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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?

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For 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: terrytao.wordpress.com/2010/10/04/… –  Willie Wong Oct 23 '10 at 17:18
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I 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. –  Ross Millikan Oct 23 '10 at 17:30
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You 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. –  Nuno Oct 23 '10 at 17:41
    
I don't understand the definition of their "interval". It seems an "interval" in their example is $[n,n+1)\times\mathbb R$. –  KennyTM Oct 23 '10 at 18:21
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@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. –  Jonas Meyer Oct 23 '10 at 18:55
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up vote 6 down vote accepted

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.

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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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Well 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. –  Nuno Oct 23 '10 at 18:54
    
Yeah, I was about to edit my post to reflect that I had noticed that. Oops. –  kahen Oct 23 '10 at 18:56
    
I think your solution to the original problem is missing one along the direction $\theta=0$ with the closed end to the left. –  Ross Millikan Oct 23 '10 at 22:43
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