set theory – Suprema of directed sets

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).