Is the image of every open set under a non-zero discontinuous linear function dense in $\mathbb{R}$? Given a normed space $V$ over $\mathbb{R}$, is it true that the image of every open set of $V$ under a non-zero discontinuous linear function $V\to\mathbb{R}$ is dense in $\mathbb{R}$?
I couldnt prove or disprove yet ,
thanks.
 A: Suppose linear $L: V \to \mathbb{R}$ is discontinuous, that is, unbounded. We'll show that the image of an open unit ball is whole $R$. Since $L$ is discontinuous,  there's a sequence $u_1, u_2, \ldots$ such that for all $n$, $||u_n|| < 1$, but $|L(u_n)| > n$. Take any $x \in \mathbb{R}$. There's some natural number $k$ such that $|x| < k$. Since $|L(u_k)| > k$, $\left|\frac{x}{L(u_k)}\right| < 1$, so $\frac{x}{L(u_k)} u_k$ is inside open unit ball in $V$. But $L(\frac{x}{L(u_k)} u_k) = \frac{x}{L(u_k)} L(u_k) = x$, so $x$ is in the image of the open unit ball under $L$.
A: Let $O$ be a nonempty open subset of $V$ and $f : V \to \mathbb{R}$ be a discontinuous functional. In particular, there exist $x \in O$ and $\rho>0$ such that $B:=B(x,\rho) \subset O$. 
Because $f$ is linear, $f(B)$ is convex (since $B$ is itself convex) so $f(B)$ is an interval. But $f(B)$ is not bounded (since $f$ is discontinuous at $x$) and is symmetric with respect to $f(x)$ (since $f$ is linear), so $f(B)= \mathbb{R}$.
You deduce that $f(O)= \mathbb{R}$.
