Translation of Open Set in Topological Vector Space is Open

Theorem
Let $\struct {X, \tau}$ be a topological vector space.

Let $U \subseteq X$ be an open set.

Let $x \in X$.

Then:


 * $x + U$ is open.

Proof
Let $T_x$ be the translation operator for $x$.

From Translation Operator on Topological Vector Space is Homeomorphism, we have:


 * $T_x$ is a homeomorphism.

So, since $U$ is open, we have:


 * $\map {T_x} U$ is open.

That is:


 * $x + U$ is open.