---
abstract: 'We find (completeness type) conditions on topological semilattices $X,Y$ guaranteeing that each continuous homomorphism $h:X\to Y$ has closed image $h(X)$ in $Y$.'
address:
- 'T. Banakh: Ivan Franko National University of Lviv (Ukraine) andJan Kochanowski University in Kielce (Poland)'
- 'S. Bardyla: Ivan Franko National University of Lviv (Ukraine)'
author:
- Taras Banakh and Serhii Bardyla
title: |
    Completeness and absolute $H$-closedness\
    of topological semilattices
---

Introduction
============

It is well-known that a topological group $X$ is complete in its two-sided uniformity if and only if for any isomorphic topological embedding $h:X\to Y$ into a Hausdorff topological group $Y$ the image $h(X)$ is closed in $Y$.

In this paper we prove a similar result for topological semilattices.  A [*topological semilattice*]{} is a topological space $X$ endowed with a continuous binary operation $X\times X\to X$, $(x,y)\mapsto xy$, which is associative, commutative and [*idempotent*]{} in the sense that $xx=x$ for all $x\in X$.

We define a Hausdorff topological semigroup $X$ to be

-   [*$H$-closed*]{} if for any isomorphic topological embedding $h:X\to Y$ to a Hausdorff topological semigroup $Y$ the image $h(X)$ is closed in $Y$;

-   [*absolutely $H$-closed*]{} if for any continuous homomorphism $h:X\to Y$ to a Hausdorff topological semigroup $Y$ the image $h(X)$ is closed in $Y$.

(Absolutely) $H$-closed topological semigroups were introduced by Stepp in [@Stepp1969] (resp. [ @Stepp1975]) who called them ([*absolutely*]{}) [*maximal*]{} topological semigroups.  More information on (absolutely) $H$-closed topological semigroups can be found in [@Ban; @BanRav2001; @Bardyla-Gutik-2012; @BardGut-2016(2); @Bardyla-Gutik-Ravsky; @Batikova2009; @ChuchmanGutik2007; @Gutik-2017; @Gutik-2014; @Gutik-Pavlyk-2003; @GutikPagonRepovs2010; @GutikPavlyk2001; @GutikPavlyk2003; @GutikRepovs2008; @Raikov1946; @Ravsky2003; @Stepp1969; @Stepp1975; @Yokoyama2013]. 

In this paper we are concentrated at the problem of detecting (absolutely) $H$-closed topological semilattices.  For discrete topological semilattices this problem has been resolved by combined efforts of Stepp [@Stepp1975] and Banakh, Bardyla [@BB17] who proved that a discrete topological semilattice $X$ is absolutely $H$-closed if and only if $X$ is $H$-closed if and only if all chains in $X$ are finite. 

A subset $C$ of a semilattice $X$ is called a [*chain*]{} if $xy\in \{x,y\}$ for all $x,y\in C$.  This is equivalent to saying that any two elements of $C$ are comparable in the [*natural partial order*]{} $\le$ on $X$, defined by $x\le y$ iff $xy=x$. Endowed with this partial order, each semilattice becomes a [*poset*]{}, i.e., a set endowed with a partial order.  In a Hausdorff topological semilattice $X$ the partial order $\{(x,y)\in X\times X:x\le y\}$ is a closed subset of $X\times X$, which means that $X$ is a [*pospace*]{}, i.e., a topological space endowed with a closed partial order.  A semilattice $X$ is [*linear*]{} if the natural partial order on $X$ is linear (i.e., $X$ is a chain in $X$). 

The following characterization of (absolutely) $H$-closed discrete topological semilattices is a combined result of Stepp [@Stepp1975] and Banakh, Bardyla [@BB17] (who proved the equivalences $(2)\Leftrightarrow(3)$ and $(1)\Leftrightarrow(3)$, respectively). 

\[t:Stepp\] For a discrete topological semilattice $X$ the following conditions are equivalent:

1.   $X$ is $H$-closed;

2.   $X$ is absolutely $H$-closed;

3.   all chains in $X$ are finite. 

The implication $(3){\Rightarrow}(2)$ in this theorem can be also derived from the following characterization proved by Banakh and Bardyla in [@BB17]. 

\[t:BB\] For a Hausdorff topological semilattice $X$ the following conditions are equivalent:

1.   all closed chains in $X$ are compact;

2.   all maximal chains in $X$ are compact;

3.   each non-empty chain $C\subset X$ has $\inf C\in\overline{C}$ and $\sup C\in\overline{C}$;

4.   each closed subsemilattice of $X$ is absolutely $H$-closed;

5.   each closed chain in $X$ is $H$-closed. 

Here by $\overline{C}$ we denote the closure of a set $C$ is a topological space $X$.

A subset $A$ of a poset $(X,\le)$ is called [*lower*]{} (resp. [* upper*]{}) if $A={\downarrow}A$ (resp.  $A={\uparrow}A$) where $${\downarrow}A=\{x\in X:\exists a\in A \;(x\le a)\}\mbox{ \ and \ }{\uparrow}A=\{x\in X:\exists a\in A \;(x\ge a)\}.$$

In 2008 Gutik and Repovš proved the following characterization of (absolutely) $H$-closed linear semilattices. 

\[t:GR\] For a linear Hausdorff topological semilattice $X$ the following conditions are equivalent:

1.   $X$ is $H$-closed;

2.   $X$ is absolutely $H$-closed;

3.   any non-empty chain $C\subset X$ has $\inf C\in\overline{{\uparrow}C}$ and $\sup C\in \overline{{\downarrow}C}$ in $X$.

In [@Yokoyama2013] Yokoyama extended Gutik-Repovš Theorem \[t:GR\] to topological pospaces with finite antichains. 

Trying to extend Gutik-Repovš characterization to all (not necessarily linear) topological semilattices we have discovered that the last condition of Theorem \[t:GR\] admits at least three non-equivalent versions, defined as follows. 

\[d:complete\] A topological semilattice $X$ is defined to be

-   [*$k$-complete*]{} if each non-empty chain $C\subset X$ has $\inf C\in \overline C$ and $\sup C\in\overline C$;

-   [*$s$-complete*]{} if each non-empty subsemilattice $S\subset X$ has $\inf S\in\overline{{\uparrow}S}$ and each non-empty chain $C\subset X$ has $\sup C\in\overline{{\downarrow}C}$;

-   [*$c$-complete*]{} if for each closed upper set $F\subset X$, each non-empty chain $C\subset F$ has $\inf C\in F$ and $\sup C\in \overline{{\downarrow}C}$ in $X$.

In Proposition \[p:diag\] we shall prove that for any topological semilattice the following implications hold: $$\mbox{$k$-complete ${\Rightarrow}$ $s$-complete ${\Rightarrow}$ $c$-complete}.$$ By Theorem \[t:BB\], a Hausdorff topological semilattice is $k$-complete if and only if all maximal chains in $X$ are compact.  Theorems \[t:Stepp\] and \[t:GR\] imply that a discrete or linear Hausdorff topological semilattice $X$ is $s$-complete if and only if $X$ is $c$-complete if and only if $X$ is (absolutely) $H$-closed. 

These completeness properties of topological semilattices will be paired with the following notions. 

We say that a pospace $(X,\le)$ is

-   [*well-separated*]{} if for any points $x<y$ there exists a neighborhood $V\subset X$ of $y$ such that $x\notin{\uparrow}\overline{{\uparrow}\overline{{\uparrow}V}}$;

-   [*down-open*]{} if for every open set $U\subset X$ the lower set ${\downarrow}U$ is open in $X$;

-   a [*topological lattice*]{} if any two points $x,y\in X$ have $\inf\{x,y\}$ and $\sup\{x,y\}$ and the binary operations $\wedge:X\times X\to X$, $\wedge:(x,y)\mapsto\inf\{x,y\}$, and $\vee:X\times X\to X$, $\vee:(x,y)\mapsto\sup\{x,y\}$ are continuous. 

In Lemma \[l:well\] we shall prove that for a Hausdorff topological semilattice the following implications hold: $$\mbox{linear ${\Rightarrow}$ topological lattice ${\Rightarrow}$ down-open ${\Rightarrow}$ well-separated.}$$

The main result of this paper is the following theorem, which will be proved in Section \[s:proof\], after some preparatory work made in Sections \[s:ul\]–\[s:tech\]. 

\[main\] Let $h:X\to Y$ be a continuous homomorphism from a topological semilattice $X$ to a Hausdorff topological semigroup $Y$. The image $h(X)$ is closed in $Y$ if one of the following conditions is satisfied:

1.   $X$ is $k$-complete;

2.   $X$ is $s$-complete and $Y$ is well-separated;

3.   $X$ is $c$-complete and $X$ or $Y$ is down-open. 

Some corollaries of Theorem \[main\] can be formulated using the notion of a $\vec{\mathcal C}$-closed topological semilattice for a category $\vec{\mathcal C}$ whose objects are topological semigroups and morphisms are continuous homomorphisms between topological semigroups. 

Given a class ${\mathcal C}$ of topological semigroups by ${\mathsf{h}{:}\mathcal C}$ and ${\mathsf{e}{:}\mathcal C}$ we denote the category whose objects are topological semigroups in the class ${\mathcal C}$ and morphisms are continuous homomorphisms and isomorphic topological embeddings of topological semigroups in the class ${\mathcal C}$, respectively. 

Let $\vec{\mathcal C}$ be a category whose objects are topological semigroups and morphisms are continuous homomorphisms between topological semigroups.  An object $X$ of the category $\vec{\mathcal C}$ is called [*$\vec{\mathcal C}$-closed*]{} if for any morphism $h:X\to Y$ of the category $\vec{\mathcal C}$ the image $h(X)$ is closed in $Y$.

In particular, for a class ${\mathcal C}$ of topological semigroup, a topological semigroup $X\in{\mathcal C}$ is called

-   [*${\mathsf{h}{:}\mathcal C}$-closed*]{} if for any continuous homomorphism $f:X\to Y\in{\mathcal C}$ the image $f(X)$ is closed in $Y$;

-   [*${\mathsf{e}{:}\mathcal C}$-closed*]{} if for each isomorphic topological embedding $f:X\to Y\in{\mathcal C}$ the image $f(X)$ is closed in $Y$.

Therefore a topological semigroup $X$ is (absolutely) $H$-closed if and only if it is ${\mathsf{e}{:}\!\mathsf{T\!S}}$-closed (resp.  ${\mathsf{h}{:}\!\mathsf{T\!S}}$-closed) for the class ${\mathsf{T\!S}}$ of Hausdorff topological semigroups. 

We shall be interested in ${\mathsf{h}{:}\mathcal C}$-closedness for the following classes ${\mathcal C}$ of topological semilattices:

-   ${\mathsf{T\!sL}}$ of all Hausdorff topological semilattices;

-   ${\mathsf{T\!sL_w}}$ of all well-separated Hausdorff topological semilattices;

-   ${\mathsf{T\!sL_o}}$ of all down-open Hausdorff topological semilattices;

-   ${\mathsf{TL}}$ of Hausdorff topological lattices. 

The results of this paper allow us to draw the following diagram, containing implications between various completeness and closedness properties of a Hausdorff topological semilattice.  The implications from this diagram will be proved in Section \[s:diagram\].  $$\xymatrix{
\mbox{compact}\atop\mbox{linear}\ar@{=>}[r]\ar@{=>}[d]&
\mbox{$s$-complete}\atop\mbox{linear}\ar@{<=>}[r]\ar@{=>}[d]&
\mbox{$c$-complete}\atop\mbox{linear}\ar@{=>}[r]\ar@{=>}[d]&
\mbox{$c$-complete}\atop\mbox{topological lattice}\ar@{=>}[dd]\\
\mbox{has compact}\atop\mbox{maximal chains}\ar@{=>}[r]\ar@{<=>}[d]&
\mbox{has $s$-complete}\atop\mbox{maximal chains}\ar@{<=>}[r]&
\mbox{has $c$-complete}\atop\mbox{maximal chains}\ar@{=>}[d]\\
\mbox{$k$-complete}\ar@{=>}[r]\ar@{=>}[d]&
\mbox{$s$-complete}\ar@{=>}[r]\ar@{=>}[d]&\mbox{$c$-complete}\ar@{=>}[d]
&\mbox{$c$-complete}\atop\mbox{down-open}\ar@{=>}[l]\ar@{=>}[d]\\
\mbox{${\mathsf h}{:}\!{\mathsf{T\!sL}}$-closed}\ar@{=>}[r]&
\mbox{${\mathsf h}{:}\!{\mathsf{T\!sL_w}}$-closed}\ar@{=>}[r]&
\mbox{${\mathsf h}{:}\!{\mathsf{T\!sL_o}}$-closed}\ar@{=>}[d]&\mbox{${\mathsf h}{:}\!{\mathsf{T\!sL}}$-closed}\ar@{=>}[l]\\
&&
\mbox{${\mathsf h}{:}\!{\mathsf{TL}}$-closed}. 
}$$

Upper and lower sets in topological semilattices {#s:ul}
================================================

In this section we prove some auxiliary results related to upper and lower sets in topological semilattices. 

Simple examples show that in general, the closure of an (upper) lower set in a pospace is not necessarily an (upper) lower set in the pospace.  However, in semilattices we have the following positive result. 

\[l:updown\] Let $X$ be a topological semilattice. 

1.   For any open set $U\subset X$ the upper set ${\uparrow}U$ is open in $X$.

2.   The interior of any upper set $U\subset X$ is an upper set in $X$.

3.   The closure $\bar L$ of any lower set $L\subset X$ is a lower set in $X$.

Given an open set $U\subset X$, for every point $x\in{\uparrow}U$ choose a point $u\in U$ with $u\le x$ and observe that $O_x=\{z\in X:uz\in U\}$ is an open neighborhood of $x$, contained in ${\uparrow}U$.

Given any upper set $P\subset X$ consider its interior $P^\circ$ in $X$ and observe that for any point $x\in P^\circ$ there exists an open set $U\subset P$ containing the point $x$. Taking into account that the upper set ${\uparrow}U$ is open, we conclude that ${\uparrow}x\subset{\uparrow}U\subset P^\circ$, which means that $P^\circ$ is an upper set in $X$.

For any lower set $L\subset X$, the complement $X\setminus L$ is an upper set, whose interior $(X\setminus L)^\circ$ is an upper set in $X$. Then the complement $X\setminus (X\setminus L)^\circ=\bar L$ is a lower set in $X$.

Applying Lemma \[l:updown\](1) to the continuous operation $(x,y)\mapsto\sup\{x,y\}$ in a topological lattice, we obtain the following simple (but important) fact. 

\[c:l=&gt;do\] Each topological lattice $X$ is down-open. 

The following proposition is a straightforward corollary of [@Ward1954 Lemma 1]. 

\[p:Ward\] Each linear pospace is a topological lattice. 

\[l:neib\] Any points $x\not\le y$ of a pospace $X$ have open neighborhoods $V_x,V_y\subset X$ such that ${\uparrow}V_x\cap{\downarrow}V_y=\emptyset$.

Since the partial order $\le$ is closed, the points $x,y$ have open neighborhoods $V_x,V_y$ such that $V_x\times V_y$ is disjoint with the partial order $\le$ in $X\times X$. Consequently, $\tilde x\not\le\tilde y$ for any $\tilde x\in V_x$ and $\tilde y\in V_y$. We claim that ${\uparrow}V_x\cap {\downarrow}V_y=\emptyset$. Assuming that this intersection contains some point $v$, we could find points $\tilde x\in V_x$ and $\tilde y\in V_y$ such that $\tilde x\le v\le \tilde y$, which contradicts the choice of $V_x$ and $V_y$.

Any points $x\not\le y$ in Hausdorff topological semilattice $X$ have open neighborhoods $V_x,V_y\subset X$ such that ${\uparrow}V_x\cap{\downarrow}V_y=\emptyset$.

Chain-closed sets in semilattices {#s2}
=================================

A subset $A$ of a poset $X$ is called

-   [*lower chain-closed in $X$*]{} if $\inf C\in A$ for any chain $C\subset A$ possessing $\inf C\in X$;

-   [*upper chain-closed in $X$*]{} if $\sup C\in A$ for any chain $C\subset A$ possessing $\sup C\in X$;

-   [*chain-closed in $X$*]{} if $A$ is lower chain-closed and upper chain-closed in $X$.

A poset $(X,\le)$ is defined to be [*chain-complete*]{} if each chain $C\subset X$ has $\inf C$ and $\sup C$ in $X$. Definition \[d:complete\] implies that each $c$-complete topological semilattice is chain-complete. 

The following lemma can be easily derived from the definitions. 

\[c:fid=&gt;cc\] Any closed upper or lower set in a $c$-complete topological semilattice is chain-closed. 

For two subsets $A,B$ of a semilattice $X$ let $AB=\{ab:a\in A,\;b\in B\}$ be their product in $X$.

The proof of the following simple fact can be found in Lemma III-1.2 of [@GHKLMS]. 

\[l:inf-c\] Let $X$ be a semilattice and $A,B\subset X$ be two sets that have $\inf A$ and $\inf B$ in $X$. Then $(\inf A)\cdot(\inf B)=\inf(A\cdot B)$.

A semilattice $X$ is called [*chain-continuous*]{} if for any chain $C\subset X$ possessing $\sup C\in X$ we get $x\cdot\sup C=\sup xC$.

\[l:sup-clo\] A Hausdorff topological semilattice $X$ is chain-continuous if $\sup C\in\overline{{\downarrow}C}$ for any chain $C\subset X$ possessing $\sup C\in X$. Consequently, each $c$-complete Hausdorff topological semilattice is chain-continuous. 

Let $C\subset X$ be a chain possessing $\sup C\in X$. It is easy to see that for any $a\in X$ the product $a\cdot\sup C$ is an upper bound for the set $aC$. To show that $a\cdot\sup C=\sup aC$, we need to check that $a\cdot\sup C\le b$ for any upper bound $b\in X$ of the set $aC$ in $X$. Taking into account that the set ${\downarrow}b$ is a closed lower set in $X$, we conclude that $\sup aC\in \overline{{\downarrow}(aC)}\subset{\downarrow}b$ and hence $a\cdot\sup C\le b$ and $a\cdot\sup C=\sup aC$.

For a subset $A$ of a poset $X$ by the [*chain-closure*]{} $\overleftrightarrow{A}$ of $A$ we understand the smallest chain-closed subset of $X$ that contain the set $A$.  It is equal to the intersection of all chain-closed subsets of $X$ containing $A$. 

\[l:sup\] Let $X$ be a chain-continuous semilattice.  For any point $a\in X$ and set $B\subset X$ we get $$a\cdot\overleftrightarrow{B}\subset \overleftrightarrow{aB}.$$

Consider the shift $s_a:X\to X$, $s_a:x\mapsto ax$. We claim that the set $P=s_a^{-1}(\overleftrightarrow{aB})$ is chain-closed in $X$. Indeed, for any chain $C\subset P$ possessing $\inf C$, Lemma \[l:inf-c\] implies that $a\cdot\inf C=\inf aC$. Since the set $\overleftrightarrow{aB}\supset s_a(P)\supset s_a(C)=aC$ is chain-closed in $X$, $s_a(\inf C)=a\cdot\inf C=\inf aC\in
\overleftrightarrow{aB}$ and hence $\inf C\in P$.

By analogy we can prove that for any chain $C\subset P$ possessing $\sup C$ in $X$, we get $\sup C\in P$. This means that the set $P$ is chain-closed, contains $B$ and hence $\overleftrightarrow{B}\subset P$, which implies $a\overleftrightarrow{B}=s_a(\overleftrightarrow{B})\subset s_a(P)\subset\overleftrightarrow{aB}$.

\[c:sup\] For any subsets $A,B\subset X$ of a chain-continuous semilattice $X$, we get $$\overleftrightarrow{A}\cdot\overleftrightarrow{B}\subset \overleftrightarrow{AB}.$$

For any point $a\in A$, by Lemma \[l:sup\], we get $a\overleftrightarrow{B}\subset\overleftrightarrow{aB}\subset\overleftrightarrow{AB}$ and hence $A\overleftrightarrow{B}\subset\overleftrightarrow{AB}$. Applying Lemma \[l:sup\] once more, we conclude that for every $b\in\overleftrightarrow{B}$, we get $$\overleftrightarrow{A}b\subset \overleftrightarrow{Ab}\subset \overleftrightarrow{A\overleftrightarrow{B}}\subset
\overleftrightarrow{\overleftrightarrow{AB}}=\overleftrightarrow{AB}$$and hence $\overleftrightarrow{A}\cdot\overleftrightarrow{B}\subset\overleftrightarrow{AB}$.

Countable chain-closures of sets in semilattices
================================================

For a subset $A$ of a chain-complete poset $X$ let $$\begin{aligned}
&\bar\downarrow A=\{\inf C:\mbox{$C$ is a countable chain in $A$}\},\\
&\bar\uparrow A=\{\sup C:\mbox{$C$ is a countable chain in $A$}\}, \mbox{ and}\\
&\bar{\uparrow}\bar{\downarrow}A=\bar{\uparrow}({\bar\downarrow}A). 
\end{aligned}$$

\[l:cccl\] For any subsets $A,B\subset G$ of a semilattice $X$ we get $\bar\downarrow A\cdot\bar\downarrow B\subset\bar\downarrow(AB)$. If the semilattice $X$ is chain-continuous, then $\bar\uparrow A\cdot\bar\uparrow B\subset\bar\uparrow(AB)$.

Given two points $a\in\bar\downarrow A$ and $b\in\bar\downarrow B$, find countable chains $\vec A\subset A$ and $\vec B\subset B$ such that $a=\inf \vec A$ and $b=\inf \vec B$. Since $\vec A$ and $\vec B$ are countable, we can find decreasing sequences $\{a_n\}_{n\in{\omega}}\subset \vec A$ and $\{b_n\}_{n\in{\omega}}\subset \vec B$ such that $a=\inf\{a_n\}_{n\in{\omega}}$ and $b=\inf\{b_n\}_{n\in{\omega}}$. It follows that $\{a_nb_n\}_{n\in{\omega}}$ is a decreasing sequence in $AB$. Moreover, for any $m,k\in{\omega}$ there exists $n\in{\omega}$ such that $a_nb_n\le a_mb_k$, which implies that $ab\le \inf\{a_nb_n\}_{n\in{\omega}}=\inf\{a_nb_k\}_{n,k\in{\omega}}=\inf (\vec A \vec B)$.

By Lemma \[l:inf-c\], for every $n\in{\omega}$, $$a_nb=a_n\cdot\inf\{b_k\}_{k\in{\omega}}=\inf\{a_nb_k\}_{k\in{\omega}}\ge \inf\{a_mb_k\}_{m,k\in{\omega}}=\inf(\vec A\vec B).$$ Applying the Lemma \[l:inf-c\] once more, we get $ab=\inf_{n\in{\omega}}a_nb\ge \inf (\vec A\vec B)$ and hence $$ab=\inf (\vec A\vec B)=\inf\{a_nb_n\}_{n\in{\omega}}\in \bar\downarrow{AB}.$$ By analogy we can prove that the chain-continuity of $X$ implies $\bar\uparrow A\cdot\bar\uparrow B\subset\bar\uparrow(AB)$.

\[l:w\] Let $\{V_n\}_{n\in{\omega}}$ be a decreasing sequence of non-empty sets in a chain-continuous chain-complete semilattice $X$ such that $V_{n+1}V_{n+1}\subset V_n$ for all $n\in{\omega}$. Then the set $L=\bigcap_{n\in{\omega}}{\bar{\uparrow}\bar\downarrow V_n}$ is a non-empty subsemilattice of $X$.

To see that the set $L$ is a subsemilattice of $X$, observe that for every positive integer $n$, Lemma \[l:cccl\] guarantees that $$LL\subset (\bar\uparrow\bar\downarrow V_n)\cdot(\bar\uparrow\bar\downarrow V_n)
\subset \bar\uparrow(\bar\downarrow V_n\cdot\bar\downarrow V_n)\subset\bar\uparrow\bar\downarrow(V_nV_n)\subset\bar\uparrow\bar\downarrow V_{n-1}$$ and hence, $LL\subset\bigcap_{n\in{\omega}}\bar\uparrow\bar\downarrow V_{n}=L$, which means that $L$ is a subsemilattice of $X$.

To see that $L$ is not empty, for every $n\in{\omega}$ choose a point $v_n\in V_{n+1}$. For any numbers $n<m$ consider the product $v_n\cdots v_m\in X$. Using the inclusions $V_kV_k\subset V_{k-1}$ for $k\in\{m,m-1,\dots,n+1\}$, we can show that $v_n\cdots v_m\in V_n$. Observe that for every $n\in{\omega}$ the sequence $(v_{n}\cdots v_m)_{m>n}$ is decreasing and by the chain-completeness of $X$, this chain has the greatest lower bound $\bar v_n:=\inf_{m>n}v_n\cdots v_m$ in $X$, which belongs to $\bar\downarrow{V_n}$.

Taking into account that $v_n\cdots v_m\le v_k\cdots v_m$ for any $n\le k<m$, we conclude that $\bar v_n\le \bar v_k$ for any $n\le k$. By the chain-completeness of $X$ the chain $\{\bar v_n\}_{n\in{\omega}}$ has the least upper bound $\bar v=\sup\{\bar v_n\}_{n\in{\omega}}$ in $X$. Since for every $k\in{\omega}$ the chain $\{\bar v_n\}_{n>k}$ is contained in $\bar\downarrow{V_k}$, the least upper bound $\bar v=\sup\{\bar v_n\}_{n\in{\omega}}=\sup\{\bar v_n\}_{n\ge k}$ belongs to $\bar\uparrow\bar\downarrow{V_k}$ for all $k\in{\omega}$. Then $\bar v\in\bigcap_{n\in{\omega}}\bar\uparrow\bar\downarrow{V_n}=L$, so $L\ne\emptyset$.

The Key Lemma
=============

In this section we shall prove a Key Lemma for the proof of Theorem \[main\]. 

\[l:super\] Let $h:X\to Y$ be a homomorphism of topological semilattices.  The subsemilattice $S=h(X)\subset Y$ is closed in $Y$ if the following conditions are satisfied:

1.   $X$ is chain-complete and chain-continuous;

2.   for any $y\in \bar S\setminus S\subset Y$ and $x\in X$ there exists a sequence $(U_n)_{n\in{\omega}}$ of neighborhoods of $y$ in $Y$ such that such that the point $x$ does not belong to the chain-closure $\overleftrightarrow{U}$ of the set $U=\bigcap_{n\in{\omega}}\bar{\uparrow}\bar{\downarrow}h^{-1}(U_n)$ in $X$.

Given any point $y\in \bar S$, we should prove that $y\in S$. To derive a contradiction, assume that $y\notin S$. By transfinite induction, for every non-zero cardinal $\kappa$ we shall prove the following statement:

-   [*for every family $\{U_{\alpha,n}\}_{(\alpha,n)\in\kappa\times {\omega}}$ of neighborhoods of $y$ in $Y$, the set $\bigcap_{\alpha\in\kappa}\overleftrightarrow{\bigcap_{n\in{\omega}}\bar\uparrow\bar\downarrow h^{-1}(U_{\alpha,n})}$ is not empty.*]{} 

To prove this statement for the smallest infinite cardinal $\kappa={\omega}$, fix an arbitrary double sequence $(U_{\alpha,n})_{(\alpha,n)\in{\omega}\times{\omega}}$ of neighborhoods of $y$. Using the continuity of the semilattice operation at $y$, construct a decreasing sequence $(W_n)_{n\in{\omega}}$ of neighborhoods of $y$ such that $W_n\subset \bigcap_{\alpha+k\le n}U_{\alpha,k}$ and $W_{n+1}W_{n+1}\subset W_n$ for all $n\in{\omega}$. Then the preimages $V_n=h^{-1}(W_n)$, $n\in{\omega}$, form a decreasing sequence $(V_n)_{n\in{\omega}}$ of non-empty sets in $X$ such that $V_nV_n\subset V_{n-1}$ for all $n>0$. By Lemma \[l:w\], the set $\bigcap_{n\in{\omega}}\bar\uparrow\bar\downarrow V_n$ is not empty.  Then the set $$\bigcap_{\alpha\in{\omega}}\overleftrightarrow
{\bigcap_{n\in{\omega}}\bar\uparrow\bar\downarrow h^{-1}(U_{\alpha,n})}\supset\bigcap_{\alpha\in{\omega}}\bigcap_{n\in{\omega}}
\bar\uparrow\bar\downarrow h^{-1}(U_{\alpha,n})\supset\bigcap_{n\in{\omega}}\bar\uparrow\bar\downarrow h^{-1}(W_n)=\bigcap_{n\in{\omega}}\bar\uparrow\bar\downarrow V_n$$ is not empty as well. 

Now assume that for some infinite cardinal $\kappa$ and all cardinals $\lambda<\kappa$ the statement $(*_\lambda)$ has been proved.  To prove the statement $(*_\kappa)$, fix any family $(U_{\alpha,n})_{(\alpha,n)\in\kappa\times{\omega}}$ of neighborhoods of $y$ in the topological semilattice $Y$.

Using the continuity of the semilattice operation on $Y$, for every $\alpha\in\kappa$ choose a decreasing sequence $(W_{\alpha,n})_{n\in{\omega}}$ of neighborhoods of $y$ such that $W_{\alpha,n}\subset U_{\alpha,n}$ and $W_{\alpha,n+1}^2\subset W_{\alpha,n}$ for all $n\in{\omega}$. By Lemma \[l:w\], the intersection $L_\alpha=\bigcap_{n\in{\omega}}\bar\uparrow\bar\downarrow{h^{-1}(W_{\alpha,n})}
\subset\bigcap_{n\in{\omega}}\bar\uparrow\bar\downarrow{h^{-1}(U_{\alpha,n})}$ is a non-empty subsemilattice in $X$. Lemma \[c:sup\] implies that $\overleftrightarrow{L_\alpha}\cdot\overleftrightarrow{L_\alpha}\subset\overleftrightarrow{L_\alpha L_\alpha}=\overleftrightarrow{L_\alpha}$, which means that $\overleftrightarrow{L_\alpha}$ is a chain-closed subsemilattice in $X$. Then for every $\beta\in\kappa$ the intersection $$L_{<\beta}=\bigcap_{\alpha<\beta}\overleftrightarrow{L_\alpha}=
\bigcap_{\alpha<\beta}\overleftrightarrow{\bigcap_{n\in{\omega}}
\bar\uparrow\bar\downarrow h^{-1}(W_{\alpha,n})}$$ is a chain-closed subsemilattice of $X$. By the inductive assumption $(*_{|\beta\times{\omega}|})$, the semilattice $L_{<\beta}$ is not empty. 

Choose any maximal chain $M$ in $L_{<\beta}$. The chain-completeness of $X$ guarantees that $M$ has $\inf M\in X$. Taking into account that $L_{<\alpha}$ is chain-closed in $X$, we conclude that $\inf M\in L_{<\alpha}$. We claim that $x_\alpha:=\inf M$ is the smallest element of the semilattice $L_{<\alpha}$. In the opposite case, we could find an element $z\in L_{<\alpha}$ such that $x_\alpha\not\le z$ and hence $x_\alpha z<x_\alpha$. Then $\{x_\alpha z\}\cup M$ is a chain in $L_{<\alpha}$ that properly contains the maximal chain $M$, which is not possible.  This contradiction shows that $x_\alpha=\inf M$ is the smallest element of the semilattice $L_{<\alpha}$.

Observe that for any ordinals $\alpha<\beta<\kappa$ the inclusion $L_{<\beta}\subset L_{<\alpha}$ implies $x_\alpha\le x_\beta$. So, $\{x_\beta\}_{\beta\in\kappa}$ is a chain in $X$ and by the chain-completeness of $X$, it has $\sup\{x_\beta\}_{\beta\in\kappa}\in \bigcap_{\beta<\kappa}L_{<\beta}\subset\bigcap_{\alpha\in\kappa}
\overleftrightarrow{\bigcap_{n\in{\omega}}\bar\uparrow\bar\downarrow h^{-1}(U_{\alpha,n})}$. So, the latter set is not empty and the statement $(*_\kappa)$ is proved. 

By condition (2), for every point $x\in X$ there exists a sequence $(U_{x,n})_{n\in{\omega}}$ of open neighborhoods of $y$ in $Y$ such that $x\notin\overleftrightarrow{\bigcap_{n\in{\omega}}\bar\uparrow\bar\downarrow h^{-1}(U_{x,n})}$. Then the set $\bigcap_{x\in X}\overleftrightarrow{\bigcap_{n\in{\omega}}\bar\uparrow\bar\downarrow h^{-1}(U_{x,n})}$ is empty, which contradicts the property $(*_\kappa)$ for $\kappa=|X|$.

Well-separated semilattices {#s:well}
===========================

We recall that a topological semilattice $X$ is [*well-separated*]{} if for any points $x<y$ in $X$ there exists a neighborhood $U\subset X$ of $y$ such that $x\notin{\uparrow}\overline{{\uparrow}\overline{{\uparrow} U}}$.

\[l:well\] A Hausdorff topological semilattice $X$ is well-separated if one of the following conditions is satisfied:

1.   $X$ admits a continuous injective homomorphism into a well-separated topological semilattice;

2.   the pospace $X$ is down-open;

3.   $X$ is a topological lattice. 

1\. Assume that $h:X\to Y$ is a continuous injective homomorphism of $X$ into a well-separated topological semilattice $Y$. Given any points $x,y\in X$ with $x<y$, consider their images $\bar x=h(x)$ and $\bar y= h(y)$ and observe that $\bar x<\bar y$ (by the injectivity of $h$).  Since $Y$ is well-separated, the point $\bar y$ has a neighborhood $V\subset Y$ such that $\bar x\notin{\uparrow}\overline{{\uparrow}\overline{{\uparrow}V}}$. By the continuity and monotonicity of $h$, the neighborhood $U:=h^{-1}(V)$ of $y$ witnesses that $X$ is well-separated as $x\notin {\uparrow}\overline{{\uparrow}\overline{{\uparrow}U}}$.

2\. Assume that the lower set of any open set in $X$ is open.  To show that $X$ is well-separated, take any points $x,y\in X$ with $x<y$. By Lemma \[l:neib\], the points $x,y$ have open neighborhoods $V_x,V_y$ in $X$ such that ${\downarrow}V_x\cap{\uparrow}V_y=\emptyset$. By our assumption, the lower set ${\downarrow}V_x$ is open in $X$ and its complement $F=X\setminus{\downarrow}V_x$ is a closed upper set.  Then for the neighborhood $V_y\subset F$ of $y$ we get $${\uparrow}\overline{{\uparrow}\overline{{\uparrow}V_y}}\subset
{\uparrow}\overline{{\uparrow}\overline{{\uparrow}F}}=F\subset X\setminus\{x\}.$$

3\. If $X$ is a topological lattice, then by Corollary \[c:l=&gt;do\], $X$ is down-open and by the preceding item, the topological semilattice $X$ is well-separated. 

Find an example of a Hausdorff topological semilattice $X$ which is not well-separated. 

Is each regular topological semilattice well-separated? 

A Main Technical Result {#s:tech}
=======================

\[t:main\] For a continuous homomorphism $h:X\to Y$ from a $c$-complete topological semilattice $X$ to a Hausdorff topological semigroup $Y$, the image $h(X)$ is closed in $Y$ if one of the following conditions is satisfied:

1.   $X$ or $Y$ is a down-open topological semilattice;

2.   $\overleftrightarrow{U}\subset\overline{{\uparrow}U}$ for any open set $U\subset X$;

3.   $Y$ is well-separated topological semilattice and $\overleftrightarrow{S}\subset{\uparrow}\overline{{\uparrow}S}$ for any subsemilattice $S\subset X$.

Observing that the closure $\bar S$ of the semilattice $S$ in the Hausdorff topological semigroup $Y$ is a semilattice, we can assume that $Y=\bar S$ is a topological semilattice. 

Being $c$-complete, the semilattice $X$ is chain-complete.  By Lemma \[l:sup-clo\], the topological semilattice $X$ is chain-continuous.  The closedness of the set $S:=h(X)$ in $Y$ will follow from Lemma \[l:super\] as soon as we check the condition (2) of this lemma. 

Given any points $y\in\bar S\setminus S\subset Y$ and $x\in X$ we need to find a sequence $(U_n)_{n\in{\omega}}$ of neighborhoods of $y$ in $Y$ such that $x\notin \overleftrightarrow{\bigcap_{n\in{\omega}}\bar\uparrow\bar\downarrow h^{-1}(U_n)}$. Depending on the relation between the points $y$ and $s:=h(x)$, we consider two cases. 

If $s\not\le y$, then by Lemma \[l:neib\], the points $s$ and $y$ have neighborhoods $V_s$ and $V_y$ in $Y$ such that ${\uparrow}V_s\cap{\downarrow}V_y=\emptyset$. By Lemma \[l:updown\](1), the upper set ${\uparrow}V_s$ is open in $Y$. Then $F:=h^{-1}(Y\setminus{\uparrow}V_s)$ is a closed lower set in $X$ that contains $h^{-1}(V_y)$ but does not contain the point $x\in h^{-1}(V_s)$. By Lemma \[c:fid=&gt;cc\], the closed lower set $F$ is chain-closed in $X$. Then for the neighborhoods $U_n:=V_y$, $n\in{\omega}$ we get $$\overleftrightarrow{\bigcap_{n\in{\omega}}\bar\uparrow\bar\downarrow h^{-1}(U_{n})}=\overleftrightarrow{\bar\uparrow\bar\downarrow h^{-1}(V_y)}=\overleftrightarrow{h^{-1}(V_y)}\subset \overleftrightarrow{F}=F\subset X\setminus \{x\}.$$

The case $s\le y$ is more complicated.  In this case $y\not\le s$ and we can apply Lemma \[l:neib\] to find two open neighborhoods $V_s,V_y\subset Y$ of $s,y$ such that ${\downarrow}V_s\cap{\uparrow} V_y=\emptyset$. By Lemma \[l:updown\](1), the upper set ${\uparrow}V_y$ is open in $Y$. The continuity of the homomorphism $h$ implies that $h^{-1}({\uparrow}V_y)$ is an open upper set in $X$.

If $X$ is down-open, then the lower set ${\downarrow}h^{-1}(V_s)$ is open in $X$ and $X\setminus{\downarrow}h^{-1}(V_s)$ is an upper closed set that contains $h^{-1}(V_y)$ and is disjoint with $h^{-1}(V_s)$.

If $Y$ is down-open, then the lower set ${\downarrow}V_s$ is open in $Y$ and the preimage $h^{-1}({\downarrow}V_s)$ is an open lower set in $X$. Then its complement $X\setminus h^{-1}({\downarrow}V_s)$ is an upper closed set that contains $h^{-1}(V_y)$ and is disjoint with $h^{-1}(V_s)$.

In both cases we have found an upper closed set $F\subset X$, containing the set $h^{-1}(V_y)$ and disjoint with the set $h^{-1}(V_s)\ni x$. By the $c$-completeness of $X$, the upper closed set $F$ is chain-closed.  For every $n\in{\omega}$ put $U_n:=V_y$ and observe that $$\overleftrightarrow{\bigcap_{n\in{\omega}}\bar\uparrow\bar\downarrow h^{-1}(U_n)}\subset \overleftrightarrow{\bar\uparrow\bar\downarrow h^{-1}(V_y)}= \overleftrightarrow{h^{-1}(V_y)}\subset  \overleftrightarrow{F}=F\subset X\setminus\{x\}.$$

If the condition (2) of the theorem holds, then we put $U_n:={\uparrow}V_y$ for all $n\in{\omega}$ and observe that $$\overleftrightarrow{\bigcap_{n\in{\omega}}\bar\uparrow\bar\downarrow h^{-1}(U_n)}=\overleftrightarrow{\bar\uparrow\bar\downarrow h^{-1}({\uparrow}V_y)}=\overleftrightarrow{h^{-1}({\uparrow}V_y)}
\subset\overline{h^{-1}({\uparrow}V_y)}\subset X\setminus h^{-1}(V_s)\subset X\setminus \{x\}.$$

Finally, assume that the condition (3) holds.  In this case we can replace $V_y$ by a smaller neighborhood and additionally assume that $s\notin {\uparrow}\overline{{\uparrow}\overline{{\uparrow} V_y}}$. By the continuity of the semilattice operation at $y$, there exists a decreasing sequence $(U_n)_{n\in{\omega}}$ of open neighborhoods of $y$ such that $U_n\subset U_nU_n\subset U_{n-1}\cap {\uparrow} V_y$ for $n\in{\omega}$. Replacing each neighborhood $U_n$ by ${\uparrow}U_n$, we can assume that $U_n={\uparrow}U_n$ is an upper set.  Then $W_n:=h^{-1}(U_n)$ is an open upper set in $X$ such that $W_nW_n\subset W_{n-1}\cap h^{-1}({\uparrow}V_y)$ for all $n\in{\omega}$. Lemma \[l:cccl\] guarantees that $(\bar\downarrow W_n)\cdot(\bar\downarrow W_n)\subset\bar\downarrow (W_nW_n)\subset \bar\downarrow W_{n-1}$ and hence $({\uparrow}\bar\downarrow W_n)\cdot({\uparrow}\bar\downarrow W_n)\subset{\uparrow}\bar\downarrow W_{n-1}$, which implies that $F=\bigcap_{n\in{\omega}}{\uparrow}\bar\downarrow W_n$ is a subsemilattice of $X$ with $F={\uparrow}F$. By Lemma \[l:w\], the semilattice $\bigcap_{n\in{\omega}}\bar\uparrow\bar\downarrow W_n\subset \bigcap_{n\in{\omega}}{\uparrow}\bar\downarrow W_n=F$ is not empty.  By our assumption, $\overleftrightarrow{F}\subset{\uparrow}\overline{{\uparrow}F}=
{\uparrow}\overline{F}$. Then $$\overleftrightarrow{\bigcap_{n\in{\omega}}\bar\uparrow\bar\downarrow h^{-1}(U_n)}\subset \overleftrightarrow{\bigcap_{n\in{\omega}}{\uparrow}\bar\downarrow W_n}\subset\overleftrightarrow{F}\subset{\uparrow}\overline{F}.$$It remains to show that $x\notin {\uparrow}\overline{F}$. Observe that $F\subset {\uparrow}\bar\downarrow h^{-1}({\uparrow}V_y)$. We claim that $\bar\downarrow h^{-1}({\uparrow}V_y)\subset{\uparrow}\overline{h^{-1}({\uparrow}V_y)}$. Indeed, for any $a\in \bar\downarrow h^{-1}({\uparrow}V_y)$, we can find a countable chain $C\subset h^{-1}({\uparrow}V_y)$ with $a=\inf C$ and observe that $C$ is a semilattice such that ${\uparrow}C\subset {\uparrow}h^{-1}({\uparrow}V_y)=h^{-1}({\uparrow}V_y)$. By our assumption, $\overleftrightarrow{C}\subset{\uparrow}\overline{{\uparrow}C}$ and hence $$a=\inf C\in\overleftrightarrow{C}\subset{\uparrow}\overline{{\uparrow}C}\subset {\uparrow}\overline{h^{-1}({\uparrow}V_y)}.$$

Therefore $\bar \downarrow h^{-1}({\uparrow}V_y)\subset {\uparrow}\overline{h^{-1}({\uparrow}V_y)}$ and $F\subset{\uparrow}\bar{\downarrow}h^{-1}({\uparrow}V_y)\subset {\uparrow}{\uparrow}\overline{h^{-1}({\uparrow}V_y)}=
{\uparrow}\overline{h^{-1}({\uparrow}V_y)}$. By the continuity and monotonicity of $h$, $$h(F)\subset h({\uparrow}\overline{h^{-1}({\uparrow} V_y)})\subset {\uparrow}h(\overline{h^{-1}({\uparrow} V_y)}\subset {\uparrow}\overline{h(h^{-1}({\uparrow} V_y))}\subset {\uparrow}\overline{{\uparrow}V_y}.$$ Then $h(\overline{F})\subset\overline{h(F)}\subset \overline{{\uparrow}\overline{{\uparrow}V_y}}$ and $h({\uparrow}\overline{F})\subset{\uparrow} \overline{{\uparrow}\overline{{\uparrow}V_y}}\subset Y\setminus\{s\}$, which implies the desired non-inclusion $x\notin{\uparrow}\overline{F}$.

Now it is legal to apply Lemma \[l:super\] to conclude that the set $S$ is closed in $Y$.

Proof of Theorem \[main\] {#s:proof}
=========================

In this section we shall prove a more general version of Theorem \[main\]. 

Let $h:X\to Y$ be a continuous homomorphism from a $c$-complete topological semilattice $X$ to a Hausdorff topological semigroup $Y$. We shall prove that the image $S:=h(X)$ is closed in $Y$ if one of the following conditions is satisfied:

1.   $X$ is $k$-complete;

2.   $X$ is $s$-complete and $Y$ is well-separated;

3.   $X$ or $Y$ is down-open;

4.   $X$ or $Y$ is a topological lattice. 

1\. Assume that $X$ is $k$-complete.  Then for each closed subset $F\subset X$ and each non-empty chain $C\subset X$ we get $\{\inf C,\sup C\}\subset\bar C\subset \bar F=F$, which means that $F$ is chain-closed.  In particular, for any open subset $U\subset X$ the closed set $\overline{{\uparrow}U}$ is chain-closed and hence contains the chain-closure $\overleftrightarrow{U}$ of the set $U$. Now we see that the condition (2) of Theorem \[t:main\] is satisfied and hence $h(X)$ is closed in $Y$.

2\. Assume that $X$ is $s$-complete and $Y$ is well-separated.  Then for any non-empty subsemilattice $S\subset X$ we get $\inf S\in\overline{{\uparrow}S}$. By the $c$-completeness of $X$, the closed upper set ${\uparrow}\inf S$ is chain-closed and hence $\overleftrightarrow{S}\subset {\uparrow}\inf S={\uparrow}\overline{S}$. Now we can apply Theorem \[t:main\](3) and conclude that $h(X)$ is closed in $Y$.

3\. If $X$ or $Y$ is down-open, then $h(X)$ is closed in $Y$ by Theorem \[t:main\](1)

4\. If $X$ or $Y$ is a topological lattice, then $X$ or $Y$ is down-open according to Corollary \[c:l=&gt;do\].  By the preceding item $h(X)$ is closed in $Y$.

${\mathsf{h}{:}\mathcal C}$-Closed topological semilattices {#s:diagram}
===========================================================

In this section we prove the implications drawn in the diagram at the end of the introduction.  These implications can be derived from Corollary \[c:l=&gt;do\] (saying that each topological lattice is down-open), Lemma \[l:well\] (establishing the embeddings of the classes ${\mathsf{T\!sL_o}}\subset{\mathsf{T\!sL_w}}\subset {\mathsf{T\!sL}}$), Theorem \[t:BB\] (implying that a Hausdorff topological semilattice is $k$-complete if and only if it has compact maximal chains) and the following two propositions. 

\[p:diag\] For any Hausdorff topological semilattice $X$ the following statements hold. 

1.   If $X$ is $k$-complete, then $X$ is $s$-complete. 

2.   If $X$ is $s$-complete, then $X$ is $c$-complete. 

3.   If $X$ has $c$-complete maximal chains, then $X$ is $c$-complete. 

1\. Suppose $X$ is $k$-complete.  To show that $X$ is $s$-complete, fix any non-empty subsemilattice $S\subset X$ and observe that $\overline{{\uparrow}S}$ is a closed subsemilattice in $X$. Using Zorn’s Lemma, choose any maximal chain $M$ in $\overline{{\uparrow}S}$. By the $k$-completeness of $X$ the chain $M$ has $\inf M\in\overline{M}\subset \overline{{\uparrow} S}$. We claim that $a:=\inf M$ is a lower bound for $S$. Given any element $s\in S$, observe that $as\le a$ and $as\in\overline{{\uparrow}S}$. By the maximality of the chain $M$, $as\in M$ and hence $a\le as$, which means that $a=as\le s$ and hence $a$ is a lower bound for the semilattice $S$. To see that $a$ is the largest lower bound for $S$, take any lower bound $b$ for $S$ and observe that $S\subset{\uparrow}b$ implies $a\in \overline{{\uparrow}S}\subset{\uparrow}b$ and hence $b\le a$. So, $\inf S=a\in \overline{{\uparrow}}S$.

On the other hand, the $k$-completeness of $X$ guarantees that each non-empty chain $C\subset X$ has $\sup C\in \overline{C}\subset \overline{{\downarrow} C}$. Hence $X$ is a $s$-complete semilattice. 

2\. Suppose that $X$ is $s$-complete semilattice.  Let $F\subset X$ is an arbitrary closed upper set and $C$ be an arbitrary chain in $F$. Observe that the upper set ${\uparrow} C\subset F$ and its closure $\overline{{\uparrow}C}\subset \overline{F}=F$ are subsemilattices in $X$. Since $X$ is $s$-complete, the semilattice ${\uparrow}C$ has $\inf {\uparrow}C\in \overline{{\uparrow}C}\subset F$. It is clear that $\inf C=\inf {\uparrow} C\in F$.

Since $X$ is $s$-complete semilattice each non-empty chain $C\subset X$ has $\sup C\in\overline{{\downarrow}C}$. Hence $X$ is $c$-complete semilattice. 

3\. Let $X$ be a semilattice with $c$-complete maximal chains.  Let $F\subset X$ be an arbitrary closed upper set and $C$ be an arbitrary chain in $F$. Using Zorn’s Lemma, extend the chain $C$ to a maximal chain $M\subset X$. Since the closure of a chain in a semilattice is a chain, the maximal chain $M$ is closed in $X$.

Consider the upper set $U:=M\cap {\uparrow}C\subset M\cap F$ in $M$ and its closure $\overline{U}\subset \overline{M\cap F}\subset \overline{M}\cap\overline{F}=M\cap F$. By Proposition \[p:Ward\], the linear pospace $M$ is a topological lattice.  Applying Lemma \[l:updown\](3) (to the continuous semilattice operation $M\times M\to M$, $(x,y)\mapsto \sup\{x,y\}$), we conclude that $\overline{U}$ is an upper set in $M$.

The $c$-completeness of the maximal chain $M$ ensures that the chain $C$ has the greatest lower bound $\inf_M C\in \overline{U}\subset F$ in $M$. We claim that $\inf_M C$ is the greatest lower bound of $C$ in $X$. Given any lower bound $b\in X$ for $C$, we conclude that $U=M\cap {\uparrow}C\subset{\uparrow}b$ and hence $\inf_M C\in \overline{U}\subset{\uparrow}b$. Then $b\le \inf_M C$, which means that $\inf C=\inf_M C\in F$.

On the other hand, the $c$-completeness of $M$ ensures that the chain $C\subset M$ has the least upper bound $\sup_M C\in\overline{M\cap{\downarrow}C}$ in $M$. We claim that $\sup_M C$ is the least upper bound for $C$ in $X$. Given any upper bound $b\in X$ for $C$, observe that $C\subset{\downarrow}b$ and hence $\sup_M C\in\overline{M\cap{\downarrow}C}\subset {\downarrow}b$, which means that $\sup_M C\le b$ and $\sup_M C$ is the least upper bound for $C$ in $X$. Then $\sup C=\sup_M C\in\overline{M\cap{\downarrow}C}\subset\overline{{\downarrow}C}$, which completes the proof of the $c$-completeness of the topological semilattice $X$.

Proposition \[p:diag\] and the general version of the Theorem \[main\], proved in Section \[s:proof\], imply the following corollary. 

\[p:diag2\] For any Hausdorff topological semilattice $X$ the following statements hold. 

1.   If $X$ is $k$-complete, then $X$ is ${\mathsf h}{:}\!{\mathsf{T\!sL}}$-closed. 

2.   If $X$ is $s$-complete, then $X$ is ${\mathsf h}{:}\!{\mathsf{T\!sL_w}}$-closed. 

3.   If $X$ is $c$-complete, then $X$ is ${\mathsf h}{:}\!{\mathsf{T\!sL_o}}$-closed. 

4.   If $X$ is $c$-complete and down-open, then $X$ is ${\mathsf h}{:}\!{\mathsf{T\!sL}}$-closed. 

Some Comments and Open Problems {#s:last}
===============================

In this section we shall discuss some results and open problems related to Main Theorem \[main\]. 

Theorems \[t:Stepp\], \[t:BB\], \[t:GR\], \[main\] suggest the following intriguing open problems. 

\[prob\] Is any (regular) absolutely $H$-closed topological semilattice $X$ $c$-complete? 

Is each $s$-complete (regular) Hausdorff topological semilattice $X$ absolutely $H$-closed? 

Is a Hausdorff topological semilattice $X$ (absolutely) $H$-closed if all maximal chains in $X$ are $c$-complete? 

Is a Hausdorff topological semilattice $X$ $s$-complete if all maximal chain in $X$ are $s$-complete? 

Is each $c$-complete Hausdorff topological semilattice $s$-complete? 

Is each absolutely $H$-closed topological semilattice chain-complete? 

The following example shows that $H$-closed topological semilattices need not be $c$-complete. 

\[Bard\] There exists a topological semilattice $X$ such that

1.   $X$ is metrizable, countable, and locally compact;

2.   $X$ is $H$-closed;

3.   $X$ is not chain-complete;

4.   $X$ contains a closed upper set $U$, which is not chain-closed in $X$, so $X$ is not $c$-complete;

5.   there exists an injective continuous homomorphism $h:X\to Z$ to a compact Hausdorff topological semilattice $Z$ whose image $h(X)$ is not closed in $Z$.

Let $\bar{\mathbb Z}=\{-\infty\}\cup{\mathbb Z}\cup\{+\infty\}$ be the set of integer numbers with attached elements $-\infty,+\infty$ such that $-\infty<z<+\infty$ for all $z\in{\mathbb Z}$. We endow $\bar{\mathbb Z}$ with the semilattice operation of minimum.  Let $\{0,1\}$ be the discrete two-element semilattice with the semilattice operation of minimum.  In the product semilattice $\bar{\mathbb Z}\times\{0,1\}$, consider the subsemilattice $$X:=(\bar{\mathbb Z}\times\{0\})\cup({\mathbb Z}\times\{1\}).$$ Endow $X$ with the topology $$\begin{aligned}
\tau=\;&\{U\subset X:(-\infty,0)\in U{\Rightarrow}(\exists n\in{\mathbb Z}\;\forall m\le n\;\; (-2m,0)\in U)\}\;\cup\\
 &\cup\{U\subset X:(+\infty,0)\in U{\Rightarrow}(\exists n\in{\mathbb Z}\;\forall m\ge n\;\; (2m,0)\in U)\}. 
\end{aligned}$$

By [@Bardyla-Gutik-2012 Theorem 4], the topological semilattice $X$ is $H$-closed.  On the other hand, the upper closed set ${\mathbb Z}\times\{1\}$ is not chain-closed in $X$ as $(-\infty,0)=\inf ({\mathbb Z}\times\{1\})\notin{\mathbb Z}\times\{1\}$. So, $X$ is not $c$-complete.  Also the set ${\mathbb Z}\times\{1\}$ has no upper bound in $X$, so the semilattice $X$ is not chain-complete. 

Next, endow $\bar{\mathbb Z}$ with the compact metrizable topology $$\begin{aligned}
\tau_k=\;&\{U\subset X:(-\infty,0)\in U{\Rightarrow}(\exists n\in{\mathbb Z}\;\forall m\le n \;\;(-m,0)\in U)\}\;\cap\\
 &\cap\{U\subset X:(+\infty,0)\in U{\Rightarrow}(\exists n\in{\mathbb Z}\;\forall m\ge n\;\; (m,0)\in U)\}. 
\end{aligned}$$ Then the identity map ${\mathrm{id}}:X\rightarrow \bar{\mathbb Z}\times\{0,1\}$ is an injective continuous homomorphism whose image ${\mathrm{id}}(X)$ is not closed in $\bar{\mathbb Z}\times\{0,1\}$.

In [@Yokoyama2013] Yokoyama asked the question: [*Is each $H$-closed pospace chain-complete?*]{}  The following example gives a negative answer to this question. 

Let $\mathbb I:=[0,1]$ be the unit interval endowed with the usual topology $\tau$ and let $L=\{1/n:n\in\mathbb N\}$. Let $\tau_{1}$ be the topology on $\mathbb I$, generated by the base $\{V\setminus L:V\in\tau\}$. On the space $X$ consider the partial order $\preceq$ in which $x\preceq y$ iff either $x=y$ or $x,y\in L$ and $x\le y$. By [@Engelking1989 3.12.5], $X=([0,1],\tau_{1})$ is an $H$-closed topological space, which implies that $(X,\preceq)$ is an absolutely $H$-closed pospace.  On the other hand, $L$ is a maximal chain in $(X,\leq)$ without lower bound in $X$.

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