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For which integers $n$ is it possible to find a subset $S$ of $\mathbb R^2$ such that every infinite line contains exactly $n$ points of $S$?

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2 Answers 2

up vote 3 down vote accepted

I’ve not seen Galvin’s argument, but here’s one that works for the finite cases.

Fix $n>1$. Let $\mathscr{L}=\{L_\xi:\xi<2^\omega\}$ be an enumeration of the lines in the plane. Suppose that $\alpha<2^\omega$ and that for each $\beta<\alpha$ we have a set $A_\beta\subseteq\mathbb{R}^2$ such that

$\qquad(0)_\beta:\qquad$ $A_\xi\subseteq A_\beta$ for every $\xi<\beta$;
$\qquad(1)_\beta:\qquad$ $|A_\beta|\le |\beta|+\omega$;
$\qquad(2)_\beta:\qquad$ no $n+1$ points of $A_\beta$ are collinear; and
$\qquad(3)_\beta:\qquad$ for each $\xi<\beta$, $|L_\xi \cap A_\beta|=n$.

Let $A_\alpha'=\bigcup\limits_{\beta<\alpha}A_\beta$. If $\alpha$ is a limit ordinal, let $A_\alpha=A_\alpha'$; clearly $(0)_\alpha-(3)_\alpha$ hold.

If $\alpha=\xi+1$, $L_\xi$ intersects $A_\alpha'$ in $k$ points for some $k\le n$. If $k=n$, let $A_\alpha=A_\alpha'$; as before, $(0)_\alpha-(3)_\alpha$ hold. Otherwise, let $B$ be the union of the lines determined by collinear $n$-point subsets of $A_\alpha'$; there are fewer than $2^\omega$ such lines, so $|L_\xi\setminus B|=2^\omega$, and we may choose a set $F_\xi \subseteq L_\xi \setminus B$ of cardinality $n-k$. If we then set $A_\xi = A_\alpha'\cup F_\xi$, $(0)_\alpha-(3)_\alpha$ are satisfied.

Now let $A = \bigcup\limits_{\xi<2^\omega}A_\xi$; then each line in the plane intersects $A$ in exactly $n$ points.

(I suppose that it should be noted that $n=0$ also works!)

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It's an open problem AFAIK whether a Borel $n$-point set exists. –  Henno Brandsma Nov 2 '11 at 7:10

According to Gelbaum and Olmsted, Counterexamples in Analysis, Chapter 10, problem 21, page 144, "F. Galvin has shown the following: if $1\lt n\le\aleph_0$ then there is a non-measurable set in the plane such that the intersection of $S$ with any line consists of precisely $n$ points." Unfortunately, no bibliographic details are given.

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