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Let $J$ be a family of parallel closed intervals in the plane $\mathbb{R}^2$, no two different of them contained in any common line. Moreover, for arbitrary three intervals belonging to $J$ there exists a line intersecting all three of them. I would like to prove that there exists a line intersecting all segments belonging to $J$! (I'd like to add that $J$ can be infinite).

It seems to be rather complicated even in the case of $4$ lines! Moreover, the statement is wrong if we replace segments by open intervals: there is no line intersecting each interval of the family $\{(0,\frac{1}{n})\times \{n\}: n\in \mathbb{N}\}$, but each triple of course can be intersected by some vertical line.

Can anybody please suggest something?

I was trying to prove some simple finite cases but failed. I think it is important to neglect specificity of finiteness: the set of points of segments can even be dense!

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Here is a proof for a finite number of parallel lines. For each segment $s_i$, let $L_i$ be the set of lines that intersect $s_i$. We may regard each line $\ell \in L_i$ as a point $(m,b) \in \mathbb{R}^2$, where $m$ is the slope of $\ell$ and $b$ is its $y$-intercept. In this way, we may regard $L_i$ as a subset of $\mathbb{R}^2$. It is easy to check that $L_i$ is convex for all $i \in I$. By hypothesis, $L_i \cap L_j \cap L_k$ is non-empty for all $i,j,k$. Thus, by Helly's theorem, $\bigcap_{i \in I} L_i$ is non-empty. That is, there is a line that passes through each $s_i$.

There is a version of Helly's theorem which applies to infinite families of convex sets, but with the additional requirement that the sets also be compact. Unfortunately for us, while $L_i$ is convex, it is not compact because the $y$-intercepts always take on all values in $(-\infty, \infty)$.

Edit. There is a suitable version of Helly's theorem for infinite families which implies the result for infinite $I$. You can find the version here, together with a proof of your result, which is referred to as the Common Transversal Theorem or the Skewer Theorem.

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    $\begingroup$ For the case of infinite $I$, we can argue as follows: Fix two distinct $i,j\in I$. Then $X=L_i\cap L_j$ is compact, and we have that $X\cap\bigcap_{k\in J}(X\cap L_k)\neq\emptyset$ for every finite $K\subseteq I$. But now all considered sets in the intersection are closed subsets of the compact set $X$, and it follows that $\bigcap_{i\in I}L_i\neq\emptyset$. $\endgroup$ Apr 13, 2015 at 1:39
  • $\begingroup$ @ThomasKalinowski Thanks for that argument! Off the top of my head I only remembered the infinite version of Helly where all sets are compact, but you are absolutely correct. $\endgroup$
    – Tony Huynh
    Apr 13, 2015 at 1:45

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