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Here I proved the following result:

Proposition: Let $(K,|\cdot|)$ be a valued field with non-trivial valuation and let $X$ be a vector space over $(K,|\cdot|)$. Two norms $p_1,p_2$ on $X$ are equivalent (i.e., they induce the same topology) iff there are constants $c_1$ and $c_2$ such that $c_1p_1\leq p_2\leq c_2p_1$

Is the proposition true when the valuation is trivial? (The valuation $|\cdot|$ is trivial when $|x|=1$, for each $x\in K^*$)

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Here's a counterexample. Let $K$ be any field with the trivial valuation and let $X$ be an infinite-dimensional vector space with basis $B$. We can define one norm $p_1$ on $X$ by $p_1(x)=1$ for all nonzero $x\in X$. We can define another norm $p_2$ by saying $p_2(x)$ is the number of nonzero coefficients there are when we write $x$ as a linear combination of elements of $B$. Then both norms induce the discrete topology, but $p_2/p_1$ is unbounded.

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  • $\begingroup$ Great counterexample! Thanks @Eric Wofsey. Now I am wondering if the proposition is true in the finite dimensional case. Please let me know if you know something about that case. $\endgroup$ – Chilote Apr 3 '16 at 21:17
  • $\begingroup$ In the link below you can find the remaining question math.stackexchange.com/q/1726562 $\endgroup$ – Chilote Apr 3 '16 at 22:06

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