Is it possible to have a normed vector space $V$ over some field $F$ that is not connected?
(I already know that if $F=\mathbb{R}$ or $F=\mathbb{C}$ then we can easily show that this is not possible. But maybe there are fields which enable us to construct such a normed vector space. Or maybe someone can explain to me that this is not a property of the corresponding field at all.)
I will consider valued field. Let me briefly recall the relevant definitions:
Valued field: Let $F$ be a field with valuation $|\cdot|:F\to\mathbb{R}$. This is, for all $x,y\in F$, $|\cdot|$ satisfies:
Then we define a norm on a vector space over a valued field in the following way:
Norm: Let $(F,|\cdot|)$ be a valued field and $V$ be a vector space over $(F,|\cdot|)$. A function $p:V\to \mathbb{R}$ is a norm iff for each $a,b\in V$ and each $k\in F$, it satisfies:
Take the field $\mathbb{F}_2=\{ 0, 1 \}$. Then we can define the absolute value $\phi: \mathbb{F}_2 \rightarrow \mathbb{R}$ by $\phi(0)=0$ and $\phi(1)=1$. Now consider $\mathbb{F}_2$ as a normed vector space over itself, then it is not connected. In fact the metric induced by this norm is the discrete metric.
If the field $F$ is connected, then any normed vector over this field is connected. Assume that $E$ is not connected and write it as $E=U\cup V$ with $U,V$ open and $U\cap V = \emptyset$ and $U,V\neq \emptyset$. Pick $x\in U$ and $y\in V$ and consider the map $f: F \rightarrow E, f(\lambda) = x+\lambda (y-x) $. Note that $f$ is continuous and injective. Now we can write $K= f^{-1}(U) \cup f^{-1}(V)$. As $f$ is continuous, we have that $f^{-1}(U), f^{-1}(V)$ are both open. Furthermore, $0\in f^{-1}(U)$ and $1\in f^{-1}(V)$. Finally, as $f$ is injective and $U\cap V= \emptyset$, we get $f^{-1}(U)\cap f^{-1}(V)=\emptyset$. This is a contradiction as we assumed that $K$ is connected.
The standard definitions of the notion of a normed vector space require the field of scalars to be $\Bbb{R}$ and $\Bbb{C}$. The notion generalises to context of a field equipped with a real-valued valuation. Alternatively, the notion can be generalised to any ordered field of scalars, e.g., the rational numbers $\Bbb{Q}$ (just using the same axioms, with the absolute value interpreted in the obvious way). A non-trivial normed $\Bbb{Q}$-vector space in this generalised sense is not topologically connected, under the topology arising from the order topology on $\Bbb{Q}$.
In more detail: if $K$ is an ordered field, $K$ in a topological field under the order topology. We can define: $$|\lambda| = \max(\lambda, -\lambda)$$ and then define a normed vector space over $K$, to be a $K$-vector space $V$ equipped with a function $\|\cdot\| : V \to K$ such that $$ \|x\| = 0 \mbox{ iff } x = 0 \\ \|x + y\| \le \|x\| + \|y\|\\ \|\lambda x\| = |\lambda|\|x\| $$ The set of open balls $B(x, \epsilon) = \{y \mid \|y - x\| < \epsilon\}$ for $x\in V$ and $0 < \epsilon \in K$ give a basis for a topology that makes $V$ a topological vector space. This generalisation is the natural one, if, for example, one is working in non-standard analysis.
Aside: $\Bbb{C}$ is not an ordered field, but the definition of a norm on a $\Bbb{C}$-vector space asks for the norm to take its values in $\Bbb{R}$.
