# Removing highly scattered points in the plane to eliminate all high-area rectangles

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).

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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 5:10
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 5:12
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 5:13
@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 5:16
@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