It is well-known that all norms on a finite dimensional -vector space are equivalent: the proof is immediate, because the unit ball is compact. A similar argument works with replaced by any locally compact field. It is less known that the same is true over any *complete* field, without assuming local compactness. The compactness argument no longer works, but the proof is still not very difficult. However, the arguments I saw in the literature so far are somewhat bulky (see, for instance, Section 144 of van der Waerden’s “Algebra”).

Recently our master student Martin Djukanovic showed me a wonderful short proof of this statement. We argue by induction in the dimension. There is nothing to prove in dimension . Now let be an -dimensional vector space over some complete field, . We fix a basis of and show that any norm on is equivalent to the sup-norm with respect to this basis:

By induction, this holds in dimension . In particular, in dimension any normed space over a complete field is itself complete, because this is obviously true for the sup-norm.

For any we have with . It remains to show that with some other . Assuming the contrary, we find an infinite sequence of non-zero vectors such that as . Taking a subsequence, we may assume that the sup-norm of each is attained at its -th coordinate, and after re-scaling we may assume that the -th coordinate (and thereby the sup-norm) of each is . Thus, each can be written as , where and .

Since , we have as . Thus, the sequence converges to . Hence it is a Cauchy sequence. But is complete by the induction, which implies that converges in . We obtain , a contradiction.