# Show that $\mathbb{R}^{n}$ ($n>2$) minus a countable number of lines is still path-connected [duplicate]

Define $C$ to be a set of countable number of lines. Let $p,q$ be a couple of points in $\mathbb{R}^{n}\setminus C$. Then if the line segment $\bar{pq}\cap C=\emptyset$, we're done. If $\bar{pq}\cap C\neq\emptyset$, the intersection is a set of countable number of points. Since $n>2$, at each point in the intersection, there exists an arc that goes around the point. Thus, there exists a path connecting $p$ and $q$ that goes along the line segment $\bar{pq}$ with arcs replacing the points in $\bar{pq}\cap C$. Therefore, $\mathbb{R}^{n}\setminus C$ is path-connected.

It's pointed out to me that this proof is not exactly correct. I don't know where it went wrong.

• You're assuming that there's an arc near the intersection point that doesn't intersect the rest of $C$, which is pretty close to what you're trying to prove. Dec 5 '17 at 16:59
• @anomaly I see. If the arc intersects $C$, can I still find another arc around the intersection point or since a line is two dimensional, I could go to "another dimension" to go around the intersection point? Dec 5 '17 at 17:11
• I'm not sure if this helps, but there's a nice dimensional reduction: Take an arbitrary 2-sphere whose surface includes $p$ and $q$. A line can intersect the sphere in at most two points. Thus, to prove that there's a path on the sphere between $p$ and $q$, you can prove that $\mathbb{R}^2$ minus a countable number of points is path-connected. Dec 5 '17 at 17:13
• The place where it goes wrong is, "there exists an arc that goes around the point." What does "goes around the point" mean?
– Neal
Dec 5 '17 at 23:24

Reduce the problem to showing that $\mathbb{R}^2$ minus a countable set of points is path-connected, using one of these methods.

1. Let $S^2 \subset \mathbb{R}^n$ be a 2-sphere that contains $p$ and $q$ on the surface. A line intersects a sphere at most twice, so $C \cap S^2$ is countable. Remove any point $x \in S^2 \setminus (C \cup \{p, q\})$ and identify $S^2 \setminus \{x\} \cong \mathbb{R}^2$ by homeomorphism, and look for a path in $S^2 \setminus (C \cup \{x\})$.

2. Take the uncountable family $\mathcal{P}$ of 2-planes that contain $p$, $q$, and one point on a fixed line not coplanar with the line $pq$. $C$ is a countable set of lines, and $\mathcal{P}$ is pairwise disjoint except at $pq$ (which is not in $C$), so some $P \in \mathcal{P}$ contains no line in $C$. $P \setminus C$ is a countable set of points, as every line in $C$ intersects $P$ at most once. We'll find a path within $P$.

Now let $C \subset \mathbb{R}^2$ be a countable set of points, and $P, Q \in \mathbb{R}^2 \setminus C$ be arbitrary. There is an uncountable family of paths in $\mathbb{R}^2$ that are all disjoint except at $P$ and $Q$ themselves. (For one such family, for each point $B$ on the perpendicular bisector of $PQ$, take the circular arc $PBQ$.) Only a countable number of such paths can contain a point of $C$, so some path avoids $C$.

• This makes sense. But why consider $\mathbb{R}^{2}$? I thought the 2-sphere lives in $\mathbb{R}^{3}$? Dec 5 '17 at 20:46
• Also, you are working under the assumption that for any couple of points, there’s a 2-sphere such that the two points live on the surface of the said sphere, right? Dec 5 '17 at 20:49
• To elaborate: Take a 2-sphere $S^2$ containing $p$ and $q$ and remove a point (besides $p$ and $q$). The remainder is homeomorphic to $\mathbb{R}^2$, and you won't spuriously introduce path-connectedness of $S^2$ minus a set, only destroy it (in the case that every path between some pair of points on the sphere goes through the point removed). And yes, there's always a 2-sphere that contains $p$ and $q$: just take the $(n-1)$-sphere of appropriate radius centered at the midpoint of $p$ and $q$, intersected with an arbitrary 3-hyperplane that also contains $p$ and $q$. Dec 5 '17 at 20:59
• – bof
Dec 5 '17 at 23:04
• You should have $S^{2}\setminus\left(C\cup\left\{x\right\}\right)$ instead of $S^{2}\setminus\left(C\cap\left\{x\right\}\right)$, don't you? Dec 5 '17 at 23:28