Convergence of a sequence in a non-Archimdean valuation for a field.

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Admittedly, this is a homework question and so please just hint me where to go with this question if it is ok, thank you!

So suppose $k$ is a field and $|.|$ is a non-Archimedean valuation. In other words, it satisfies the following conditions:

  • $|x|\geq 0$ and $|x|=0\iff x=0$
  • $|xy|=|x||y|$
  • $|x+y|\leq\max\{|x|,|y|\}$

Q: The question is that if $a_n$ converges to $a$ in $k$ and $a$ is not zero, then there exists $N$ such that for all $n\geq N$, $|a_n|=|a|$.

My attempt thus far:

So using the deinition of $a_n\to a$ and reverse triangle inequality, I have managed to show that $|a_n|\to|a|$ (in the real sense.)

Thus by definition, for all $\epsilon>0$, there exists $N$ such that for all $n\geq N$, $|a_n|\in(|a|-\epsilon,|a|+\epsilon)$.

But now I am stuck, this reminds me of discreteness of a valuation map. If I could show that this is discrete then we are done. However, after some research, it does not seem to me that such valauation map has to be discrete. Thus I am wondering how should I approach this question?

Thank you so much in advance!

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$$|x|> |y| \implies |x+y| = |x|$$

Proof: if $|x+y|< |x|$ then $$|x| =|x+y-y| \le \max(|x+y|,|y|)<|x|$$