Let $(X, \le)$ be a partially ordered set. We call a subset $S \subseteq X$...


* ... a *chain* if each two elements in $S$ are comparable with respect to $\le$ (in other words, $S$ is linearly ordered with respect to $\le$).
* ... *directed* if for all $x,y \in S$ there exists $z \in S$ that dominates $x$ and $y$.


Obviously, every chain is directed.


**Question.** Assume that every chain in $X$ has a supremum. Does it follow that every directed set in $X$ has a supremum?


**Remarks.**


(1) By Zorn's lemma, $X$ has a maximal element (in fact, every element of $X$ is dominated by a maximal element of $X$).


(2) Since the empty set is a chain and thus has a supremum, it follows that $X$ has a smallest element (though this doesn't seem to be particularly relevant to the question).


(3) Let $D \subseteq X$ be directed. We cannot apply Zorn's lemma directedly to $D$ since the supremum of a chain in $D$ might not be in $D$. What we can do is to add the set of all supremuma of subsets of $D$ (whenever they exist) to $D$, and thus obtain a new set $\tilde D$. Then $\tilde D$ is closed with respect to taking suprema, but I cannot see if (and why) $\tilde D$ is directed.


Actually, the answer to the question is *yes* if and only if this set $\tilde D$ is always directed: the implication "$\Rightarrow$" is trivial, and the implication "$\Leftarrow$" follows from applying Zorn's lemma to $\tilde D$ and from the fact that a maximal element in a directed set is always the supremum of this set.


(4) In general, a directed set does not necessarily contain a co-final chain. For instance, let $\mathcal{F}$ denote the set of all finite subsets of $\mathbb{R}$, ordered by set inclusion. Obviously, $\mathcal{F}$ is directed; but every union of a chain of finite sets if at most countable, so $\mathcal{F}$ does not contain a co-final chain.


(5) Let $D \subseteq X$ be directed. We can apply Zorn's lemma to the set $\mathcal{D}$ of all directed subsets of $D$ that have a supremum in $X$, or to the set $\mathcal{S}$ of all subsets of $D$ that have a supremum in $X$; so $\mathcal{D}$ and $\mathcal{S}$ both have a maximal element $D\_{\max}$ and $S\_{\max}$, respectively. But I see no way to show that $D\_{\max}$ or $S\_{\max}$ is co-final in $D$ (and thus equal to $D$).


(6) If $X$ is a *lattice* (i.e., every (non-empty) finite subset of $X$ has a supremum), then the answer to the question is *yes*: Indeed, let $D \subseteq X$ be directed, and let $\mathcal{S}$ denote the set of all subsets of $D$ that have a supremum in $X$. Then $\mathcal{S}$ contains all finite subsets of $D$, and $\mathcal{S}$ is stable with respect to monotone unions (i.e., unions of chains). This implies that $\mathcal{S}$ equals the power set of $D$, so in particular, $D \in \mathcal{S}$.


**Motivation.** In a preprint of mine I briefly considered a similar question in the context of ordered vector spaces, and I remarked that I do not know the answer in this specific vector space setting. Now, I'm about to submit a revision of this preprint, and I noted that I do not even know the answer for general partially ordered sets (without any vector space structure).



Suprema of directed sets
Yes, a poset that has suprema of all chains also has suprema of all directed sets. This is known, and I vaguely recall seeing it attributed to Solovay. The proof consists of showing, by induction on cardinals $\kappa$, that having suprema of all chains implies having suprema for all directed set of size $\leq\kappa$.


The case of finite $\kappa$ is trivial, since a finite directed set has a top element. The case of $\kappa=\aleph\_0$ is almost trivial, since a countable directed set contains a cofinal chain. So assume from now on that $\kappa$ is uncountable and that $D$ is a directed subset of $X$ with $|D|=\kappa$.


Write $D$ as the union of an increasing (with respect to $\subseteq$) transfinite sequence of subsets $D\_i$, each of cardinality $<\kappa$. (If $\kappa$ is a regular cardinal, then this sequence necessarily has length $\kappa$, but if $\kappa$ is singular then the sequence can be shorter.) We may assume that each $D\_i$ is also directed; just choose for each pair in $D$ an upper bound in $D$, and close each $D\_i$ under the "chosen upper bound" function. (The cardinality of the closure will be at most the maximum of $|D\_i|^2$ and $\aleph\_0$, so it is still $<\kappa$.)


So, by induction hypothesis, each $D\_i$ has a supremum $s\_i$ in $X$, and, since the $D\_i$ form an $\subseteq$-increasing sequence, their suprema $s\_i$ form a $\leq$-increasing sequence in $X$. By hypothesis, the sequence of $s\_i$'s has a supremum $x$ in $X$, and it is easy to check that this $x$ also serves as the supremum of $D$.


