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Let $S$ be a set of points in a plane $P$, having the following property: for any point $X \in P$ there is at least one point $M \in S$ so that the distance $|XM|$ is rational. Find the minimum cardinal of such a set.

First, let's notice there is such a set, by taking $S=P$. Also if $S$ is a line in $P$. Someone claimed there is finite set $S$ having the required property. Does anyone have an idea how to solve this?

UPDATE

A related question here.

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  • Why the set of points of a line having rational coordinates has the desired property? – timon92 Sep 13 '15 at 10:57
  • @timon92 I don't have a prove, but I was told it is true because of the density of $\mathbb{Q}$ in $\mathbb{R}$. Anyway, I remove it –  Sep 13 '15 at 11:15
  • The proposed duplicate is about removing all points with irrational distances from one of the given points. That's a rather different problem. – hmakholm left over Monica Sep 13 '15 at 12:36

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$S$ must have cardinality $\mathfrak c$.

The assumed property is that $$ P = \bigcup_{(M,d)\in S\times\mathbb Q} \bigl\{ X\in P\bigm| |XM|=d \bigr\} $$ The sets in the union on the right are circles, and each circle intersects the $x$-axis in at most two points. Since the $x$-axis has $\mathfrak c$ points in it, it can only be totally within the union if there are at least $\mathfrak c$ circles. However if $S$ is infinite (and finitely many circles certainly won't cover the plane), then $|S\times\mathbb Q|=|S|\times\aleph_0 = |S|$, so $|S\times\mathbb Q|\ge\mathfrak c$ implies $|S|\ge\mathfrak c$.

On the other hand, there are only $\mathfrak c$ points that can be in $S$.

In particular, "the points of a line having rational coordinates" won't do -- there are too few of them. (Not to mention that some lines, such as $x+y=\sqrt2$, contain no points with rational coordinates).

hmakholm left over Monica
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  • OK, but what about S being all the points of rationale coordinates? –  Sep 13 '15 at 11:17
  • Also, the $P$ set looks maximal to me, or, at least doesn't look minimal –  Sep 13 '15 at 11:21
  • @EugenCovaci: (1) No, there are not enough points with rational coordinates. (2) I don't get what you mean by "$P$ looks maximal" -- it's your own letter that stands for the entire plane. – hmakholm left over Monica Sep 13 '15 at 12:33