= Topology determines norm equivalence
Two <norms> $p,q$ on a <vector space> give the same <topology> if and only if $c p(x)\le q(x)\le C p(x)$ for some positive constants. Indeed, continuity of the identity at zero gives $p(x)<\delta\Rightarrow q(x)<1$. For $x\ne0$, apply this to $\delta x/(2p(x))$ to obtain $q(x)<2p(x)/\delta$. Reverse the roles of the <norms> for the lower bound. Conversely the two inequalities compare every translated <open ball>. This proof uses <absolute homogeneity of a norm>, not finite dimension.
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