Two notions of equivalent norms? I have two statements in mind that taken without further caution could seem contradictory:


*

*all norms are equivalent in finite dimension 

*there are infinitely many non-equivalent norm over the rationals (Ostrowski)


So I should be missing a point. Is it that the first statement is only valid for the reals or complexes? (however I have the impression that the proof still holds over the rationals)
Or is it rather than the two notions of equivalence (one with bounds, continuity of the identity; the other with equality up to a certain power) are different? Case in which: why are these two natural, what motivates one in some cases and the other one in others?
 A: A norm on a field is not the same as a norm on a field as a vector space over itself.
The definitions are:
Let $F$ be a field. A function $|\cdot|: F \times F \to F$ is a norm on $F$ if for all $x,y \in F$ we have


*

*$\left|x\right| \ge 0 \text{ and } \left|x\right| = 0 \iff x = 0$

*$\left|xy\right| = \left|x\right|\left|y\right|$

*$\left|x+y\right| \le \left|x\right| + \left|y\right|$


$|\cdot|$ is more commonly called a valuation on $F$ and $(F, \left|\,\cdot\,\right|)$ is a valued field.
Let $V$ be a vector space over $F$ with a fixed valuation $|\cdot|$. A function $\|\cdot\| : V \times V \to F$ is a norm on $V$ if for all $x,y \in V$ and scalars $\alpha \in F$ holds:


*

*$\|x\| \ge 0 \text{ and } \|x\| = 0 \iff x = 0$

*$\|\alpha x\| = \left|\alpha\right|\|x\|$

*$\|x+y\| \le \|x\| + \|y\|$


For example, the $p$-adic absolute value $\left|\,\cdot\,\right|_p$ is a valuation on $\mathbb{Q}$ but it is not a norm on the vector space $\mathbb{Q}$ over the valued field $(\mathbb{Q}, |\cdot|)$ where $|\cdot|$ is the standard absolute value on $\mathbb{Q}$.
The claim that "all norms on a finite-dimensional space are equivalent" holds only for norms on a vector space, not valuations (as Ostrowski's theorem shows).
Furthermore, all norms on a finite-dimensional vector space over a valued field $(F, \left|\,\cdot\,\right|)$ do not have to be equivalent if $(F, \left|\,\cdot\,\right|)$ is not complete as a metric space with the metric $(x,y) \mapsto \left|x-y\right|$.
Consider the vector space $\mathbb{Q}^2$ over over the valued field $(\mathbb{Q}, |\cdot|)$ where $|\cdot|$ is the standard absolute value.
Define two norms on the vector space $\mathbb{Q}^2$ as $\|(x,y)\|_1 = \left|x\right| + \left|y\right|$ and $\|(x,y)\|_2 = |x + \sqrt{2}y|$.
They are not equivalent. Namely, let $(x_n)_n$ be a sequence of rational numbers such that $x_n \to \sqrt{2}$. Then
$$\|(x_n,-1)\|_2 = |x_n - \sqrt{2}| \xrightarrow{n\to\infty} 0$$
but $\|(x_n,-1)\|_1 \ge 1$.
This example is taken from this question. 
