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

Question

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 $n$.

Answer 1

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 $T$.