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.


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