# The cofinality of the poset $[\kappa]^{<\kappa}$ for a singular cardinal $\kappa$

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?

• @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. – Taras Banakh Oct 21 '18 at 10:41
• @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. – Taras Banakh Oct 21 '18 at 10:55

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.

• Thank you for the answer. In the first line it is better write that the cofinality of $[\kappa]^{<\kappa}$ is $>\kappa$ (not just cardinality). – Taras Banakh Oct 22 '18 at 7:47
• By the way, your answer resolves also the other my MO-problem: mathoverflow.net/questions/313402/… – Taras Banakh Oct 22 '18 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 $$\mathrm{cf}([\kappa]^{< \kappa})=\kappa^{<\kappa} > \kappa.$$.

• You can see the definition at zero-sharp. what is used here is Jensen's covering lemma – Mohammad Golshani Oct 21 '18 at 12:18
• @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. – Taras Banakh Oct 21 '18 at 14:11