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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!

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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
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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
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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 math.columbia.edu/algebraic_geometry/stacks-git/sets.pdf –  Michael Greinecker Jan 30 '12 at 2:25

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up vote 4 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.

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