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Let $E$ a Topological Vector Spaces, $T: E \longrightarrow \mathbb{K}$, linear. If for some $x \in E$, $Tx \neq 0$. Then, $T$ is open.

I think that it is sufficient to prove that $T(G) \subset \mbox{Int}(T(G))$, $G \in \tau_E$

Any help is appreciated


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1 Answer 1

Factor $T = \bar{T}P$ as follows: $E \xrightarrow{P} E/\ker{T} \xrightarrow{\bar{T}} \mathbb{K}$.

The quotient map $P$ is open because for open $U \subset E$ the set $$P^{-1}P(U) = U + \ker{T} = \bigcup_{x \in \ker{T}} (U + x)$$ is open, so $P(U)$ is open by definition of the quotient topology. $\bar{T}$ is open as a bijection of finite-dimensional spaces. Every composition of open maps is open.

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How about the converse, i.e. if $T$ were open, would $\bar{T}$ be open if we considered general topological vector spaces? – Libertron Sep 7 '13 at 21:25

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