I'm trying to prove that if $X$ is a topological vector space over $\mathbb{R}$ and $f:X\to \mathbb{R}$ is a nonzero continuous linear map, then $f$ is open in the sense that if $G$ is open so is $f(G)$. Here's my idea so far. Let $G$ be open in $X$, and let $\alpha\in f(G)$. Then $\alpha=f(x)$ for some $x\in G$. Since $G$ is open, there exists an open neighborhood $V$ of the origin such that $V+x\subset G$. This gave me the idea that I should really be proving that if $V$ is an open neighborhood of the origin in $X$, then $f(V)$ is an open neighborhood of the origin in $\mathbb{R}$, since if this were true, we would have that $f(V+x)=f(V)+\alpha$ is an open subset of $f(G)$ containing $\alpha$, so we'd be done. Is this the right direction to go? I just want a hint, please don't solve the problem for me. (I also wonder if the surjectivity of $f$ comes into play here) Thank you!
EDIT: I have an idea. We have a lemma that says that any nonempty balanced subset of $\mathbb{R}$ is a symmetric interval, and also that if $V$ is a neighborhood of the origin, then there exists a balanced neighborhood $U$ of the origin such that $U\subset U+U\subset V$. So if $V$ is an open neighborhood of $0$, find a balanced open neighborhood $U\subset V$. Then $f(U)$ is a nonempty balanced subset of $\mathbb{R}$ by linearity of $f$, so it is a symmetric interval by the lemma, say with endpoints $\pm\epsilon$. This shows $(-\epsilon,\epsilon)\subset f(U)\subset f(V)$, so $f(V)$ is an open neighborhood of $0$. Does this work?
