Proof
(related to Proposition: A Field with an Absolute Value is a Metric Space)
- By hypothesis, $(F,+,\cdot)$ is a field with an absolute value $|\cdot|:F\to\mathbb R.$
- Set $d(x,y):=|x-y|$ for all $x,y\in F.$
- We have to verify, that $d$ is a metric.
- If $d(x,y)=0,$ then $|x-y|=0,$ and by definition of absolute value $x=y.$
- Vice versa, if $x=y,$ then $x-y=0$ and $d(x,y)=|0|=0.$
- Obviously, $d(x,y)=d(y,x)$ is symmetric.
- Also, both, $d$ fulfills the triangle inequality, because $|\cdot|$ does.
- It follows that $(F,d)$ is a metric space.
∎
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