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Fix an integer $n\ge 2$ and suppose that ${\cal L}$ is a set of lines in $\mathbb{R}^n$. Is there a set $M\subseteq \mathbb{R}^n$ with the following properties?

  1. $M$ intersects all the elements of ${\cal L}$, but
  2. for all $m\in M$, the set $M\setminus\{m\}$ no longer intersects all the elements of $\cal L$.

In fact, I do not even know the answer for $n=2$.

Ali Taghavi
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Dominic van der Zypen
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  • I might have misunderstood this, but it seems that it suffices to choose one point from each line in a such way that none of the chosen points is intersection of this line with some other line in $\mathcal L$. (In the other words, for each $l\in\mathcal L$ we choose a point from $l \setminus \bigcup_{k\in \mathcal L} k$.) – Martin Sleziak Aug 07 '17 at 07:09
  • This idea may not work if a line is covered by other lines in the collection. For n=2 and finitely many lines, one can take all intersections (or representatives in the case of no intersections), but for infinitely many lines (e.g. all of them) it is unclear to me, even with a well ordering of the lines. Gerhard "Isn't Meeting With A Solution" Paseman, 2017.08.07. – Gerhard Paseman Aug 07 '17 at 07:19
  • @Martin I think this works if for every line there is a point that is only covered by that line. But ${\cal L}$ may be such that this is not the case. – Dominic van der Zypen Aug 07 '17 at 07:22
  • Ok, so I missed that $\mathcal L$ is not necessarily finite. So the above works if $\mathcal L$ is countable (and it seems that in fact $|\mathcal L|<\mathfrak c$ suffice). – Martin Sleziak Aug 07 '17 at 07:22
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    Actually, when n is 2 take for M the set of x and y axes. This works when the set of lines is rich enough, and for thinner sets of lines a subset of this M might be constructed. Gerhard "Is Working On Getting Thinner" Paseman, 2017.08.07. – Gerhard Paseman Aug 07 '17 at 07:23
  • @MartinSleziak we almost wrote the same thing :) – Dominic van der Zypen Aug 07 '17 at 07:23
  • @GerhardPaseman looks good to me - thanks...! – Dominic van der Zypen Aug 07 '17 at 07:24
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    Indeed, more generally for n=2, pick a line l, and let M be a subset of that line containing all intersections with non parallel lines to l, union a set of representatives of points, one for each line parallel to and different from l. This should work, and you may be able to project down to two dimensions in general. Gerhard "Has Met A Solution Now" Paseman, 2017.08.07. – Gerhard Paseman Aug 07 '17 at 07:40
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    And if you don't like projection, well order the lines and proceed as follows: for the next unselected line, add to M all intersection points of that line with other unselected lines in the set of lines, and then throw all those lines out of the set and continue. Gerhard "Ensure They're All Lined Up" Paseman, 2017.08.07. – Gerhard Paseman Aug 07 '17 at 07:47

2 Answers2

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If we have a set $\mathcal L$ of lines in $\mathbb R^n$ such that $|\mathcal L|=\mathfrak c$, we can get the set $M$ with the desired properties using transfinite induction.

  • Take any well-ordering of the set of lines $\mathcal L=\{l_\alpha; \alpha<\mathfrak c\}$ (such that $\alpha\ne\beta$ implies $l_\alpha\ne l_\beta$.)
  • By transfinite induction we define $M_\alpha$ for $\alpha<\mathfrak c$ which is either singleton or empty set. If $l_\alpha$ intersects $\bigcup\limits_{\beta<\alpha} M_\beta$, we put $M_\alpha=\emptyset$. (We add nothing if $l_\alpha$ already contains some points.) Otherwise we put $M_\alpha=\{m_\alpha\}$ where $m_\alpha \in l_\alpha \setminus \bigcup\limits_{\beta<\alpha} l_\beta$. (This set is non-empty, since $l_\alpha$ has cardinality $\mathfrak c$ and the intersection $l_\alpha \cap \bigcup\limits_{\beta<\alpha} l_\beta$ has cardinality smaller than $\mathfrak c$; each of the lines $l_\beta$ can intersect $l_\alpha$ at most in one point.)
  • Now simply put $M=\bigcup\limits_{\alpha<\mathfrak c} M_\alpha$.

The set $M$ has the desired properties. Clearly, $M$ intersects each $l_\alpha$. Moreover, if $m\in M$ and $m=m_\alpha$, then $m$ is the only point on the line $l_\alpha$. (If $m$ was added in the $\alpha$'s step, then $l_\alpha$ does not contain any of the points from $\bigcup\limits_{\beta<\alpha} M_\beta$, i.e., the points added in the preceding steps. And the construction is done in such way that no new point on $l_\alpha$ can be added in the steps after $\alpha$.)

If $|\mathcal L|<\mathfrak c$ then we can simply choose one point from each set $l \setminus \bigcup\limits_{\substack{k\ne l\\ k\in \mathcal L}} k$. This set is non-empty - it has cardinality $\mathfrak c$.

I should point out that Gerhard Paseman suggested in a comment another solution using transfinite induction. And for $n=2$ he suggested in another comment a solution which does not use transfinite induction.

Proof of the existence of Mazurkiewicz two-point set is in somewhat similar vein, see also this question: Subset of the plane that intersects every line exactly twice.

Martin Sleziak
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For $n=2$ we define $M$ as follows:

$M$ is the union of the following sets:

1)The intersection with $x\_$ axis for lines not parallel to this axis.

2)The intersection with $y\_$axis for lines perpenedecular to this axis.

Ali Taghavi
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