Given a field $F$ and an absolute value $|\ |$ on $F$, define the distance $d(x,y)$ between two elements $x,y\in F$ by $$ d(x,y) = |x - y|. $$ I just worked through the proofs that $d$ defines a metric on $F$ and that addition and multiplication in $F$ are continuous with respect to $d$. Now I'm trying to show that inversion is continuous with respect to $d$. What I've done so far is as follows:
Let $x_0\in F\setminus \{0\}$ and let $\varepsilon > 0$. We want to find $\delta > 0$ such that $\forall x\in F$ with $d(x,x_0) < \delta$, $x\neq 0$ and $d(1/x, 1/x_0) < \varepsilon$. We first show that $\forall x\in F: d(x,x_0) < |x_0|$, $x\neq 0$. Assume that $d(x,x_0) < |x_0|$ and $x = 0$. Then \begin{align*} d(x,x_0) = |x - x_0| = |x_0| < |x_0|, \end{align*} a contradiction, so we conclude that $x\neq 0$. Define $\delta_1 := |x_0|$.
Now pick any $\delta\in\mathbb R$ with $0 < \delta < \delta_1$, and pick $x\in F$ with $d(x,x_0) < \delta$. Then $x\neq 0$, and \begin{align*} d\left(\frac{1}{x}, \frac{1}{x_0} \right) = \left| \frac{1}{x} - \frac{1}{x_0} \right| = \left| \frac{x_0 - x}{xx_0} \right| = \frac{|x_0 - x|}{|x| |x_0|} = \frac{d(x_0,x)}{|x||x_0|} < \frac{\delta}{|x| |x_0|} . \end{align*} Now $|x_0| = |(x_0 - x) + x| \leq |x_0 - x| + |x| < \delta + |x|$, so $|x| > |x_0| - \delta > 0$. Then, continuing, we have \begin{align*} d\left(\frac{1}{x}, \frac{1}{x_0} \right) < \frac{\delta}{|x| |x_0|} < \frac{\delta}{(|x_0| - \delta) |x_0| } . \end{align*} I can't figure out where to go from here. Any advice?
Update: After reading Chris Cutler's comment, I realized one possible resolution would be to just solve the inequality \begin{align*} \frac{\delta}{(|x_0| - \delta) |x_0| } < \varepsilon \end{align*} for $\delta$. However, Hagen von Eitzen's solution is much nicer.
If you demand $\delta<\frac12|x_0|$, then the denominator is $>\frac12|x_0|^2$. I you additionally demand $\delta<\frac12|x_0|^2\epsilon$, you are done.