$\mathbb R^3$ minus a line is connected. Let $S\subseteq \mathbb R^3$ be homeomorphic to $\mathbb R$. Prove that $\mathbb R^3 \setminus S$ is connected.
I haven't been able to solve this, although my topology skills are pretty weak. My friend told me he managed to prove this using results from his "dimension topology" class. Although I think this should be solvable using more mainstream stuff like homology or something. But I'm not sure.
Regards.

A short answer using transversality and its properties:
Any path $\gamma$ joining two points $p$ and $q$ in $\mathbb{R}^3\setminus S$ is homotopic to a path $\gamma'$ (with fixed endings) which is tranverse to $S$.
But $\gamma'\pitchfork S$ (thanks to codimension $2$ of the two submanifolds) means $\gamma'\cap S=\varnothing$.
