Cofinal subset and maximal elements in a poset 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!
 A: 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.
A: It seems to me that the answer and the comment are rather imprecise - one also needs to properly define what it means for a set $D \subseteq P$ to be undounded (cofinal) in a chain $L \subseteq P$. For instance, if we define a chain as a liearly ordered subset of a poset, then obviously every $\{p\}$ is a chain, therefore the claim that a set $D$ is cofinal in every chain becomes vacuous. In particular, containing every maximal element of $P$ follows from the condition of being cofinal in every chain.  
Let's define the following notion of domination:
Let $(P, \leq)$ be a poset, and let $D \subseteq P$, $X \subseteq P$ be subsets of the poset.
We will say that $D$ dominates $X$, if:
$$\forall x \in X \: \exists d \in D \: x \leq d.$$
Now the following equivalence holds for any $D \subseteq P$:
$D$ is cofinal in $P$ iff $D$ dominates every maximal chain in $P$. 
Proof.
If $D$ is cofinal, then by definition it dominates every subset of the poset. 
If $D$ dominates every maximal chain, let $p \in P$. By the Hausdorff Maximal Principle there is a maximal chain $L \ni p$. $D$ dominates $L$, and thus there is $d \in D$ s.t. $p \leq d$. $\Box$
Now a remark: $D$ dominating every maximal chain is equivlaent to the conjunction of the following two conditions: $D$ containing every maximal element of $P$ and $D$ dominating every chain of $P$ without the greatest element and this is probably what the comment and the answer meant. 
Further, one cannot replace the definition of domination with a weaker condition (that could be read as: $D$ is unbounded over $X$:
$$ \forall x \in X \: \exists d \in D \: d \not\leq x.$$
For a counterexample define $$P = \{(0,n): n \in \omega\} \cup \{(1, \pi)\}$$
and let the ordering be given as follows:
$$(i, j) \leq (k,l) \text{   iff   }  (i=k \: \wedge j \leq l).$$
The set $D:=\{(0,n): n \in \omega\}$ is unbounded over every maximal chain, but it is not cofinal, since there is no element $d \in D$ that would be greater than $(1, \pi)$.
Also, one cannot require the condition of domination to be restricted to $X$ itself (hence the ambiguity of the expression "$D$ is unbounded (cofinal) in a chain"), i .e. we could not say that $D$ dominates $X$ iff:
$$\forall x \in X \: \exists d \in D \cap X \: x \leq d,$$
since then look at the following example: let
$$ P = \{(0, n): n \in \omega\} \cup \{(1, k): k \in \omega\}$$
and define:
$$(i,n) \leq (j, k) \text{   iff   }  (i=j=0 \: \wedge n \leq k) \: \vee \: (i<j \: \wedge \: n \leq k ).$$
Now the set $\{(1, k): k \in \omega\}$ is clearly cofinal in $P$ (and contains all the maximal elements of $P$), but it is not unbounded (cofinal) in every (maximal) chain, since it does not have any element in a maximal chain $L = \{(0,n): n \in \omega)\}$ (also, it obviously dominates every maximal chain).
