I would have thought that this is a good candidate for an inductive proof, but I have searched for one and failed. Is there such a proof, and if not why not ? Here's how far I got.
- It's easy to establish that equivalence is transitive so it's only necessary to prove that any norm is equivalent to a norm of choice, e.g. $ ||.||_1$
- In any finite space any norm satisfies $||.|| \le M ||.||_1$. Prove this by picking a basis and applying the triangle inequality.
- That leaves to prove the other half of the equivalence inequality that $||.||_1 \le m ||.||$
- Base case for induction: $Dim(V) = 1$. For a basis vector $e_1$ in $V$ any vector $v = \nu_1 e_1$ and $||v|| = ||\nu_1 e_1|| = |\nu_1|.||e_1|| = ||v||_1.||e_1||/||e_1||_1$ So, $||v||_1 \le m||v||$ where $m = ||e_1||_1/||e_1||$
But, I can't prove the inductive step that if true for $Dim(V) = n$ then true for $Dim(V) = n+1$.
(I have some references for the topological proof - it's specifically an inductive proof that I'm looking for),