Cofinal subset and maximal elements in a poset

Question

Let $(P;\leq)$ be a poset. A subset $A$ of $P$ is said to be cofinal in $P$ if for every $x$ in $P$ there is a $y$ in $A$ such that $x \leq y $.

I was wondering if it is true that a subset of $P$ is cofinal iff it contains all maximal elements in $P$? This is how I understand cofinal, but I am afraid that this statement might miss something.

Thanks and regards!

Answer 1

Slightly revised and corrected 11 November 2023.

A subset $C$ of $P$ is cofinal in $P$ iff for each $x\in P$ there is a $y\in C$ such that $x\le y$. If $x$ is maximal in $P$, then the only $y\in P$ such that $x\le y$ is $y=x$, so a cofinal subset of $P$ must contain every maximal element of $P$. However, a subset of $P$ can contain every maximal element of $P$ without being cofinal.

A family of examples that I have found useful in thinking about such things consists of suborders of the partial order $\langle P,\le\rangle$ given by $P=\mathbb{R}^2$ and $\langle x_0,y_0\rangle\le\langle x_1,y_1\rangle$ iff $x_0\le x_y$ and $y_0\le y_1$. Clearly a subset of $P$ is cofinal iff it is unbounded to the northeast, so to speak.

As an example of what you can get by looking at suborders of $P$, let $$P_0=[0,1)^2\cup\{\langle 1,0\rangle,\langle 0,1\rangle\}\;.$$ The points $\langle 1,0\rangle$ and $\langle 0,1\rangle$ are maximal in $P_0$, so they must belong to any cofinal subset of $P_0$, but they clearly aren’t enough: for any $\langle x,y\rangle\in(0,1)^2$, $\langle x,y\rangle\not\le\langle 0,1\rangle$ and $\langle x,y\rangle\not\le\langle 1,0\rangle$. It’s not hard to see that $A\subseteq P_0$ is cofinal in $P_0$ iff $\{\langle 1,0\rangle,\langle 0,1\rangle\}\subseteq A$ and $A$ contains a sequence converging to $\langle 1,1\rangle$ in $\mathbb{R}^2$.

Note that this also shows that Michael Greinecker’s characterization in the original comments is incorrect: as PatrickR points out in a comment below,

$$\left(\left[0,\frac12\right)\times\{0\}\right)\cup\left(\left\{\frac12\right\}\times[0,1)\right)$$

is a maximal chain in $P_0$ that is disjoint from many cofinal subsets of $P_0$, e.g., the cofinal set $\left\{\langle x,x\rangle\in P_0:\frac23\le x<1\right\}$.

Here’s an easy way to see that a poset with no maximal elements can have any infinite cofinality. Let $\kappa$ be any infinite cardinal, let $P=\kappa\times\mathbb{N}$, and define the order $\preceq$ by $\langle \alpha,m\rangle\preceq\langle \beta,n\rangle$ iff $\alpha=\beta$ and $m\le n$. Clearly any cofinal subset of $P$ must be cofinal in each copy of $\mathbb{N}$, so every cofinal subset of $P$ must have cardinality $\kappa\cdot\omega=\kappa$. If the poset is linearly ordered, however, its cofinality must be a regular cardinal. Thus, for example, you can have a poset whose cofinality is $\omega_\omega$, but it can’t be linearly ordered.