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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!

share|improve this question
    
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

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