Every topological vector space over a connected field is connected? I think this is true but when I tried searching it I didn't find anything.  Let $V$ be a linear topological space over a connected field $\mathbb{K}$.  Then for every $x\in V$, the map $a \mapsto ax$ is continuous and hence has connected image $[x]=\{ax$ $|$ $a \in \mathbb{K}\}$.  Since $V=\cup_{x\in V}[x]$, and each $[x]$ share the point $0$, $V$ is connected.  Am I missing something?      
 A: To clarify the definitions, I will only assume the following:


*

*$\mathbb K$ and $V$ are topological spaces.

*There are elements $0,1\in\mathbb K$ and ${\bf 0}\in V$.

*There is a continuous function $s\colon \mathbb K\times V\to V$ (where $\mathbb K\times V$ is given the product topology) satisfying $s(0,v)={\bf 0}$ and $s(1,v)=v$ for all $v\in V$.

*$\mathbb K$ is connected.


In particular, when $V$ is a topological vector space over a connected topological field $\mathbb K$ and $s(k,v)$ is the scalar multiplication function, these conditions are satisfied.
I will now show that conditions $(1)$ through $(4)$ force $V$ to be connected as well.
First, observe that for all $v\in V$ the map $\mathbb K\to \mathbb K\times V$ which sends $k\in\mathbb K$ to $(k,v)$ is continuous. (This holds in general for the product topology.) Composing this map with the scalar multiplication $s$, it follows that for all $v\in V$, the set
$$
s(\mathbb K,v):=\{s(k,v)\colon k\in\mathbb K\}
$$
is the continuous image of a connected space, and hence connected. (Wikipedia calls this the "main theorem of connectedness".)
Since $s(1,v)=v$ for all $v\in V$, we have that
$$
V=\bigcup_{v\in V}s(\mathbb K,v).
$$
On the other hand, since $s(0,v)={\bf 0}$ for all $v\in V$, the intersection of these sets is non-empty. The union of connected sets with non-empty intersection is connected, as explained here, and therefore (since each $s(\mathbb K,v)$ is connected) so is $V$ itself.
