A complete join-semilattice is a join-semilattice $L$ such that for any subset $A\subseteq L$ , $\bigvee A$ , the arbitrary join operation on $A$ , exists. Dually, a complete meet-semilattice is a meet-semilattice such that $\bigwedge A$ exists for any $A\subseteq L$ . Because there are no restrictions placed on the subset $A$ , it turns out that a complete join-semilattice is a complete meet-semilattice, and therefore a complete lattice. In other words, by dropping the arbitrary join (meet) operation from a complete lattice, we end up with nothing new. For a proof of this, see here. The crux of the matter lies in the fact that $\bigvee$ ($\bigwedge$ ) applies to any set, including $L$ itself, and the empty set $\varnothing$ , so that $L$ always contains has a top and a bottom.

Variations. To obtain new objects, one looks for variations in the definition of ``complete''. For example, if we require that any $A\subseteq L$ to be countable, we get what is a called a countably complete join-semilattice (or dually, a countably complete meet-semilattice). More generally, if $\kappa$ is any cardinal, then a $\kappa$ -complete join-semilattice is a semilattice $L$ such that for any set $A\subseteq L$ such that $|A|\le \kappa$ , $\bigvee A$ exists. If $\kappa$ is finite, then $L$ is just a join-semilattice. When $\kappa=\infty$ , the only requirement on $A\subseteq L$ is that it be non-empty. In [1], a complete semilattice is defined to be a poset $L$ such that for any non-empty $A\subseteq L$ , $\bigwedge A$ exists, and any directed set $D\subseteq L$ , $\bigvee D$ exists.

Example. Let $A$ and $B$ be two isomorphic complete chains (a chain that is a complete lattice) whose cardinality is $\kappa$ . Combine the two chains to form a lattice $L$ by joining the top of $A$ with the top of $B$ , and the bottom of $A$ with the bottom of $B$ , so that

• if $a\le b$ in $A$ , then $a\le b$ in $L$
• if $c\le d$ in $B$ , then $c\le d$ in $L$
• if $a\in A$ , $c\in B$ , then $a\le c$ iff $a$ is the bottom of $A$ and $c$ is the top of $B$
• if $a\in A$ , $c\in B$ , then $c\le a$ iff $a$ is the top of $A$ and $c$ is the bottom of $B$

Remark. Although a complete semilattice is the same as a complete lattice, a homomorphism $f$ between, say, two complete join-semilattices $L_1$ and $L_2$ , may fail to be a homomorphism between $L_1$ and $L_2$ as complete lattices. Formally, a complete join-semilattice homomorphism between two complete join-semilattices $L_1$ and $L_2$ is a function $f:L_1\to L_2$ such that for any subset $A\subseteq L_1$ , we have $$f(\bigvee A)=\bigvee f(A)$$ where $f(A)=\lbrace f(a)\mid a\in A\rbrace$ . Note that it is not required that $f(\bigwedge A)=\bigwedge f(A)$ , so that $f$ needs not be a complete lattice homomorphism.

To give a concrete example where a complete join-semilattice homomorphism $f$ fails to be complete lattice homomorphism, take $L$ from the example above, and define $f:L\to L$ by $f(a)=1$ if $a\ne 0$ and $f(0)=0$ . Then for any $A\subseteq L$ , it is evident that $f(\bigvee A)=\bigvee f(A)$ . However, if we take two incomparable elements $a,b\in L$ , then $f(a\wedge b)=f(0)=0$ , while $f(a)\wedge f(b)= 1\wedge 1=1$ .

Bibliography

1

G. Gierz, K. H. Hofmann, K. Keimel, J. D. Lawson, M. W. Mislove, D. S. Scott, Continuous Lattices and Domains, Cambridge University Press, Cambridge (2003).

2

P. T. Johnstone, Stone Spaces, Cambridge University Press (1982).