# Sorgenfrey Line is not a topological vector space

Show that Sorgenfrey line is not a topological vector space.

My attempt:

We know that if $X$ is a topological vector space, then the map $$x\mapsto \alpha x$$ is a homeomorphism for each scalar $\alpha \neq 0.$ Hence, open sets under this map are mapped to open sets.

In Sorgenfrey line, the set $A=[1,2)$ is open. But $-2A=(-4,-2]$ is not open. Therefore, Sorgenfrey line is not a topological vector space.

Am I correct?

Yes, you are correct. I would have used $\alpha=-1$ instead of $\alpha=-2$, but that's a matter of taste.