Using line segments to cover $\mathbb{R}^3$ Prove that $\mathbb{R}^3$ can be covered with pairwise disjoint closed line segments of length $1$.
Any advice/hints on how to approach this problem is appreciated. 
 A: There is a solution using the axiom of choice. Choose a well-ordering of $\mathbb R^3$ using a bijection $\varphi:\omega\to \mathbb R^3$, where $\omega$ is the smallest ordinal whose cardinality is that of $\mathbb R^3$. Then inductively define a collection of closed length $1$ line segments as follows: for each ordinal $\alpha<\omega$, if $\varphi(\alpha)$ is not covered by a previously  chosen line segment, choose some line segment with one endpoint at $\varphi(\alpha)$ which intersects no other line segment already chosen. 
To see this is possible, consider the sphere of radius $1$ centered at $\varphi(\alpha)$, and project all of the previously chosen line segments onto this sphere. This sphere has $2^{\aleph_0}$ longitudes. Since there are fewer than $2^{\aleph_0}$ previously chosen line segments (all initial segments of $\omega$ have cardinality less than that of $\omega$), at least one of these longitudes, $\ell$ will not have any line segments lying along it. Each line segment intersects $\ell$ in at most two places, so some point $p$ on $\ell$ is not contained in any line segment. This means that the closed segment between $p$ and $\varphi(\alpha)$ will not intersect any previously chosen segment. 
This uncountable collection of line segments is disjoint (by transfinite induction), and also exhausts $\mathbb{R}^3$, so we are done.
