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. So can the plane be decomposed into unit open intervals? closed intervals?
Decomposing the plane into intervals
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0For closed intervals, you cannot do it with countably many. Just take a slice of the plane, the intersection of your covering with the slice gives a covering of the real line by closed intervals, which is impossible: http://terrytao.wordpress.com/2010/10/04/covering-a-non-closed-interval-by-disjoint-closed-intervals/ – 2010-10-23
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1I also can't do it with countably many because it won't cover the area. We need uncountably many just to cover the unit square. – 2010-10-23
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2You cannot cover the plane with countably many open/closed intervals, since each of them has measure zero in $\mathbb{R}^2$. For the general case, I'd be surprised if it's possible. – 2010-10-23
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0I don't understand the definition of their "interval". It seems an "interval" in their example is $[n,n+1)\times\mathbb R$. – 2010-10-23
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4@KennyTM: An interval here is a line segment of positive finite length, open if it doesn't contain the endpoints, closed if it does, and half-open if it contains one endpoint but not the other. If you take the union of half-open intervals [n,n+1) x {y} as y ranges over R, you get [n,n+1) x R. – 2010-10-23
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0@Nuno: that was exactly my question. The problem is the corners. Closed ones get in the way and open ones can't cover. We need continuum many, but I always expected that as my "close" answer was already that many. – 2010-10-25
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0@Ross: Sorry if I didn't point that out clearly. I was referring to Willie Wong's answer. I think the measure zero argument is easier to understand than that we cannot cover a non-closed interval with countably many disjoint closed intervals. About your question, my intuition says that we cannot decompose the plane into closed/open intervals. Since you didn't get any answers here, what about posting it on MO? – 2010-10-25
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0I posted it to MO. Anton Petrunin gave me a link to a previous problem that solves it for R^3: http://mathoverflow.net/questions/28647 That problem used circles, but works for either kind of interval. It does require a well-order of R. – 2010-10-26
2 Answers
I posted this to Math Overflow and Jeff Strom gave the following answer:
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 787–800.
Seems pretty easy. We're going to cover the complex plane instead since that's obviously equivalent to covering $\mathbb{R}^2$. Start with the family of unit length line segments $I_\theta = \{a\cdot e^{i\theta} \mid a \in (0,1] \}$, $\theta \in (0,2\pi)$. Then define $I_{\theta,n}$, $n \in \mathbb{N}_0$ to be $I_\theta$ translated a length of $n$ in the direction of $\theta$. This collection covers the entire complex plane except the ray $[0,\infty)$ which we cover with the half-open intervals $[k,k+1)$.
Armed with this idea, it should be easy to see how to for any $k$ fill $\mathbb{R}^k$ with intervals of unit length that "point in all directions".
FYI: \mathbb{N}_0
breaks stuff. You have to escape the underscore: \mathbb{N}\_0
.
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3Well he isn't asking how to solve the problem stated in that link. He's trying to decompose the plane into closed or open intervals. – 2010-10-23
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0Yeah, I was about to edit my post to reflect that I had noticed that. Oops. – 2010-10-23
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0I think your solution to the original problem is missing one along the direction $\theta=0$ with the closed end to the left. – 2010-10-23