Finite dimensional norm equivalence over a complete valued field

The norm axioms imply $\mathrm{\parallel }\sum a_i{v}_i\mathrm{\parallel }\le (\sum \mathrm{\parallel }{v}_i\mathrm{\parallel })\max \mathrm{\mid }{a}_i\mathrm{\mid }$ when n is positive. In dimension zero completeness and comparison are immediate. In dimension one $\mathrm{\parallel }a{v}_1\mathrm{\parallel }=\mathrm{\mid }a\mathrm{\mid }\mathrm{\parallel }{v}_1\mathrm{\parallel }$ gives both bounds and completeness.

Proceed by finite induction. Assume the result for smaller dimensions. Every coordinate hyperplane $H_i$ is complete in its restricted norm and therefore closed: a point in its closure is approached by a sequence within distance 1/n, a Cauchy sequence whose limit in $H_i$ equals that point. Put $c_i={\inf }_{h\in H_i}\mathrm{\parallel }{v}_i-h\mathrm{\parallel }>0$, since the complement of the closed hyperplane is open. Translation and scaling by a nonzero scalar give ${\inf }_{h\in H_i}\mathrm{\parallel }{a}_i{v}_i-h\mathrm{\parallel }=\mathrm{\mid }{a}_i\mathrm{\mid }{c}_i$; the assertion is also valid for $a_i=0$.

For $v=\sum a_j{v}_j$, subtract its other coordinates to obtain $\mathrm{\mid }{a}_i\mathrm{\mid }{c}_i\le \mathrm{\parallel }v\mathrm{\parallel }$. Thus $\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}\le ({\max }_i{c}_i^{-1})\mathrm{\parallel }v\mathrm{\parallel }$, completing the induction. A Cauchy sequence has coordinatewise limits in F and converges by the upper bound, so V is complete. Every subspace, being finite-dimensional, is complete by the same argument and hence closed. No compactness of a unit sphere or nontrivial scalar valuation was assumed.