absolute value i have a basic question on completions.
Define the trivial absolute value $|\cdot|_0$ on $\mathbb{Q}$ by $|x|_0 = 1$ if $x\neq 0$ and by $|0|_0=0$.
Is it true that the completion of $\mathbb{Q}$ under $|\cdot|_0$ is again $\mathbb{Q}$? If so, how can you prove it?
My attempt: We want to show that $\mathbb{Q}$ is complete under the trivial absolute value, so take any Cauchy sequence of rationals... then i don't see why its limit is again in $\mathbb{Q}$.
Thanks in advance!
 A: The idea is to see what a Cauchy sequence is under the trivial absolute value. Indeed, note that $|\cdot|$  takes only two values, $1$ and $0$. Therefore, if $a_n$ is Cauchy, there exists $N$ such that $n,m > N \implies |a_n - a_m| < \frac 12$ so $a_n  - a_m = 0$ is forced i.e. $a_n = a_m$.
Hence, all Cauchy sequences are eventually constant i.e. there is some $N$ such that after $N$ terms the sequence looks like $a,a,a,a,a,a...$ for some $a$. Now it is obvious what the limit of such a Cauchy sequence is : it is $a$ itself, and since $a$ belongs to the sequence it belongs to the rational numbers, rendering $\mathbb Q$ complete.
Of course, nothing special about $\mathbb Q$ : the trivial norm is complete on any (non-empty) set, since the same argument will apply.
A: Because there is a $p\in\mathbb N$ such that$$(\forall m,n\in\mathbb{N}):m,n\geqslant p\implies|x_m-x_n|<1\implies x_m=x_n.$$That is, the sequence becomes constant after a certain point and therefore it converges to that constant.
