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What I want to know is: the preimage of an integer by a valuation map is an open set?
Can we cover a valuation field by open sets with elements with fixed valuation?

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up vote 2 down vote accepted

If $v:K^\times \to \mathbb Z$ is a discrete valuation, $|\cdot|$ the induced absolute value on $K$ and $x$ an element of $K$ then all $y\in K$ with $|x-y| < |x|$ satisfy $|y|=|x|$: $$|y| \leq \max\{|x-y|,|x|\} = |x|,\\ |x| \leq \max\{|x-y|,|y|\} = |y|$$ (where the second equality holds because $|x|\leq |x-y|$ would contradict $|x-y|<|x|$).

So every point $x \in K$ has an open neighbourhood $U = B_{|x|}(x)$ such that all elements $y\in U$ have the same valuation as $x$. Hence the preimage of any fixed valuation is open in $K$.

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