TITLE: Nonconstant linear functional on a topological vector space is an open mapping
QUESTION [12 upvotes]: In the middle of another proof (Theorem 3.4, p. 60) in his Functional Analysis book, Rudin says that
"every nonconstant linear functional on $X$ (topological vector space) is an open mapping." 
Is this obvious? Could someone please show me how the proof goes?

REPLY [15 votes]: Consider arbitrary linear functional $\Lambda : X\mapsto \mathbb{R} $, pick $x^*\in X, \Lambda(x^*)=1$ (this can be done as we know $\Lambda$  is linear and $\exists x_0, \Lambda(x_0)=c\neq 0$, we can then pick $x^*=c^{-1}x_0$). We claim: $\forall V$ open subset of $X$, $\Lambda(V)$ is open in $\mathbb{R}$. Suppose $y\in \Lambda(V)$, we know $\exists x\in V, \Lambda(x)=y$. From the continuity of multiplication we know $\exists \delta>0, \forall |r| < \delta, x+ rx^* \in V, y+r=\Lambda(x+rx^*)\in f(V)$ and thus $B_\delta(y) = (y-\delta, y+\delta)\subseteq \Lambda(V) $. Therefore, $\Lambda(V)$ is open in $\mathbb{R}$ and $\Lambda$ is an open mapping.