Problem. Let $X$ be a topological vector space and $f:X\to\mathbb{K}$ a linear mapping. Prove that if $f$ is discontinuous, then $f(A)=\mathbb{K}$ for all nonempty open set $A\subset X$.
I'd like to know if the solution below is correct.
Solution.
Let $A\subset X$ be a nonempty open set and $a\in A$. Then, $A=a+V$ where $V$ is a neighborhood of $0$. Let $W$ be a balanced neighborhood of $0$ such that $W\subset V$.
Since $f$ is discontinuous, there exist $\varepsilon>0$ such that $f(U)\not\subset B(0,\varepsilon)$ for all neighborhood $U$ of $0$.
Claim 1: $f(W)$ is unbounded.
Proof: If $f(W)$ is bounded, then there exist $r>0$ such that $f(W)\subset B(0,r)$. Taking $U=\frac{\varepsilon}{r} W$ we obtain
$f(U)= \frac{\varepsilon}{r}f(W)\subset \frac{\varepsilon}{r}B(0,r)= B(0,\varepsilon),$
a contradiction.
Claim 2: $\mathbb{K}\subset f(W)$.
Proof: Take $\alpha\in\mathbb{K}$. From Claim 1, there exist $w\in W$ such that $|f(w)|>|\alpha|$. Taking $\beta=\frac{\alpha}{f(w)}$ we conclude that $f(\beta w)=\alpha$ and $\beta w\in W$ (because $|\beta|<1$ and $W$ is balanced). Thus $\alpha\in f(W)$.
It follows from Claim 2 that
$$\mathbb{K}=f(a)+\mathbb{K}\subset f(a)+f(W)\subset f(a)+f(V)=f(a+V)=f(A).$$
Thanks.