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Problem. Is there a subset $X$ in the Euclidean plane such that $X$ is not contained in a line and for any points $a,b,c,d\in X$ with $a\ne b$ and $c\ne d$, the intersection $X\cap\overline{ab}$ is infinite and the intersection $X\cap\overline{ab}\cap\overline{cd}$ is not empty.

Here $\overline {uv}$ denotes the (unique) line containing distinct points $u,v$ in the Euclidean plane.

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  • $\begingroup$ Do you know the answer to the following question: Let $F$ be the field with two elements, and let $FP^2$ be the Fano plane. Let $\bar F$ be the algebraic closure of $F$. Take any embedding $FP^2\to \mathbb R^2$. Does it extend to an embedding of $\bar FP^2$? $\endgroup$
    – Mikhail Katz
    Commented Aug 5, 2023 at 20:03
  • $\begingroup$ @MikhailKatz No, I do not know the answer to this problem. I even do not see why the Fano plane does embed into the Euclidean plane. $\endgroup$
    – Taras Banakh
    Commented Aug 5, 2023 at 20:47
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    $\begingroup$ No finite projective plane can be embedded in $\mathbb R^2$. $\endgroup$
    – Keith Kearnes
    Commented Aug 5, 2023 at 20:49
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    $\begingroup$ As to why no finite projective plane can be embedded in the real projective plane: this is because the projective plane structure recovers the field structure (once fixed arbitrary points on a line called $0$, $\infty$, $1$), and evidently $\mathbb{R}$ has no finite subfield. $\endgroup$
    – Gro-Tsen
    Commented Aug 6, 2023 at 10:19

1 Answer 1

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Let $\ P^2(\mathbb Q)\ $ and $\ P^2(\mathbb R)\ $ be the projective planes over rationals and reals. Let $\,\ L\subseteq P^2(\mathbb R)\,\ $ be a straight line in the real plane such that

$$ L\cap P^2(\mathbb Q)\,\ =\,\ \emptyset, $$

and let projective map $\,\ F: P^2(\mathbb R)\to P^2(\mathbb R)\,\ $ map $\ L\ $ onto the infinite line. Then

$$ X\,\ :=\,\ F(P^2(\mathbb Q)) $$

is the required set (see the OP Question).

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    $\begingroup$ Great Answer! Thank you. Your answer implies that the axioms of projective geometry are consistent with the axioms of ordered geometry, and the existence of parallels does not follow from axioms of ordered geometry (but do follow from axioms including the congruence and possibility to construct a straight angle). $\endgroup$
    – Taras Banakh
    Commented Aug 6, 2023 at 4:07

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