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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.

user7090
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  • Think about filling a box with matchsticks, all pointing the same way. – ChocolateAndCheese May 16 '14 at 16:54
  • @ChocolateAndCheese So we need to partition $\mathbb{R}^3$ in to disjoint closed line segments of length 1. So they form some sort of equivalence class. Will the Axiom of Choice come in to play? – user7090 May 16 '14 at 16:59
  • No need for axiom of choice or even thinking of equivalence classes. Think about lines parallel to the $z$-axis. For every $(x,y)\in \mathbb{R}^2$, you have a line in $\mathbb{R}^3$ parallel to the $z$-axis. This line is a copy of $\mathbb{R}$. How do you cover $\mathbb{R}$ with these disjoint intervals? Now take a union of all such covering over all points $(x,y)\in \mathbb{R}^2$. – ChocolateAndCheese May 16 '14 at 17:03
  • @CC are you sure? They are segments. Can you cover $\mathbb{R}$ with such segments? – PA6OTA May 16 '14 at 17:05
  • Yes, make the segments half-open: $\mathbb{R} = \cup_{z\in\mathbb{Z}} (z, z+1]$ – ChocolateAndCheese May 16 '14 at 17:05
  • @ChocolateAndCheese, How would I show that such a cover of $\mathbb{R}$ is possible using half-open line segments? I see how the solution would follow easily once this is shown. – user7090 May 16 '14 at 17:07
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    Also, the line segments need to be closed. – user7090 May 16 '14 at 17:10
  • I don't think this is possible. $\mathbb{R}^3$ is a what is called normal space, so disjoint closed sets give rise to disjoint open sets. If $\mathbb{R}^3$ is a union of disjoint closed sets, then it is a union of disjoint open sets, and is therefore not connected. But $\mathbb{R}^3$ is connected, so this is not possible. – ChocolateAndCheese May 16 '14 at 17:21
  • See this post:http://math.stackexchange.com/questions/556876/can-closed-sets-in-real-line-be-written-as-a-union-of-disjoint-closed-intervals – ChocolateAndCheese May 16 '14 at 17:25
  • @CC. Hmm, this question is being asked for an introductory set theory course. – user7090 May 16 '14 at 17:42

1 Answers1

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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.

Mike Earnest
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