# Continuity of product in absolute value

Let $$F$$ be a field, and $$\| \cdot\|$$ be an absolute value (or norm). I want to prove that with this norm, multiplication is continuous map on $$F$$ in the sense:

for $$x_0,y_0\in F$$, and any $$\varepsilon>0$$, there is $$\delta>0$$ such that $$\|x-x_0\|<\delta \mbox{ and } \|y-y_0\|<\delta \mbox{ implies } \|xy-x_0y_0\|<\varepsilon.$$

Of course, the general method of this is to make adjustment in $$\|xy-x_0y_0\|$$:

$$\| xy-x_0y_0\| = \|x\|\|y-y_0\| + \|y_0 \| \|x-x_0\|.$$ Here $$\|y_0\|$$ is a given fixed number, so second term on right side can be made arbitrarily small; but in the first term, I didn't get how to control $$\|x\|$$? Any hint?

I think you can use $$\|x\| \le \|x - x_0\| + \|x_0\|$$ to help bound the first term.