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Is it possible to cover $\mathbb{R}^2$ with uncountably many disjoint non-degenerate line segments?

If a formal definition is necessary, let's define a line segment as a set $\{(x, mx+c): x \in [a, b]\}$ for some fixed constants $m, c, a, b \in \mathbb{R}$ with or a set $\{(u, x): x \in [a, b]\}$ for some fixed constants $u, a, b \in \mathbb{R}$. We say a line segment so defined is non-degenerate if $a \neq b$, i.e. the line segment is not a point.

This question was vaguely motivated by the observation that it's possible to cover $\mathbb{R}$ with uncountably many disjoint non-degenerate points. YuvalFilmus points out that the answer is negative in the one-dimensional case.

donburi
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  • This question seems related but (1) there's no answer and (2) it's in $\mathbb{R}^3$, which might plausibly be different. – donburi Oct 17 '14 at 07:10
  • When you say "cover", you mean "with no overlap"? I don't think that changes the answer (yes), but it is considerably harder. – Eric Stucky Oct 17 '14 at 07:19
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    The one-dimensional version is impossible. Given a partition of $\mathbb{R}$ into non-degenerate closed intervals, choose two of them $I_1 = [x_1,y_1]$, $I_2 = [x_2,y_2]$ such that $y_1 < x_2$. Then $(y_1,x_2)$ must be a union of disjoint non-degenerate closed intervals. Each of them contains some rational so there can be at most countably many of them, and an appeal to http://terrytao.wordpress.com/2010/10/04/covering-a-non-closed-interval-by-disjoint-closed-intervals/ finishes the proof. – Yuval Filmus Oct 17 '14 at 07:23
  • Whoops, @YuvalFilmus - you're absolutely right - there's also an answer here on StackOverflow with a similar proof addressing the one-dimensional case. Had a brain fart there. – donburi Oct 17 '14 at 07:34
  • @EricStucky: yup, I meant with no overlap. – donburi Oct 17 '14 at 07:35
  • @EricStucky The answer is trivially true for the one-dimensional case if you admit overlap ($\bigcup_{n \in \mathbb{Z}} [n,n+1]$), whereas it's false if you forbid them. So I can definitely see it changing the answer. – Najib Idrissi Oct 17 '14 at 07:42
  • @Najib: I agree in principle but I think I see a construction. Take the open square, and cut it into nine parts. The center can be removed, leaving four open corners and four half-open sides. The sides can be removed as well, and the process can recurse. I don't believe this leaves any extra points. (Of course, it would if we kept the endpoints when we divide a line in thirds; this is a Cantor-like construction. But this construction avoids that pitfall) – Eric Stucky Oct 17 '14 at 08:28

2 Answers2

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The answer is YES, you can cover the plane by non-degenerate line segments.

It is clear one can cover

  • any $1 \times 1$ closed squares (i.e those isometric to $[0,1] \times [0,1]$ ) by line segments of length $1$.
  • any $1 \times n$ quarter-open rectangles ( i.e. those isometric to $[0,1) \times [0,n]$ ) by line segments of length $n$.

You then proceed to cover $\mathbb{R}^2$ by

  1. first placing a $1 \times 1$ closed square in the center.

  2. surround the $1 \times 1$ closed square by four $1 \times 2$ quarter-open rectangles. The "open" sides of the rectangles will be facing "inwards" and the five quadrilaterals together form a $3\times 3$ closed square.

  3. surround the $3 \times 3$ closed square by four $1 \times 4$ quarter-open rectangles. The "open" sides again facing "inwards" and the nine quadrilaterals together form a $5 \times 5$ closed square.

  4. Just repeat this procedure. If you have a $(2k-1) \times (2k-1)$ closed square, you surround it by four $1 \times 2k$ rectangles to form a $(2k+1) \times (2k+1)$ closed square.

Following is a picture illustrating the arrangement of the squares/rectangles of line segments.

  • The line segments are represented by color bars.
  • The black lines indicate the the boundaries of the squares/rectangles.

Cover plane by line segments

achille hui
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  • @EricStucky I don't understand the construction in your comment but it is very likely to be equivalent. In this sort of construction, a picture definitely worth more than thousand words ;-p – achille hui Oct 17 '14 at 08:37
  • Actually, I realized after saying it that the constructions are quite different. In particular, if we ignore the center, yours has rotational symmetry, but while mine has the suggestion of reflective symmetry, I don't think it actually achieves it. – Eric Stucky Oct 17 '14 at 08:53
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Slice $\mathbb R^2$ into uncountably many copies of $\mathbb R^1$.

Scaramouche
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