Completion of Valued Field

Theorem
Let $\left({k, \left|{\cdot}\right|}\right)$ be a valued field.

Then there exists a unique completion $k'$ of $k$ as a metric space and an absolute value $\left|{\cdot}\right|'$ such that $\left({k', \left|{\cdot}\right|'}\right)$ 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 $\left|{\cdot}\right|' = d' \left({x, 0}\right)$ is an absolute value on $k'$.

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

Suppose $\left|{\cdot}\right|'$ were not an absolute value on $k'$.

This would imply discontinuity of the metric.

This would contradict 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.