Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
What if $P$ has no maximal elements? (Take, for example, $\mathbb{Z}$ or $\mathbb{R}$ with the usual order.) – Qiaochu Yuan Jan 30 '12 at 1:42
@QiaochuYuan: Thanks! If a poset does not have a maximal element, what are its cofinals then? Are they the unions of all upper subsets? Does the poset have a cofinality? – Tim Jan 30 '12 at 1:45
Any set not bounded above is cofinal in $\mathbb{Z}$ and $\mathbb{R}$. – Michael Greinecker Jan 30 '12 at 1:46
It should include all maximal elements and be unbounded in every chain. I don't think one can say much more. – Michael Greinecker Jan 30 '12 at 1:51
If it has no maximal element, it is infinite and must have at least countable cofinality. There is no upper bound on how large such a set can be in general. See Proposition 7.2. at – Michael Greinecker Jan 30 '12 at 2:25
up vote 5 down vote accepted

As Michael Greinecker pointed out in the comments, a subset of $P$ is cofinal iff it contains all maximal elements of $P$ and is unbounded (cofinal) in every chain in $P$.

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 $\mathbb{R}^2$.

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 $m\le n$. 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.