$\mathbb R^3$ 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: First: it is not reasonable to prove this with homology. Let's start with a story.
Suppose $X$ is a subset of the sphere homeomorphic to $D^k$. Then Hatcher proves in proposition 2B.1 that $S^n \setminus X$ has trivial reduced homology. (He also proves that if $X \cong S^k$, then $S^n \setminus X$ has the same reduced homology as $S^{n-k-1}$. This won't be as important to us.) The way the proof goes, roughly, is to use Mayer-Vietoris on the complement of $I^{k-1} \times [0,1/2]$ and $I^{k-1} \times [1/2, 1]$. If the homology isn't what we said above, we show the same is true for one of these sub-intervals, keep making the sub-intervals smaller, and then show that the homology of these complements ultimately reduces to the homology of $I^{k-1} \times \{p\}$ (all, again, a sketch - see the actual proof.) 
It was essential that we used $I$ here, or else we wouldn't have access to the Mayer-Vietoris sequence - the complement of these sets is open! The same is not true for a copy of $\Bbb R$. If the $\Bbb R$ had closed image, you'd be able to access the problem with similar techniques (most easily by taking the one-point-compactification and thinking of this as the complement of an $S^1$ in $S^3$ and using the above technique). You might want to say "OK, well there they took a nested intersection of intervals and looked at their complement. We know the result works for embeddings of $[0,1]$. Why don't we try to take a nested union instead?" This does not work - just think about the union of discs on the $xy$-plane of increasing radius! Clearly their complements are path-connected but their increasing union is not. So there's something special going on about the 1-dimensionality of the line. (Note, by the way, that FINITE unions are easy to deal with via the Mayer-Vietoris sequence. Indeed it's not hard to show using the above facts that the complement of any $n-2$-dimensional finite simplicial complex embedded in $\Bbb R^n$ is path-connected.) Even the most souped-up duality theorems, like Alexander duality in Cech cohomology, need to assume compactness (even if they can drop local contractibility!) After all, consider the embedding of a copy of $\Bbb R$ inside the 2-sphere that winds around an annulus, with limit set $|z| = 1$ and $|z| = 3$. Then it has three path-components, which no form of duality could expect.
OK, so how do we deal with this? We, like your friend said, use dimension theory. I don't know this stuff very well, so I'm going to cite it away. (I don't have access to Hurewicz-Wallman right now, but when I do, I'll offer a more precise reference.) This MO comment says that the space of paths from $x$ to $y$ not in some compact 1-dimensional subspace of $\Bbb R^3$ is open and dense in the space of paths from $x$ to $y$, and this answer implies that the image of any unit interval in $\Bbb R^3$ is 1-dimensional. Now apply the Baire Category theorem to the subset $U_n \subset \mathcal P(x,y)$, where $U_n$ is the set of maps that miss $[n,n+1]$ in your embedding of $\Bbb R$. Because $\mathcal P(x,y)$ is a complete metric space with the supremum metric, it is a Baire space, and hence $\bigcap_{n \in \Bbb Z} U_n$ is nonempty (and, in fact, dense). Thus we conclude.
A: It is sufficient to show that $X=\Bbb R^3\setminus S$ is path-connected. Now for any couple of points $p,q$ in $X$ let $\overline{pq}$ be the segment joining them, then this is a path that. If $\overline{pq}\cap S=\emptyset$, there is no problem. In the case that $\overline{pq}\cap S\neq \emptyset$, take a third point $r$, different to the previous and such that $\overline{pr}\cap S=\emptyset$ and $\overline{rq}\cap S=\emptyset$, and then $\overline{pr}\cup\overline{rq}$ is a path joining $p$ with $q$. 
A: Here's a nice hint: The zero-th homology $H_0(\mathbb{R}^3 - S)$ must be a direct sum of $n$ copies of $\mathbb{Z}$, where $n$ denotes the number of connected components of $\mathbb{R}^3 - S$. 
A: Assume that $\mathbb{R}^3\setminus S$ is disconnected. So there are nontrivial open, disjoint sets $U,V$ with 
$$\mathbb{R}^3\setminus S=U\sqcup V$$  $$\mathbb{R}^3=U\sqcup V\sqcup S$$
Since $U,V$ are nontrivial, $\partial U,\partial V\neq\emptyset$. 
Since $U,V$ are open, $U\cap\partial U=V\cap\partial V=\emptyset$.
We must have that $\partial U\cap V=\emptyset$ or else there is $x\in \partial U\cap V$. This would mean an open ball $W$ exists surrounding $x$ that is contained within $V$. And there would be points in $W\setminus\{x\}$ that are also in $U$, which would contradict $U$ and $V$ being disjoint.
This implies that $\partial U\subset S$. And similarly, $\partial V\subset S$.
Suppose for a moment that $S\setminus(\partial U \cup\partial V)$ is nonempty. Let $y\in S\setminus(\partial U \cup\partial V)$. Well, there is a small enough ball $Y$ surrounding $y$ that manages to neither intersect $U$ nor $V$, since $y$ is not in their boundaries. Therefore $Y$, a ball, is contained in $S$, homeomorphic to $\mathbb{R}$. This is a contradiction.
So $S\setminus(\partial U \cup\partial V)$ must be empty, and given the containment established earlier, therefore $S=\partial U \cup\partial V$. Now use the topology on $S$ induced by its homeomorphism to $\mathbb{R}$. $\partial U$ and $\partial V$ are still closed (since they are nonempty), but they union to $S$ which is open in this topology. Therefore they must have a nontrivial intersection.
Consider $z\in\partial U\cap\partial V$ and an open ball $Z$ containing $z$, with $Z$ so small that $\partial Z\cap S$ consists of two points $\{a,b\}$ and within $Z$, $S$ is homotopy equivalent to the union of the line segments $\overline{az}$ and $\overline{zb}$ while fixing $a$, $z$, and $b$. This is possible because $S$ is homeomorphic to $\mathbb{R}$.
Now, $Z\setminus S$ is connected owing to the simple form of $Z\cap S$. ($Z\setminus S$ is homotopy equivalent to a cylinder.) And intersecting the sides of the first equation above with $Z$, we have the disjoint union
$$Z\setminus S=(Z\cap U)\sqcup (Z\cap V)$$
However $z$ is in both $\partial U$ and $\partial V$, so neither set on the right is empty. This contradicts $Z\setminus S$ being connected.
So it all falls apart. The original assumption must have been false, and so $\mathbb{R}^3\setminus S$ is connected.
A: 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 $\gamma'\cap S=\varnothing$.
