# Counting tranversals in an intersecting family with bounded number of connected components

Let $$X$$ be a set and $$\mathcal{F}$$ be a finite family of subsets of $$X$$. We call a set $$T\subseteq X$$ a tranversal set for $$\mathcal{F}$$ if it intersects every set in $$\mathcal{F}$$.

Suppose that $$X=\mathbb{R}$$ and $$\mathcal{F}$$ is a finite family of intervals such that any two intersect. Then it is easy to prove that $$\mathcal{F}$$ admits a transversal set with only one element, namely $$\cap\mathcal{F}\neq \emptyset$$.

Suppose that every set in $$\mathcal{F}$$ is union of at most $$n$$ points and intervals in $$\mathbb{R}$$, and again that every two sets in $$\mathcal{F}$$ intersect. Is there a simple combinatorial argument to show whether $$\mathcal{F}$$ admits a finite transversal set of a certain cardinality?

My guess here being that one such family might always admit a transversal set of cardinality $$n$$.

• 1). "an intervals" --> "and intervals". 2). is it "[at most $n$ points] and intervals" or "at most $n$ [points and intervals]"? Jul 16, 2019 at 21:36
• @mathworker21 1)Sure 2) $n$ [points and intervals] Jul 19, 2019 at 8:33

There is the following counterexample to your guess. Given a number $$n$$ let $$\mathcal F$$ be a family of all unions of at most $$n$$ closed intervals with endpoints of the from $$p/q$$ where $$p\le q\le 4n(2n-1)$$ be non-negative integers (and $$q>0$$) and of total length $$1/2$$. Connectedness of $$[0,1]$$ easily implies that each two members of $$\mathcal F$$ intersects. On the other hand, let $$T$$ be any $$2n-2$$-point subset of a segment $$[0,1]$$. Then $$T$$ splits $$[0,1]$$ into at most $$2n-1$$ open intervals, so the total length of $$n$$ longest of them is al least $$\tfrac{n}{2n-1}$$. It is easy to show that the union of these intervals contains a member of $$\mathcal F$$ disjoint from $$T$$.
• I don't get it. You found a bad $T$. How does this show there is no good $T$? Jul 17, 2019 at 11:34
• @mathworker21 I showed that there is no good $T$ of size $2n-2$. Jul 17, 2019 at 15:50
• I'm assuming $I=[0,1]$ in your reply? Jul 24, 2019 at 9:54