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I posted this on Stack Exchange and got a lot of interest, but no answer.

A recent Missouri State problem stated that it is easy to decompose the plane into half-open intervals and asked us to do so with intervals pointing in every direction. That got me trying to decompose the plane into closed or open intervals. The best I could do was to make a square with two sides missing (which you can do out of either type) and form a checkerboard with the white squares missing the top and bottom and the black squares missing the left and right. That gets the whole plane except the lattice points. This seems like it must be a standard problem, but I couldn't find it on the web. Question: So can the plane be decomposed into unit open intervals? closed intervals?

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    $\begingroup$ I is easy in 3-space --- it can be done the same way as here: mathoverflow.net/questions/28647 $\endgroup$
    – Anton Petrunin
    Commented Oct 26, 2010 at 2:00
  • $\begingroup$ @Anton: I do not understand how you do that? You need to cover two things: $R^3$ (by half-intervals) and the unit sphere $S^2$ (by points). In the case of circles, the situation was much easier because the "directions" of circles (vectors, perpendicular to the planes where the circles are) did not need to cover $S^2$. $\endgroup$
    – user6976
    Commented Oct 26, 2010 at 3:20
  • $\begingroup$ @Mark: "Why" --- because I can do this and I did not see the simpler solution 1 hour ago :) $\endgroup$
    – Anton Petrunin
    Commented Oct 26, 2010 at 3:46
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    $\begingroup$ Anton's solution is a good one for the closed/open problem in R^3, where I did not call for all directions. We still don't have anything for R^2. $\endgroup$
    – Ross Millikan
    Commented Oct 26, 2010 at 4:35
  • $\begingroup$ It seems people are answering the Missouri state problem, not yours. You should maybe state it precisely after a bold face "question" label. $\endgroup$
    – Benoît Kloeckner
    Commented Oct 26, 2010 at 10:49

6 Answers 6

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Conway and Croft show it can be done for closed intervals and cannot be done for open intervals in the paper:

Covering a sphere with congruent great-circle arcs. Proc. Cambridge Philos. Soc. 60, 1964, pp787–800.

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  • $\begingroup$ Got to love figure 1 in this paper. $\endgroup$
    – t3suji
    Commented Oct 26, 2010 at 21:35
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Start with the collection of half-open intervals of the form $[a,a+1) \times 0$ where $a \geq 0$ is an integer. This decomposes the positive $x$-axis into half-open intervals. Now, for every value of $0 < \theta < 2\pi$, decompose the ray whose angle with the positive $x$-axis is $\theta$ into half-open intervals with the open end of the interval placed at the endpoint nearest the origin.

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  • $\begingroup$ As an answer to the original problem, I believe that misses an interval parallel to the x axis with the closed end to the right. But I was asking about filling the plane with closed or open intervals. $\endgroup$
    – Ross Millikan
    Commented Oct 26, 2010 at 2:10
  • $\begingroup$ Which interval is that? This should hit everything. It might be good to clarify the question since two people have answered the original question. I assumed you were looking for any decomposition, including using closed or open intervals. $\endgroup$
    – Jeremy West
    Commented Oct 26, 2010 at 2:16
  • $\begingroup$ From Mark's answer I see that you mean I am missing a direction. I wasn't aware that the direction was oriented in this manner; I assumed it meant unoriented direction, i.e., slope. In this case, I am even more interested to know the answer to both questions. $\endgroup$
    – Jeremy West
    Commented Oct 26, 2010 at 2:26
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    $\begingroup$ You can rescue it by making a horizontal band, say of width 1. Make the top half of right-pointing half-open intervals and the bottom half of left-pointing ones. Then use your fan above and below. A similar construction will rescue Mark's. $\endgroup$
    – Ross Millikan
    Commented Oct 26, 2010 at 5:05
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Here is a solution to the half-open interval problem:

  1. Start with the interval from $(0,0)$ to $(1,0)$ that contains the endpoint $(0,0)$.
  2. Fill in the closed unit disc minus the point $(1,0)$ using half-open intervals that point inward.
  3. Add the interval from $(1,0)$ to $(1,1)$ that contains the endpoint $(1,0)$.
  4. Fill in the closed disc of radius $\sqrt{2}$ minus the point $(1,1)$ using half-open intervals that point along inward tangents. Now we have segments in all directions.
  5. Fill in the ray $\{ (a,a) \mid a \geq \sqrt{2} \}$ using outward pointing radial segments.
  6. Fill in the rest of the plane using inward pointing radial segments arranged in concentric annuli (with slits at $\arg z = \frac{\pi}{4}$).
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  • $\begingroup$ I agree this answers the half-open all directions problem. $\endgroup$
    – Ross Millikan
    Commented Oct 26, 2010 at 4:39
  • $\begingroup$ Oops, in step 5, I should have put $a \geq 1$. $\endgroup$
    – S. Carnahan ♦
    Commented Oct 26, 2010 at 7:37
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Decompose a line without a point into a union of disjoint half-open intervals. Put copies of this line on the plane so that the distinguished point is $(0,0)$ and the lines point in all possible directions. You have covered the plane without one point, $(0,0)$. Now take one of these lines and replace it by a line completely covered by unit intervals. You get the whole plane covered. The same works in all dimensions.

Edit 1 This solution does miss one direction. If closed intervals are allowed, then instead of the last step, replace one half-open interval on one of the lines by a closed one, so that you cover $(0,0)$.

Edit 2 I just noticed that I was answering a wrong question. Fortunately the correct question has been answered already.

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    $\begingroup$ As an answer to the original problem, I think you are missing one direction along the line you used to cover (0,0). But I was asking about filling the plane with closed or open intervals. $\endgroup$
    – Ross Millikan
    Commented Oct 26, 2010 at 2:14
  • $\begingroup$ Oops. Yes, you are right. But if you are allowed closed intervals, just replace one half-interval on one of the lines by a closed one. $\endgroup$
    – user6976
    Commented Oct 26, 2010 at 2:18
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Let me answer for closed intervals. (It is well-known, but I do have only Russian references in mind.) We may decompose closed rectangle (easy). Then, if we have rectangle $R_1=a\times b$ already decomposed, then we double it, make $R_2=2R_1=a\times 2b$ and cover $R_2\setminus R_1$. So, doubling in different directions initial rectangle, we get whole plane.

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  • $\begingroup$ Are you using unit intervals? (if intervals of arbitrary length are allowed, both problems are relatively easy) $\endgroup$
    – t3suji
    Commented Oct 26, 2010 at 14:36
  • $\begingroup$ Certainly I can make a strip of unit width, taking $[0,1] X {y}$ for $y \in \mathbb{R}$ Then I can cover a strip in $y$, say taking ${x} X [0,1]$ for $x \in (1,\inf)$ But then I don't see how to get the inner corner, say above and right of (1,1). $\endgroup$
    – Ross Millikan
    Commented Oct 26, 2010 at 15:50
  • $\begingroup$ Oh, I missed the world "unit", sorry. But for intervals, I do not see solution even if they are not unit... $\endgroup$
    – Fedor Petrov
    Commented Oct 26, 2010 at 18:24
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If you consider upper semicontinuous decompositions on compact connected sets, then, in this paper it is proved that it is not possible to fill any euclidean space in such a way.

There is a paper by Roberts where he proves the two dimensional result and also gives an example of an upper semicontinuous decomposition of the plane into cellular curves (they may be not simple, but they are a decreasing intersection of disks).

I know this does not respond the question entirely, since the upper semicontinuous hypothesis is strong, however, many times desirable.

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