Path Connectedness of a Set 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.
 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.
A: 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 $A$.
A: 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 $\Bbb R^3\setminus A$.
