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complete semilattice
A complete joinsemilattice is a joinsemilattice $L$ such that for any subset $A\subseteq L$, $\bigvee A$, the arbitrary join operation on $A$, exists. Dually, a complete meetsemilattice is a meetsemilattice 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 joinsemilattice is a complete meetsemilattice, 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 joinsemilattice (or dually, a countably complete meetsemilattice). More generally, if $\kappa$ is any cardinal, then a $\kappa$complete joinsemilattice is a semilattice $L$ such that for any set $A\subseteq L$ such that $A\leq\kappa$, $\bigvee A$ exists. If $\kappa$ is finite, then $L$ is just a joinsemilattice. When $\kappa=\infty$, the only requirement on $A\subseteq L$ is that it be nonempty. In [1], a complete semilattice is defined to be a poset $L$ such that for any nonempty $A\subseteq L$, $\bigwedge A$ 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 $a\leq b$ in $A$, then $a\leq b$ in $L$

if $c\leq d$ in $B$, then $c\leq d$ in $L$

if $a\in A$, $c\in B$, then $a\leq c$ iff $a$ is the bottom of $A$ and $c$ is the top of $B$

if $a\in A$, $c\in B$, then $c\leq a$ iff $a$ is the top of $A$ and $c$ is the bottom of $B$
Now, $L$ can be easily seen to be a $\kappa$complete lattice. Next, remove the bottom element of $L$ to obtain $L^{{\prime}}$. Since, the meet operation no longer works on all pairs of elements of $L^{{\prime}}$ while $\vee$ still works, $L^{{\prime}}$ is a joinsemilattice that is not a lattice. In fact, $\bigvee$ works on all subsets of $L^{{\prime}}$. Since $L^{{\prime}}=\kappa$, we see that $L^{{\prime}}$ is a $\kappa$complete joinsemilattice.
Remark. Although a complete semilattice is the same as a complete lattice, a homomorphism $f$ between, say, two complete joinsemilattices $L_{1}$ and $L_{2}$, may fail to be a homomorphism between $L_{1}$ and $L_{2}$ as complete lattices. Formally, a complete joinsemilattice homomorphism between two complete joinsemilattices $L_{1}$ 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 $f(A)=\{f(a)\mid a\in A\}$. 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 joinsemilattice 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 $a\neq 0$ 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 $f(a)\wedge f(b)=1\wedge 1=1$.
References
 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).
Mathematics Subject Classification
06A12 no label found06B23 no label found Forums
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