What is the cofinality of the co-infinite subsets of ${\bf N}$?

Question

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 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 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 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.

REMARK: Let $\ \Gamma\ $ be the family of all co-infinite sets in $\ \mathbb N\ $ (I'm avoiding \mathcal). Then there is an explicit map $\ \gamma:\Gamma\to(0;1]\ $ such that $\ \gamma(Q)>\gamma(S)\ $ for every $\ Q\subset S\ $ (sharp inclusion, not $\subseteq),\ $ namely $\ \gamma(S)\ :=\ \sum_{n\in\mathbb N\setminus S}\ 2^{-n}.$ — Wlod AA, Feb 11 '23 at 18:49
@WlodAA That map does not interact well with cofinality, since a countable family of sets is never cofinal in the order, but the image under $\gamma$ could converge to 0. But also, why turn it upside down? I would find it more natural to use $\delta(S)=\sum_{n\in S} 1/2^n$, so that $Q\subset S\to \delta(Q)< \delta(S)$. But still, this maps sends non-cofinal families in the order to cofinal sets in the interval $[0,1)$. — Joel David Hamkins, Feb 14 '23 at 12:48
This is not so constructive as should be to insert [Math Processing Error] into the question. Fro example reference to the disambiguation page for Collection will help in the case of subcollection much more. In the disambiguation set and multiset are both included as synonmys for subcollection. But every A in A is ... is not well deductive math style as well. — Steffen Jaeschke, Feb 18 '23 at 15:07
Instead of revise the question reference to answers like Cofinality. The short definition over there at wikepedia is "In mathematics, especially in order theory, the cofinality cf(A) of a partially ordered set A is the least of the cardinalities of the cofinal subsets of A." Least opens up many understandings google: least — Steffen Jaeschke, Feb 18 '23 at 15:20
From the already cited source Cofinality the answer is: A subset of the natural numbers $\mathbb {N}$ is cofinal in $\mathbb {N}$ if and only if it is infinite, and therefore the cofinality of $\aleph _{0}$ is $\aleph _{0}$. Thus $\aleph _{0} $is a regular cardinal. — Steffen Jaeschke, Feb 18 '23 at 15:27
@TerryTao sorry to ask here I made some progress on the a great conjecture see https://math.stackexchange.com/questions/4783557/spiderman-and-the-cantors-function and https://math.stackexchange.com/questions/4789813/prove-magnitude-of-zeros-of-polynomial-le-frac-sqrt512/4790443#4790443 — DesmosTutu, Oct 21 '23 at 09:36

Joel David Hamkins · Accepted Answer · 2023-02-11T15:01:03.353

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 $\mathcal{A}'$.

Thanks! This will allow the final result (Corollary 5.2) in a recent paper https://arxiv.org/abs/2301.10303 of Tamar Ziegler and myself to be sharpened (from "uncountable" to "cardinality of the continuum"). — Terry Tao, Feb 11 '23 at 15:21
Glad to hear it. The same argument works with almost inclusion instead of full inclusion. — Joel David Hamkins, Feb 11 '23 at 15:29

What is the cofinality of the co-infinite subsets of ${\bf N}$?

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