To show that the complement of the kernel of an unbounded linear functional is path connected The kernel of a linear functional is either closed (if it is continuous) or dense (if not).  In the latter case, apparently the complement of the kernel is path connected.  How does one see this?
Yes Daniel, this is the point.  The question was on last years masters exam at my university.  The first part was to show the density of the kernel in the case the real-valued linear functional from a Banach space is discontinuous, which has been discussed elsewhere on this forum.  If the linear function is continuous then the kernel separates its negative values from its positive values, so then the complement of the kernel is not path connected, so the answer depends crucially on the fact that the functional is discontinuous.  If a point is in the kernel then the ray through that point is in the kernel.  If a point is in the complement of the kernel then the ray except the origin is in the complement, so one can push everything onto the unit sphere.  But I still don't see how to get there.  

Since $f$ must be surjective, we know that there must be some $x_0 \in X$ such that $f(x_0) = 1$. Let $z \in X$ such that $f(z) \neq 0$.


*

*If $f(z) \notin \mathbb{R}$, consider the line $t \mapsto tz + (1-t)x_0$. Note that
$$
f(tz + (1-t)x_0) = tf(z) + (1-t) \neq 0
$$
Hence that line connects $z$ to $x_0$.

*If $f(z) \in \mathbb{R}$, note that $f(iz) = if(z) \notin \mathbb{R}$, so it suffices to find a path from $z$ to $iz$. Again, $t \mapsto tz + (1-t)iz$ works since
$$
f(tz + (1-t)iz) = tf(z) + (1-t)f(z)i \neq 0 \quad\forall t\in [0,1]
$$
It is likely this proof is not the most efficient, but it seems to work.
