Let $\kappa$ be an infinite cardinal and as usual denote by $[\kappa]^\mu$ the set of all subsets of $\kappa$ having cardinality $\mu$:

Call a family $\mathcal{F} \subset [\kappa]^\omega$ sparse if for all $\mathcal{G} \in [\mathcal{F}]^{\omega_1}$ the set $\bigcup \mathcal{G}$ is uncountable.

This is a pretty natural notion that has been rediscovered by various authors and is the combinatorial core of various seemingly unrelated problems in topology, analysis and algebra (see the list of references at the end).

Getting a sparse family isn't that much of a deal. The problem is often getting a big one, or more precisely, a cofinal one, with respect to containment.

It's easy to see that there is a sparse cofinal family in $([\aleph_n]^\omega, \subseteq)$), for every $n<\omega$. This is a simple consequence of the fact that $cf([\aleph_n]^\omega, \subseteq)=\aleph_n$.

QUESTION: Is there (in ZFC) a cardinal $\kappa$ such that $cf([\kappa]^\omega, \subseteq) > \kappa$ and there is a sparse cofinal family on $[\kappa]^\omega$?

This cardinal, if it exists, must be larger than $\aleph_\omega$, because the existence of a sparse cofinal family of countable subsets of $\aleph_\omega$ can be proved to be independent of ZFC, modulo very large cardinals. Indeed (see Todorcevic's book or Blass's article for the proofs):

  • If the Chang's Conjecture variant $(\aleph_{\omega+1}, \aleph_\omega) \twoheadrightarrow (\aleph_1, \aleph_0)$ holds then there is no sparse cofinal family of countable subsets of $\aleph_\omega$. The consistency of this Chang's Conjecture variant has been proven from (slightly less than) a 2-huge cardinal by Levinski, Magidor and Shelah.

  • If $\square_{\aleph_\omega}+cf([\aleph_\omega]^\omega, \subseteq)=\aleph_{\omega+1}$ holds then there is a sparse cofinal family of subsets of $\aleph_\omega$.


  1. Blass, Andreas, On the divisible parts of quotient groups, Göbel, Rüdiger (ed.) et al., Abelian group theory and related topics. Conference, August 1-7, 1993, Oberwolfach, Germany. Providence, RI: American Mathematical Society. Contemp. Math. 171, 37-50 (1994). ZBL0823.20057.
  2. Kojman, Menachem; Milovich, David; Spadaro, Santi, Noetherian type in topological products, Isr. J. Math. 202, 195-225 (2014). ZBL1302.54008.
  3. Peter Nyikos, Generalized Kurepa and MAD families and Topology, preprint.

  4. Spadaro, Santi, On two topological cardinal invariants of an order-theoretic flavour, Ann. Pure Appl. Logic 163, No. 12, 1865-1871 (2012). ZBL1269.03047.

  5. Todorcevic, Stevo, Walks on ordinals and their characteristics, Progress in Mathematics 263. Basel: Birkhäuser (ISBN 978-3-7643-8528-6/hbk). vi, 324 p. (2007). ZBL1148.03004.

  • $\begingroup$ Does $\mathsf{GCH}+\square$ (or some similar hypothesis) imply that $([\kappa]^\omega,\subseteq)$ has a sparse cofinal family for every $\kappa$? $\endgroup$
    – Will Brian
    Feb 4, 2020 at 13:45
  • 1
    $\begingroup$ OK, I think the answer is yes. You can prove it fairly quickly using the existence of high Davies trees, which are defined in the Soukups' paper arxiv.org/abs/1705.06195, and whose existence follows from $\mathsf{GCH}+\square$. $\endgroup$
    – Will Brian
    Feb 4, 2020 at 13:55
  • $\begingroup$ Yes, it does, and if $\kappa$ has uncountable cofinality, you don't even need $\square$. $\endgroup$ Feb 4, 2020 at 14:14

1 Answer 1


Your problem is related to the generalized Chang conjectures of the form $(\kappa^+,\kappa) \twoheadrightarrow (\aleph_1,\aleph_0)$.

Observation: If $(\kappa^+,\kappa) \twoheadrightarrow (\aleph_1,\aleph_0)$ holds, and if $\mathrm{cf}([\kappa]^{\omega},\subseteq) > \kappa$, then $([\kappa]^{\omega},\subseteq)$ has no sparse cofinal family.

So if we'd like to answer your question in the negative (or restrict our search for a positive answer), we should ask when it's possible to have lots of instances of $(\kappa^+,\kappa) \twoheadrightarrow (\aleph_1,\aleph_0)$ holding simultaneously.

A paper relevant to this question is "Global Chang's Conjecture and singular cardinals" by Monroe Eskew and Yair Hayut (available here). In it, they articulate the following conjecture:

Singular Global Chang's Conjecture: For all infinite $\mu < \kappa$ of the same cofinality, $(\kappa^+,\kappa) \twoheadrightarrow (\mu^+,\mu)$.

The word "conjecture" is used here to allude to Chang's Conjecture. No one is suggesting the SGCC as a potential theorem of $\mathsf{ZFC}$, and in fact it's not hard to show that the negation of the SGCC is consistent with $\mathsf{ZFC}$. The suggestion is that the SGCC might be consistent with $\mathsf{ZFC}$, relative to large cardinals. My point is that if this conjecture is consistent, then the answer to your question is no. Based on the observation above,

If this conjecture holds, then there are no cardinals $\kappa$ with the property described in your question. In fact, SGCC implies that the answer to your question is no even when $\omega$ is replaced by an arbitrary infinite cardinal $\mu$.

So the conjecture gives a conditional answer to your question.

In the same paper (Theorem 32), the authors show that a fragment of this conjecture is consistent. This fragment is enough to show, using the observation above, that

If $\alpha < \omega_1$, then it is consistent (relative to two supercompact cardinals) that the answer to your question is no for every $\kappa < \aleph_\alpha$. That is, it is consistent that if $\kappa < \aleph_\alpha$ and $\mathrm{cf}([\kappa]^{\omega},\subseteq) > \kappa$, then $[\kappa]^\omega$ has no sparse cofinal family.

A proof of the observation above:

Recall that $(\kappa^+,\kappa) \twoheadrightarrow (\mu^+,\mu)$ is an abbreviation for the following statement:

$\bullet \ \ $ For every model $M$ for a countable language $\mathcal L$ that contains a predicate $A \subseteq M$, if $|M| = \kappa^+$ and $|A| = \kappa$ then there is an elementary submodel $M'$ of $M$ such that $|M'| = \mu^+$ and $|M' \cap A| = \mu$.

To prove the observation above, suppose $(\kappa^+,\kappa) \twoheadrightarrow (\aleph_1,\aleph_0)$, and suppose that $\mathrm{cf}([\kappa]^{\omega},\subseteq) > \kappa$.

Let $\mathcal C$ be any cofinal family in $[\kappa]^\omega$. Using the Lowenheim-Skolem theorem in the usual way, let $M$ be an elementary submodel of $H(\theta)$ (for some sufficiently big regular cardinal $\theta$) such that $\mathcal C \in M$, $\kappa \subseteq M$, and $|M| = |M \cap \mathcal C| = \kappa^+$. Let $\phi$ be a bijection $M \rightarrow \mathcal C \cap M$. (Note that this $\phi$ cannot be a member of $M$.)

Consider the language consisting of a single relation symbol and a single function, and consider the model $$(M,\in,\phi,\kappa)$$ in this language. Applying our hypothesis $(\kappa^+,\kappa) \twoheadrightarrow (\aleph_1,\aleph_0)$, there is some $M' \subset M$ such that $(M',\in,\phi,\kappa \cap M')$ is an elementary substructure of $(M,\in,\phi,\kappa)$ with $|M'| = \aleph_1$ and $|M' \cap \kappa| = \aleph_0$.

Let $\mathcal C' = M' \cap \mathcal C$. The restriction of $\phi$ to $M'$ is (by elementarity) a bijection $M' \rightarrow \mathcal C'$. Therefore $|\mathcal C'| = |M'| = \aleph_1$.

If $X \in \mathcal C'$, then $X$ is a countable set in $M'$. Because $M'$ is a model of (enough of) $\mathsf{ZFC}$, every countable set in $M'$ is a subset of $M'$. So $X \subseteq M'$. But $X \subseteq \kappa$ as well, so every $X \in \mathcal C'$ is a subset of $M' \cap \kappa$.

Therefore $\bigcup \mathcal C' = M' \cap \kappa$. Because $|M' \cap \kappa| = \aleph_0$, this means that $\mathcal C'$ is an uncountable subset of $\mathcal C$ whose union is countable. So $\mathcal C$ is not sparse.

Because $\mathcal C$ was an arbitrary cofinal family in $[\kappa]^\omega$, this shows that $[\kappa]^\omega$ does not admit a sparse cofinal family.

  • $\begingroup$ Thanks. Obviously I knew about your Observation above (see my remark about Chang's Conjecture for $\aleph_\omega$ negating sparse cofinal families of countable subsets of $\aleph_\omega$), but I wasn't aware of the paper by Eskew and Hayut. $\endgroup$ Feb 12, 2020 at 14:06

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