A *covering* of a non-empty set $X$ is a collection ${\cal U} \subseteq ({\cal P}(X)\setminus\{\emptyset\})$ such that $\bigcup {\cal U} = X$. If ${\cal U}$ is a covering of $X$ then a function $f:{\cal U}\to X$ is called a *choice function* if $f(A)\in A$ for all $A\in {\cal U}$. A *marriage* is an injective choice function.


We say that ${\cal U}$ is *critial* if


1. there is a marriage $f:{\cal U}\to X$, and
2. every marriage $f:{\cal U}\to X$ is surjective.


Is there a critical covering of $\omega$ consisting of infinite sets only? 



Critical coverings of $\omega$
The answer is no.


If $\mathcal{U}$ is a covering of $\omega$ such that a marriage exists, then $\mathcal{U}$ is countable. Set $\mathcal{U}=\{U\_i: i\in\omega\}$, and suppose each $U\_i$ is infinite. Then (for $k$ fixed) let


* $a\_0$ be the least element of $U\_0$ greater than $k$, and
* $a\_{n+1}$ be the least element of $U\_{n+1}$ which is greater than $a\_n$.


The map $U\_i\mapsto a\_i$ is a marriage whose range does not intersect $\{0, 1, . . ., k\}$, and in particular is not surjective.


A similar trick can be done to avoid an infinite set of elements: replace "least" with "second least" in the above.


