If a normed space is finite dimensional, the unit sphere is compact

Let $(V, \Vert\cdot\Vert)$ be a normed vector space with dimension n over a field $K$. Consider the set $S:= \{v \in V : \Vert v \Vert =1\}$

Then, I know that this set is compact if the field is real or complex, which in the real case can be proven by picking a basis $E$ and defining a norm on this basis such that we get an isometry between $\mathbb{R^n}$ and $V$, which obviously is continuous and sends the compact sphere (easy to see it is closed and bounded) in $\mathbb{R^n}$ to the sphere in the space V, and hence S is compact as continuous image of a compact set.

For a complex field, we can apply the same trick, and prove that the sphere is compact in $\mathbb{C^n}$ by considering the continuous identification with $\mathbb{R^{2n}}$

My question now, is, if the theorem is true for other fields other than the real or the complex field? For example, is the theorem true for the field of rational numbers?

• What do "normed vector space" and "compact" mean if you have a field other than $\mathbb{R}$ or $\mathbb{C}$? – Eric Wofsey Feb 20 '18 at 22:07
• Take for example a space over Q or a field over the p adic numbers with the non archimedian absolute value. Then define the norm in the usual way? – user370967 Feb 20 '18 at 22:19

It isn't true for a field which isn't complete, e.g. $\Bbb{Q}.$ Every normed vector space produces a metric space, by defining $d(x,y)=||x-y||.$ In a metric space, a subset is compact if it is complete and totally bounded. In $\Bbb{Q},$ the vectors of magnitude $1$ are $\{-1,1\},$ and in this situation the set is compact. However, in $\Bbb{Q}^2,$ the set of points $(x,y)$ where $x^2+y^2=1,$ is not compact, since it is not complete.
• Correct, although you don't need totally ordered. For example $\Bbb{C}$ lacks a total order. Take, for example, the $p-$adic numbers, $\Bbb{Q}_p$ where $p$ is prime. These are complete fields, and are distinct from $\Bbb{R}$ or $\Bbb{C}.$ – Chickenmancer Feb 20 '18 at 22:17