Decomposing the plane into intervals 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?
 A: 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. 
A: Here is a solution to the half-open interval problem:


*

*Start with the interval from $(0,0)$ to $(1,0)$ that contains the endpoint $(0,0)$.

*Fill in the closed unit disc minus the point $(1,0)$ using half-open intervals that point inward.

*Add the interval from $(1,0)$ to $(1,1)$ that contains the endpoint $(1,0)$.

*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.

*Fill in the ray $\{ (a,a) \mid a \geq \sqrt{2} \}$ using outward pointing radial segments.

*Fill in the rest of the plane using inward pointing radial segments arranged in concentric annuli (with slits at $\arg z = \frac{\pi}{4}$).

A: 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. 
A: 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.
A: 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.
A: 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.
