Dense disjoint open intervals

Question

Is it possible to find a collection $\mathcal{A}$ of disjoint open intervals of reals, which is dense in the sense that for any two intervals $(a, b), (e, f) \in \mathcal{A}$ with $b < e$, there is an interval $(c, d) \in \mathcal{A}$ such that $b < c$ and $d < e$?

Answer 1

A funny definition. For a non-trivial example with countably many, uniformly bounded intervals, take $C$ to be the standard $1/3$-Cantor set and $F=\cup_{n\in{\Bbb Z}} (C+n)$. Then $A=F^c$ verifies the wanted properties.

$A$ is a countable union of intervals at given levels. At level 1 you just have $A_1=(1/3,2/3)\;$ (+ ${\Bbb Z}$ translates). Then at level 2 you get $A_2=(1/9,2/9) \cup (7/9,8/9)\;$ (+ ${\Bbb Z}$ translates). At each new level you get intervals between the previous. And no endpoints on an interval in $A$ is isolated which implies the density-property.