# Is every vector space a topological space

To clarify the definitions, I will only assume the following:

1. $\mathbb K$ and $V$ are topological spaces.
2. There are elements $0,1\in\mathbb K$ and ${\bf 0}\in V$.
3. 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 $v\in V$.
4. $\mathbb K$ 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.

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