Cofinal under inclusion finite subsets of $\mathbb{N}$

Question

I have come across the notion of a cofinal under inclusion collection of finite subsets of $\mathbb{N}$. I think I am not understandinf what cofinal means, because I don't see how much an object can exist, or what it contains.

From Wikipedia article on cofinality, for $A$ and a set with binary relation $\leq$, we say $\subset A$ is cofinal in $A$ if for every $a\in A, \exists b\in B $ such that $a\leq b$.

Now when one considers the collection of finite subsets of $\mathbb{N}$, then what would be the 'cofinal under inclusion' set of this. Clearly, any finite subset is contained in a larger one. So we would have to include the larger one in our cofinal set, and could ditch the smaller one. But then if we keep going like this, we are pushing back to subsets of $\mathbb{N}$ of infinite cardinality, which are not in the set of finite subsets at all.

Answer 1

Basically, there's a type confusion here.

You're right that no individual finite set is cofinal in this sense, but what's being looked at is infinite collections of finite sets. Those totally can be cofinal - e.g. consider the infinite collection of finite sets $\mathfrak{I}=\{\{0,1,2,...,n\}: n\in\mathbb{N}\}$. Clearly every finite set of natural numbers is contained in some element of $\mathfrak{I}$. Also clearly no individual member of $\mathfrak{I}$ (or rather, no $\{X\}$ for $X\in\mathfrak{I}$) is cofinal.

They can also fail to be cofinal. E.g. $\mathfrak{K}=\{\{1,2,3,4,...,n\}:n\in\mathbb{N}\}$ is not cofinal since $\{0\}$ is not a subset of any element of $\mathfrak{K}$. Preempting another possible type confusion, note that $$\mathfrak{M}=\{\{0\},\{1\},\{2\},\{3\}, ...\}$$ is not cofinal either: although every finite set is a subset of the union of some elements of $\mathfrak{M}$, the set $\{1,2\}$ (for example) is not a subset of any single element of $\mathfrak{M}$.

Answer 2

A collection $\mathscr{A}$ of finite subsets of $\Bbb N$ is cofinal with respect to inclusion if it has the following property: for each finite $F\subseteq\Bbb N$ there is an $A\in\mathscr{A}$ such that $F\subseteq A$. For example, let

$$\mathscr{E}=\{E\subseteq\Bbb N:E\text{ is finite and }|E|\text{ is even}\}\;,$$

the set of finite subsets of $\Bbb N$ of even cardinality; then $\mathscr{E}$ is cofinal with respect to inclusion. To see this, suppose that $F$ is a finite subset of $\Bbb N$, say $F=\{k_1,k_2,\ldots,k_n\}$. If $n$ is even, then $F\subseteq F\in\mathscr{E}$. If $n$ is odd, let $\ell=1+\max F$, and let $E=F\cup\{\ell\}$. Clearly $\ell\notin F$, so $|E|=n+1$ is even, and hence $F\subseteq E\in\mathscr{E}$. In every case, therefore, there is an $E\in\mathscr{E}$ such that $F\subseteq E$, and that is precisely what is needed in order for $\mathscr{E}$ to be cofinal with respect to inclusion in the family of finite subsets of $\Bbb N$.

The family of finite subsets of $\Bbb N$ of odd cardinality and the family of finite subsets of $\Bbb N$ with prime cardinality (among many others) are also cofinal with respect to inclusion. Yet another example is the family of proper initial segments of $\Bbb N$: if we let $S_n=\{0,1,\ldots,n\}$ for each $n\in\Bbb N$, then $\{S_n:n\in\Bbb N\}$ is cofinal with respect to inclusion: if $F$ is a finite subset of $\Bbb N$, it has a maximum element $m(F)$, and clearly $F\subseteq S_{m(F)}$.

Note that it is the collection of sets that is cofinal in the family of finite subsets of $\Bbb N$, not the individual members of the collection.