Timeline for Is it always possible to "encircle" exactly $n$ points in an infinite subset of $\mathbb{R}^d$ without limit points?
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| Jan 10, 2014 at 2:16 | comment | added | Will Sawin | Take a disc that encloses much more than n points. Find a point in $\mathbb Q^d$ very close to the center of the disc which is generic with respect to all the points of the disc. Then choose a radius such that only n points in the disc are in that radius. If the new center is sufficiently close to original center, all points in the new disc will be in the old disc, because the new radius is smaller. | |
| Jan 10, 2014 at 0:29 | comment | added | Douglas Zare | @Michael: No, $\mathbb{Q}^d$ is countable. Take an enumeration of the points, and eliminate each one by adding a pair of points to $S$ at the same distance from that point. These pairs can be chose so that there is no limit point of $S$, say by making the distance from the origin be at least $i$ for each point of the $i$th pair added. Some other argument may work to enclose $n$ points, but you can't just choose a generic center and expand a sphere about this center. | |
| Jan 10, 2014 at 0:18 | comment | added | Stefan Kohl♦ | Thanks. -- That answers the first question (and also shows that one probably shouldn't ask questions which are too far off one's area of expertise!). | |
| Jan 9, 2014 at 23:24 | comment | added | Michael | Does the argument hold in $Q^d$? That is, how one would show that it's always possible to pick $p\in Q^d$ non-equidistant from $x,y\in S$? | |
| Jan 9, 2014 at 22:15 | history | answered | André Henriques | CC BY-SA 3.0 |