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The following question came up at tea today, and none of us managed to come up with an answer. I was wondering if anyone had any ideas.

Does there exist a subset $X$ of $\mathbb{R}^2$ with the following two properties.

  1. If $p,q \in X$ are distinct, then the distance from $p$ to $q$ is at least $1$.
  2. There exists some $c \in \mathbb{R}$ such that if $R \subset \mathbb{R}^2 \setminus X$ is any closed rectangle (possibly "tilted", i.e. with its sides not necessarily parallel to the coordinate axes), then the area of $R$ is at most $c$.

Of course, $X$ must be infinite. As a weak guess, I would wager that no such $X$ exists, but I have no idea how to prove it.

EDIT : The rectangles in condition 2 include their interiors (so points in $X$ cannot occur in the interiors of the rectangles).

Adam Smith
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  • I'm not entirely sure what you mean by "closed". Do you allow points of $X$ to lie in the interior of your rectangle? – Austin Mohr Oct 19 '11 at 05:10
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    In the closed upper half-plane , for $n\in \mathbb{Z}$ let $X={n,0}\cup{n,1}\cup{(2n+1)/2,2}\cup{(4n+1)/4,3}\cup{(4n+3)/4,4}\cup \ldots $ Does this eliminate all the big rectangles? – Ross Millikan Oct 19 '11 at 05:12
  • No, I don't allow points of $X$ to lie in the interior of my rectangles, though I guess I should have been clearer there. I meant closed in the point-set topological sense (so my rectangles include their boundary), though this doesn't really affect the problem. – Adam Smith Oct 19 '11 at 05:13
  • @RossMillikan : No, it doesn't. For instance, for any $k$ you have a rectangle $[0,k] \times [1/4,3/4]$, which has area $k/2$. Plus you still have the lower half plane... – Adam Smith Oct 19 '11 at 05:16
  • How about four copies of Ross' pattern, rotated by multiples of $\pi/2$ about the origin? You may have to start out with a smaller step width. – joriki Oct 19 '11 at 05:26
  • @joriki : I'm not sure what you are getting at. Can you spell it out more precisely? – Adam Smith Oct 19 '11 at 05:33
  • Sorry, I'd temporarily forgotten about the first condition -- never mind... – joriki Oct 19 '11 at 06:03
  • @AdamSmith: I didn't spell out my pattern well enough. I was going to follow with lines at n+1/8, n+3/8, n+5/8, n+7/8, n+1/16, etc. So you can't have too wide a rectangle anywhere. – Ross Millikan Oct 19 '11 at 12:52
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    @Ross: But how will this satisfy both conditions? You've already got points at $n,0$ and $n,1$, so no more points will fit in $\mathbb R\times[0,1]$ without violating the first condition, so you can't eliminate the rectangles that Adam mentioned? – joriki Oct 19 '11 at 14:16
  • @AdamSmith: This appears to be a well-know open problem, see discussion at http://mathoverflow.net/questions/3307/can-a-discrete-set-of-the-plane-of-uniform-density-intersect-all-large-triangles. – Moishe Kohan Apr 04 '14 at 00:25

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