Let $X$ be a p.o. Consider the topology on $X$ generated by

$$U_{x}^{-}:=X\setminus (x\uparrow),\quad U_{x}^{+}:=X\setminus (x\downarrow), \quad x\in X$$

Throughout this discussion I shall refer to this as the *order-topology*. (I am aware, that there are many possible natural topologies on po-sets.)

Now, it is clearly the not the case, that the topology is automatically Hausdorff, since we can consider the following trivial counterexample:

**Proposition.** Let $X$ be consist of infinitely many incomparable points. Then $X$ is not Hausdorff under the above topology.

**Proof.** Let $x,y\in A$ with $x\neq y$. Then for all $z\in X$ one has $x,y\in U_{z}^{-}\cap U_{z}^{+}$ by mutual incomparibility. Hence every open set generated by the basis contains $x,y$. Thus the points cannot be separated.
**QED**

By a Theorem by [ES Wolk (1958)] one has: if $X$ contains no infinite subset, $A$, of mutually incomparable elements, then $X$ is Hausdorff under the above topology. The converse might not hold, as the relative topology and the order topology on some subspaces $A\subseteq X$ do not in general coincide (since some elements in $A$ may have common upper/lower bounds in $X\setminus A$). Moral of the story: it’s not so clear cut, whether a p.o. endowed with an order topology is Hausdorff, and looking at subspaces doesn't necessarily help.

**QUESTION 1.** Now let $X:=D(\mathbb{P})$ the Dedekind-MacNeille completion$^{\dagger}$ of a p.o. $\mathbb{P}$. Is it the case, that $D(\mathbb{P})$ is Hausdorff? Or if not, under what conditions? Does the fact that $D(\mathbb{P})$ is a (bounded, complete) lattice bear any significance?

I also read the following claim, but did not see it proved:

**Claim.** Let $X$ be a p.o. endowed with some topology $\tau$. Then the following are equivalent

- $\leq$ is closed in $(X,\tau)\times (X,\tau)$
- for all $x\nleq y$ in $X$ there exist disjoint open sets $U\ni x$, $V\ni y$ so that $U$ is an up-set and $V$ a down-set. In particular, $(X,\tau)$ is Hausdorff.

Supposing this claim be true (of which I’m not certain), one *could* use this as a sufficient condition. Then it would suffice to prove that $\leq$ is closed under the order topology. Hence

**QUESTION 2.** Is the above claim true? If so, how does one prove the non-trivial direction (1 $\Rightarrow$ 2)?
**QUESTION 3.** In $D(\mathbb{P})$ is $\leq$ closed under the order topology?

$^{\dagger}$That is, a complete p.o. together with an order-preserving embedding $\varphi:\mathbb{P}\to D(\mathbb{P})$, so that $\mathbb{P}$ is ‘dense’ in the sense that $a=\bigvee\{\varphi(x)\mid x\in\mathbb{P},\varphi(x)\leq a\}=\bigwedge\{\varphi(x)\mid x\in\mathbb{P},\varphi(x)\geq a\}$ for all $a\in D(\mathbb{P})$.