# Is an orthocomplemented lattice automatically complete?

I am reading a paper where I find the following definition:

Definition: Let $L$ be an orthocomplemented lattice. A function $\mu: L \rightarrow [0,1]$ is called a probability measure on $L$ if

1. $\mu(0) = 0$, $\mu(1) = 1$,
2. $\mu(a_1 \lor a_2 \lor \ldots) = \sum_{k=1}^\infty \mu(a_k)$ for mutually disjoint $a_i$.

A function $\nu: B(\mathbb{R}) \rightarrow L$, where $B(\mathbb{R})$ is the family of Borel sets on $\mathbb{R}$, is called a $L$-valued measure on $B(\mathbb{R})$ if

1. $E \cap F = \emptyset$ implies $\nu(E) \perp \nu(F)$ for any $E, F \in B(\mathbb{R})$,
2. $\nu(E_1 \cup E_2 \cup \ldots) = \nu(E_1) \lor \nu(E_2) \lor \ldots$ for any family $E_i \in B(\mathbb{R})$ such that $E_i \cap E_j = \emptyset$ for $i \neq j$,
3. $\nu(\emptyset) = 0, \nu(\mathbb{R}) = 1$.

There is one thing that I don't understand in this definition. On both lists, in item 2, there appears a join of potentially infinitely many elements of $L$. Is this well-defined? I thought that for such a thing to be well-defined, the lattice needs to be complete. Is an orthocomplemented lattice automatically complete?

An orthocomplemented lattice need not be complete.

Consider the system of all subsets of $\mathbb N$ that are either finite or with finite complement. This is a sub-Boolean algebra of the powerset of $\mathbb N$, hence a orthocomplemented lattice; a non-complete one, since the system of all finite sets containing only the even numbers has no supremum in it.

The Definition in question needs at least "Let $L$ be a $\aleph_0$-complete orthocomplemented lattice" to make sense.

Perhaps there is a small sentence, something like "unless otherwise stated, all lattices are complete in this paper" in the preliminaries section of the paper.

• Its the small sentences that always get you in the end... – goblin GONE Sep 18 '13 at 2:05

I'd assume they meant that the identity in item 2 should hold when $a_1 \vee a_2 \vee ...$ is defined.

I haven't really studied orthocomplemented lattices before, so I'm just working from the Wikipedia definition, but how about the following for a non-complete orthocomplemented lattice (apologies for the strange notation):

$0<1<2<...< \infty -2 < \infty - 1 < \infty$,

$0<1'<2'<...< (\infty -2)' < (\infty - 1)' < \infty$,

with anything exclusively from line 1 neither bigger nor smaller than anything exclusively from line 2.

This is a bounded lattice (both strands are totally ordered), with lower bound $0$ and upper bound $\infty$, and for $n \geq 1$ $(\infty - n)'$ can be chosen as the orthocomplement for $n$, $n'$ as the orthocomplement for $(\infty - n)$, $(\infty - n)$ as the orthocomplement for $n'$, and $n$ as the orthocomplement for $(\infty - n)'$. $\{0,1,2...\}$ has no least upper bound in this lattice.

• Perhaps it would be better to reverse the second chain and then just say that $x\mapsto x'$ is the orthocomplementation. – Gejza Jenča Apr 10 '14 at 21:09