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
    $\begingroup$ +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). $\endgroup$ – Noah Schweber Nov 25 '16 at 0:41
  • $\begingroup$ Yeah, I was inspired by almost disjoint subsets of $\omega$ when asking this question. $\endgroup$ – xyzzyz Nov 25 '16 at 3:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.