For a cardinal $\kappa$ let $[\kappa]^{<\kappa}$ denote the family of subsets of cardinality $<\kappa$ in $\kappa$. The family $[\kappa]^{<\kappa}$ is endowed with the partial order of inclusion. A simple diagonal argument shows that for any infinite cardinal $\kappa$ the poset $[\kappa]^{<\kappa}$ has cofinality $\ge\kappa$. It is easy to see that for a regular cardinal $\kappa$ this is an equality: the poset $[\kappa]^{<\kappa}$ has cofinality $\kappa$.

Problem. What is known about the cofinality of the poset $[\kappa]^{<\kappa}$ for a singular cardinal $\kappa$?

Can it be equal to $\kappa$ for some special singular cardinals $\kappa$?

What is the cofinality of the poset $[\kappa]^{<\kappa}$ for $\kappa=\aleph_\omega$?

I hope that the answers should be known to specialists but I cannot find anything relevant in the books.

Added in Edit: In this answer @YCor proved that for any singular cardinal $\kappa$ the poset $[\kappa]^{<\kappa}$ has cofinality $>\kappa$. From this fact we can conclude that a cardinal $\kappa$ is regular if and only if $\mathrm{cof}([\kappa]^{<\kappa})=\kappa$. I suspect that this characterization (answering this MO-question) should be known. Does anybody know a suitable reference?

  • $\begingroup$ @YCor No, it is (more-or-less) obvious that $\mathrm{cof}([\kappa]^{<\kappa})\ge\mathrm{cf}(\kappa)$. But I doubt that $\mathrm{cof}([\kappa]^{<\kappa})=\mathrm{cf}(\kappa)$ is true for all singular $\kappa$. Let us wait for experts. $\endgroup$ Oct 21, 2018 at 10:41
  • $\begingroup$ @YCor I have just realized that a simple diagonal argument shows that $\mathrm{cof}([\kappa]^{<\kappa})\ge\kappa$, so my second question always has negative answer. Now I will make corresponding changes in the problem. $\endgroup$ Oct 21, 2018 at 10:55

2 Answers 2


If $\kappa$ is singular, the cofinality of the poset $[\kappa]^{<\kappa}$ is $>\kappa$.

Indeed, $\kappa$ singular means that there is a limit ordinal $\alpha<\kappa$ and an increasing family $(\lambda_\xi)_{\xi<\alpha}$ such that $\lambda_\xi<\kappa$ for each $\xi$ and $\sup_{\xi<\alpha}\lambda_\xi=\kappa$.

Suppose by contradiction that $[\kappa]^{<\kappa}$ has cofinality $\le\kappa$. So there is a family of subsets $(A_i)_{i<\kappa}$, with $A_i\subset\kappa$, $|A_i|<\kappa$ for all $i<\kappa$, and such that for every $A\subset\kappa$ such that $|A|<\kappa$ there exists $i<\kappa$ such that $A\subset A_i$.

For $\xi<\alpha$ define $$B_\xi=\bigcup_{i\le \lambda_\xi,|A_i|\le\lambda_\xi}A_i.$$

For each $B_\xi$ is union of $\le\lambda_\xi$ sets of cardinal $\le\lambda_\xi$, hence has cardinal $<\kappa$. (This is an adjustment of Mohammad Golshani's initial argument: we have to restrict the cardinal in the union to ensure that $B_\xi$ has cardinal $<\kappa$.)

Then $(B_\xi)_{\xi<\alpha}$ is cofinal in $[\kappa]^{<\kappa}$. If $A\subset\kappa$ and $|A|<\kappa$, there exists $j<\kappa$ such that $A\subset A_i$. There exists $\eta<\kappa$ such that $j\le\lambda_\eta$. In turn, there is $\xi$ with $\eta\le\xi<\kappa$ such that $|A_{\lambda_\eta}|\le\lambda_\xi$. Then $A\subset A_j\subset A_{\lambda_\eta}$. Also we have $\max(\lambda_\eta,|A_{\lambda_\eta}|)\le\lambda_\xi$, so $A_{\lambda_\eta}\subset B_\xi$. Hence $A\subset B_\xi$.

This proves that $[\kappa]^{<\kappa}$ has cofinality $<\kappa$, which is a contradiction to the obvious diagonal argument.

The idea of the proof is that the cofinality of a poset "shouldn't" be a singular cardinal; nevertheless this does not work under further assumptions. For instance, if $\kappa$ is any infinite cardinal, then $[\kappa]^{<\aleph_0}$ has cofinality (and cardinal) $\kappa$, which can be singular.

  • $\begingroup$ Thank you for the answer. In the first line it is better write that the cofinality of $[\kappa]^{<\kappa}$ is $>\kappa$ (not just cardinality). $\endgroup$ Oct 22, 2018 at 7:47
  • $\begingroup$ By the way, your answer resolves also the other my MO-problem: mathoverflow.net/questions/313402/… $\endgroup$ Oct 22, 2018 at 7:49

In general, one can prove the following:

1) if $2^{< \kappa} \leq \lambda,$ then $\mathrm{cf}([\lambda]^{< \kappa})=\lambda^{<\kappa}$.

2) If $cf(\lambda) < cf(\kappa),$ then $\mathrm{cf}([\lambda]^{< \kappa})\geq \lambda^+$. Furthermore if $0^\sharp$ does not exist, then we have the equality.

Proof of (1). If $\lambda^{<\kappa}=\lambda,$ this is clear, so suppose $\lambda < \lambda^{<\kappa}$ and suppose by contradiction that $cf([\lambda]^{< \kappa}) < \lambda^{<\kappa}$, as witnessed by a set $X$.

Then for each $x \in X,$ there are at most $2^{|x|} \leq \lambda$ many subsets of $x$, and so

$\lambda^{< \kappa} = |[\lambda]^{< \kappa}|\leq |X| \cdot \lambda < \lambda^{< \kappa}$, which is impossible.

Proof of (2). Suppose by contradiction that $\{x_\alpha : \alpha < \lambda \}$ is cofinal in $[\lambda]^{< \kappa}$. Let also $(\lambda_\xi: \xi < \mathrm{cf}(\lambda))$ be increasing and cofinal in $\lambda$. For each $\xi< \mathrm{cf}(\lambda)$ pick some $y_\xi \in [\lambda]^{< \kappa}$ which is not covered by $\{x_\alpha : \alpha < \lambda_\xi\}$. Then $y= \bigcup_{\xi < \mathrm{cf}(\lambda)}y_\xi \in [\lambda]^{< \kappa}$ and is not covered by $\{x_\alpha : \alpha < \lambda \}$. A contradiction.

If $0^\sharp$ does not exist, one can use Jensen's covering lemma to get the equality.

In particular, it follows from (1) that if $\kappa$ is a singular strong limit cardinal, then $\mathrm{cf}([\kappa]^{< \kappa})=\kappa^{<\kappa} > \kappa.$.

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    $\begingroup$ You can see the definition at zero-sharp. what is used here is Jensen's covering lemma $\endgroup$ Oct 21, 2018 at 12:18
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    $\begingroup$ @MohammadGolshani It is not clear to me why the set $y$ has cardinality $<\kappa$. In order to have $|y|<\kappa$, we need to assume that $\mathrm{cf}(\lambda)<\mathrm{cf}(\kappa)$ by we have merely that $\mathrm{cf}(\lambda)<\kappa$. So, the proof (and maybe also the formulation) of (2) should be adjusted. $\endgroup$ Oct 21, 2018 at 14:11

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