Let ${\mathcal A}$ be the family of subsets $A$ of the natural numbers ${\mathbf N}$ which are co-infinite (i.e., their complement is infinite). We partially order this family by set inclusion. A [cofinal](https://en.wikipedia.org/wiki/Cofinal_(mathematics)) subset of ${\mathcal A}$ is a subcollection ${\mathcal A}'$ such that every $A$ in ${\mathcal A}$ is contained in some $A' \in {\mathcal A}'$. The [cofinality](https://en.wikipedia.org/wiki/Cofinality) of ${\mathcal A}$ is the minimum cardinality of a cofinal subset ${\mathcal A}'$ of ${\mathcal A}$. ${}{}{}$


An easy application of the Cantor diagonal argument shows that the cofinality of ${\mathcal A}$ is uncountable, and the cofinality is of course dominated by the cardinality of the continuum; thus on the continuum hypothesis the cofinality is equal to the cardinality of the continuum. In general, what are the possible values of this cofinality?


Many years ago I [asked a similar question](https://mathoverflow.net/questions/29624/how-many-orders-of-infinity-are-there) about $\omega^{\omega}$, but the poset ${\mathcal A}$ seems to have a rather different structure (it is not closed under joins, for instance), and so the answer to this question may be rather different.



What is the cofinality of the co-infinite subsets of ${\bf N}$?
Every such cofinal family $\mathcal{A}'$ must have size continuum. The reason is that there is an almost disjoint family $\mathcal{D}$ of size continuum, a family of infinite co-infinite sets $A\subseteq\mathbb{N}$ for which any two have finite intersection. To construct such an almost disjoint family, label the nodes of the infinite binary tree with distinct natural numbers, and take the sets of labels arising along any branch of the tree. This tree has continuum many branches, and any two of them have finite intersection in the tree, so the family of label sets will be almost disjoint.


Now, if we have an almost disjoint family $\mathcal{D}$ of size continuum, your dominating family $\mathcal{A}'$ will have to contain covers of the complements of these sets, that is, covering $\mathbb{N}-A$ for every $A\in \mathcal{D}$. But if a set $X$ covers the complement of $A$, then the complement of $X$ is contained in $A$. Therefore no coinfinite set $X$ can cover the complements of two different $A,B\in\mathcal{D}$, since the complement of $X$ would be contained in $A\cap B$, which is finite. So you will need a different covering set for every $\mathbb{N}-A$ for $A\in \mathcal{D}$. Thus, you will need continuum many sets in $\mathcal{A}'$.


