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First some definitions.

A semilattice is a commutative semigroup consisting of idempotents (i.e., elements such that $xx=x$). A typical example of a semilattice is the unit interval endowed with the operation of maximum or minimum.

Each semilattice $X$ carries a natural partial order $\le$ defined by $x\le y$ iff $xy=x$.

A subset $C\subset X$ is called a chain if $xy\in\{x,y\}$ for any $x,y\in C$ (which means that $x,y$ are comparable in the partial order $\le$).

A topologized semilattice is a semilattice endowed with a topology. A topologized semilattice $X$ is called a (semi)topological semilattice if the semilattice operation $X\times X\to X$, $(x,y)\mapsto xy$, is (separately) continuous.

A topologized semilattice $X$ is called complete if each non-empty chain $C\subset X$ has $\inf C\in\bar C$ and $\sup C\in\bar C$. Here by $\bar C$ we denote the closure of $C$ in $X$.

It is clear that for any Hausdorff topological semilattice $X$ the partial order $\{(x,y)\in X\times X:xy=x\}$ is a closed subset of $X\times X$.

Problem 1. Is the partial order $\le$ of a complete Hausdorff semitopological semilattice $X$ closed in $X\times X$?

Remark 1. This problem was considered here and here. In these two papers it was shown that the answer to Problem 1 is affirmative if the complete Hausdorff semitopological semilattice $X$ satisfies one of the following topological conditions:

1) the square $X\times X$ is sequential;

2) $X$ regular or functionally Hausdorff;

3) $X$ satisfies the separation axiom $T_{2\delta}$, which means that for any distinct points $x,y\in X$ and a neighborhood $O_x\subset X$ of $x$ there exists a countable family $\mathcal U$ of closed neighborhoods of $x$ such that $\bigcap\mathcal U\subset O_x\setminus\{y\}$;

4) $X$ satisfies the separation axiom $\vec T_{2\delta}$, which means that $X$ admits a continuous injective map into a topological space satisfying the separation axiom $T_{2\delta}$.

Remark 2. In fact, the condition (4) implies the conditions (2) and (3).

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