Completion of Valued Field

Theorem
Let $(k,|\cdot|)$ be a valued field.

Then there exists a unique completion $k'$ of $k$ as a metric space and an absolute value $|\cdot|'$ such that $(k', |\cdot|')$ is a valued field.

Proof
By the completion theorem for metric spaces $k$ has a unique completion $k'$ equipped with a metric $d'$ that restricts to $d$ on $k$.

Therefore we need only show that $|\cdot|' = d'(x,0)$ is an absolute value on $k'$.

But $|\cdot|'$ is an absolute value on $k$ and a metric on $k'$.

Therefore, if $|\cdot|'$ were not an absolute value on $k'$, this would imply discontinuity of the metric, contradicting Metric is Continuous Function.

Examples

 * The completion of $\Q$ with respect to the usual absolute value is $\R$
 * The completion of $\Q$ with respect to the $p$-adic absolute value is known as the field of $p$-adic numbers and denoted $\Q_p$
 * By Ostrowski's_Theorem there are no other completions of $\Q$ as a valued field