# Family of almost disjoint subsets of $\mathbb{R}_+$ (where non-disjointness is bounded)

Let $\mathbb{R}_+ = [0, \infty)$. I'm looking for a family $\mathcal{U}$ of $2^\mathfrak{c}$ subsets of $\mathbb{R}_+$, such that any member of $\mathcal{U}$ is unbounded in $\mathbb{R}_+$, but the intersection of any two members of $\mathcal{U}$ is bounded. Does such family exists?

It would also be nice if every member $X$ of $\mathcal{U}$ satisfied $|X \cap [n, \infty)| = \mathfrak{c}$ for all natural $n$.

The idea behind the problem is that I obviously cannot have $2^\mathfrak{c}$ disjoint subsets of $\mathbb{R}_+$, but I want the non-disjointness to be packed in a bounded segment. It seems like it should be possible, as a bounded segment of $\mathbb{R}_+$ doesn't look much different from an unbounded segment, from a set-theoretical point of view.

Let $\mathcal S$ be the set of all unbounded countable subsets of $[0,\infty).$
If each member of $\mathcal U$ is unbounded then each member of $\mathcal U$ contains an unbounded countable set, so there is a function $f:\mathcal U\to\mathcal S$ such that $f(X)\subseteq X$ for all $X\in\mathcal U.$
If the pairwise intersections of $\mathcal U$ are bounded, then $f$ is injective and $|\mathcal U|\le|\mathcal S|=\mathfrak c.$
• +1 very nice! For the OP, note that this relies on the fact that there are "small" cofinal subsets of $\mathbb{R}$; in particular, this fails in the natural numbers, where we can have a size-$2^{\vert\mathbb{N}\vert}$ family of almost disjoint sets (see e.g. this question). – Noah Schweber Nov 25 '16 at 0:41
• Yeah, I was inspired by almost disjoint subsets of $\omega$ when asking this question. – xyzzyz Nov 25 '16 at 3:42