I am finding it very difficult to prove or disprove the following statement.

If $A$ is a family of countably many lines in $\mathbb{R}^3$ then $\mathbb{R}^3\setminus A$ is path connected.

I would appreciate it if somebody could give me an elementary proof of this, perhaps using standard geometry and linear algebra.

Thanks for any help.


Let $x\in\Bbb R^3\setminus A$. For each line $\ell\subseteq A$ there is a unique plane $P_\ell$ containing $x$ and $\ell$. The set of such planes $P_\ell$ is only countable, so there is a plane $P$ containing $x$ that does not contain any of the lines in $A$. Each line in $A$ therefore intersects $P$ in at most one point, so $P\cap A$ is countable.

Let $y$ be any other point of $\Bbb R^3\setminus A$; by the same reasoning there is a plane $Q$ containing $y$ such that $Q\cap A$ is countable. Moreover, we can choose $Q$ so that it is not parallel to $P$. Let $\ell=P\cap Q$. For each $z\in\ell$ consider the path consisting of the line segments $\overline{xz}$ and $\overline{zy}$. Since $P\cap A$ is countable, and $\ell$ is uncountable, there are only countably many $z\in\ell$ such that $\overline{xz}\cap A\ne\varnothing$; let $B_P=\{z\in\ell:\overline{xz}\cap A\ne\varnothing\}$. Similarly, there are only countably many $z\in\ell$ such that $\overline{zy}\cap A\ne\varnothing$, and we let $B_Q=\{z\in\ell:\overline{zy}\cap A\ne\varnothing\}$. $B_P\cup B_Q$ is countable, and $\ell$ is uncountable, so pick any $z\in\ell\setminus(B_P\cup B_Q)$; then $\overline{xz}\cup\overline{zy}$ is a path from $x$ to $y$ in $\Bbb R^3\setminus A$.


Choose a line $L$ that is not coplanar with any of the lines in $A$. Such an $L$ exists, because the directions prohibited by any one line in $A$ form a great circle on the unit sphere of directions in $\mathbb R^3$, and countably many circles don't cover the whole sphere. Now given any two points $p,q\in\mathbb R^3\setminus A$, try to join $p$ to $q$ by a path that goes from $p$ to a point $x$ on $L$ and then from $x$ to $q$. Some choices of $x$ won't work, because the path hits one of the lines in $A$. But, thanks to the non-coplanarity in the choice of $L$, any particular line in $A$ causes trouble for only (at most) two $x$'s, one by blocking the segment from $p$ to $x$ and one by blocking the segment from $x$ to $q$. So all but countably many $x$'s on $L$ don't run into any trouble.

  • $\begingroup$ @StefanH.: Two lines are coplanar if there is a plane of which both are subsets. $\endgroup$ – dfeuer Aug 24 '13 at 19:22
  • $\begingroup$ @dfeuer: But they need not be parallel? $\endgroup$ – Stefan Hamcke Aug 24 '13 at 19:24
  • $\begingroup$ @StefanH.: No. Parallel lines are always coplanar, but the converse does not hold. $\endgroup$ – dfeuer Aug 24 '13 at 19:25
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    $\begingroup$ Specifically, lines are coplanar (in Euclidean space, anyway) iff they are parallel or they intersect. $\endgroup$ – dfeuer Aug 24 '13 at 19:27
  • $\begingroup$ @dfeuer: Thanks, I see now why non-parallel would not suffice in this solution. $\endgroup$ – Stefan Hamcke Aug 24 '13 at 19:30

Take a family of circular arcs on a sphere, indexed by some continuum set (like $[0,1]$), connecting these two points. Each line can intersect at most two lines.

Expansion by dfeuer:

Let $x,y \in \Bbb R^3\setminus A$.

Let $S$ be the hollow sphere with diameter $\overline{xy}$.

Let $\mathcal H$ be the set of semicircles in $S$ connecting $x$ to $y$. If you imagine $S$ as the surface of the Earth with $x$ at the north pole and $y$ at the south pole, then $\mathcal H$ is the set of all the lines of longitude. Note that any two elements of $\mathcal H$ will intersect only at the poles $x$ and $y$.

Any line $L\subseteq A$ must intersect the sphere in $0$, $1$, or $2$ points. Since $x,y\notin A$, the line can't intersect the sphere at one of the poles, so each place it hits the sphere can only have one element of $\mathcal H$ running through it. Thus the line can hit at most two elements of $\mathcal H$. Since there are uncountably many lines of longitude, there must be at least one that doesn't hit any line in $A$.

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    $\begingroup$ Beautifully simple! $\endgroup$ – dfeuer Aug 24 '13 at 19:40
  • $\begingroup$ One minor suggestion: explicitly consider the family of semicircles of great circles of the sphere whose diameter connects the two points. $\endgroup$ – dfeuer Aug 24 '13 at 19:43
  • $\begingroup$ I tried to make it easier for someone with less experience to understand, but don't want to steal any exp points based on your answer. If you don't like my changes, feel free to revert. $\endgroup$ – dfeuer Aug 24 '13 at 23:16
  • $\begingroup$ Thanks. I like changes. I didn't make them myself, because after your suggestion I tried to construct concise explicit formulas, but couldn't and then gave up. $\endgroup$ – njguliyev Aug 25 '13 at 9:46
  • $\begingroup$ @dfeuer, Thanks for the bounty! $\endgroup$ – njguliyev Aug 28 '13 at 21:09

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