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).

including the empty set, so: directed+inhabited), and IIUC, it follows from this paper that “chain-complete ⇒ directed-complete”does nothold constructively. $\endgroup$ – Gro-Tsen Dec 11 '20 at 10:17