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 $\kappa$.



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> **Problem.** What is known about the cofinality of the poset $[\kappa]^{<\kappa}$ for a singular cardinal $\kappa$?
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> Can it be equal to $\kappa$ for some special singular cardinals $\kappa$?
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> What is the cofinality of the poset $[\kappa]^{<\kappa}$ for $\kappa=\aleph\_\omega$?
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I hope that the answers should be known to specialists but I cannot find anything relevant in the books.




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**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](https://mathoverflow.net/questions/313402/a-possible-characterization-of-regular-cardinals)) should be known. Does anybody know a suitable reference?



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




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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.


