# Linear Application that is open in a TVS

Let $T: E \to F$ be a linear map between topological vector spaces $E$, $F$. If for each nonempty open set $G$, the interior of $T(G)$ is non-empty, then, $T$ is open.

Proof: $$\mathrm{Int}(T(G))= \bigcup_{U \subset T(G), U\in\tau_F} \mathcal{U}\neq \emptyset$$

Then, exist $\mathcal{U}$ such that $\mathcal{U} \subset T(G)$...

How should I proceed? Any help is appreciated. Thanks!

-
So what is the question? –  Norbert Dec 3 '12 at 20:05
@Norbert, my proof is correct? –  P. M. O. Dec 3 '12 at 20:13
Where did you use the vector space structure? –  Sigur Dec 3 '12 at 20:16