Dilation of Open Set in Topological Vector Space is Open

Theorem

Let $K$ be a topological field.

Let $X$ be a topological vector space over $K$.

Let $\lambda \in K \setminus \set {0_K}$.

Let $U$ be a open set in $X$.

Then $\lambda U$, the dilation of $U$ by $\lambda$ is open.

Proof

Let $c_\lambda$ be the dilation by $\lambda$ mapping.

From Dilation Mapping on Topological Vector Space is Homeomorphism, $c_\lambda$ is a homeomorphism.

Hence the image $c_\lambda \sqbrk U$ is open.

That is, $\lambda U$ is open.

$\blacksquare$