A complete lattice is a poset that is closed under arbitrary joins and meets. A complete lattice, being closed under arbitrary joins and meets, is closed in particular under binary joins and meets; a complete lattice is thus a specific type of lattice, and hence satisfies associativity, commutativity, idempotence, and absorption of joins and meets.

Complete lattices can be equivalently formulated as posets which are closed under arbitrary joins; it then follows that complete lattices are closed under arbitrary meets as well.

Proof

Suppose that $P$ is a poset which is closed under arbitrary joins. Let $A \subseteq P$. Let $A^L$ be the set of lower bounds of $A$, i.e. the set $\{ p \in P \mid \forall a \in A. p \leq a \}$. Since $P$ is closed under joins, we have the existence of $\bigvee A^L$ in $P$. We will now show that $\bigvee A^L$ is the meet of $A$.

First, we show that $\bigvee A^L$ is a lower bound of $A$. Let $a \in A$. By the definition of $A^L$, $a$ is an upper bound of $A^L$. Because $\bigvee A^L$ is less than or equal to any upper bound of $A^L$, we have $\bigvee A^L \leq a$. $\bigvee A^L$ is therefore a lower bound of $A$.

Now we will show that $\bigvee A^L$ is greater than or equal to any lower bound of $A$. Let $p \in P$ be a lower bound of $A$. Then $p \in A^L$. Because $\bigvee A^L$ is an upper bound of $A^L$, we have $p \leq \bigvee A^L$.

Complete lattices are bounded

As a consequence of closure under arbitrary joins, a complete lattice $L$ contains both $\bigvee \emptyset$ and $\bigvee L$. The former is the least element of $L$ satisfying a vacuous set of constraints; every element of $L$ satisfies a vacuous set of constraints, so this is really the minimum element of $L$. The latter is an upper bound of all elements of $L$, and so it is a maximum. A lattice with both minimum and maximum elements is called bounded, and as this discussion has shown, all complete lattices are bounded.

Basic examples

Finite Lattices

The collection of all subsets of a finite lattice coincides with its collection of finite subsets. A finite lattice, being a finite poset that is closed under finite joins, is then necessarily closed under arbitrary joins. All finite lattices are therefore complete lattices.

Powersets

For any set $X$, consider the poset $\langle \mathcal P(X), \subseteq \rangle$ of $X$’s powerset ordered by inclusion. This poset is a complete lattice in which for all $Y \subset \mathcal P(X)$, $\bigvee Y = \bigcup Y$.

To see that $\bigvee Y = \bigcup Y$, first note that because union contains all of its constituent sets, for all $A \in Y$, $A \subseteq \bigcup Y$. This makes $\bigcup Y$ an upper bound of $Y$. Now suppose that $B \in \mathcal P(X)$ is an upper bound of $Y$; i.e., for all $A \in Y$, $A \subseteq B$. Let $x \in \bigcup Y$. Then $x \in A$ for some $A \in Y$. Since $A \subseteq B$, $x \in B$. Hence, $\bigcup Y \subseteq B$, and so $\bigcup Y$ is the least upper bound of $Y$.

The Knaster-Tarski fixpoint theorem

Suppose that we have a poset $X$ and a monotone function $F : X \to X$. An element $x \in X$ is called $F$-consistent if $x \leq F(x)$ and is called $F$-closed if $F(x) \leq x$. A fixpoint of $F$ is then an element of $X$ which is both $F$-consistent and $F$-closed.

Let $A \subseteq X$ be the set of all fixpoints of $F$. We are often interested in the maximum and minimum elements of $A$, if indeed it has such elements. Most often it is the minimum element of $A$, denoted $\mu F$ and called the least fixpoint of $F$, that holds our interest. In the deduction system example from monotone function examples, the least fixpoint of the deduction system $F$ is equal to the set of all judgments which can be proven without assumptions. Knowing $\mu F$ may be first step toward testing a judgment’s membership in $\mu F$, thus determining whether or not it is provable. In less pedestrian scenarios, we may be interested in the set of all judgments which can be proven without assumption using possibly infinite proof trees; in these cases, it is the greatest fixpoint of $F$, denoted $\nu F$, that we are interested in.

Now that we’ve established the notions of the least and greatest fixpoints, let’s try an exercise. Namely, I’d like you to think of a lattice $L$ and a monotone function $F : L \to L$ such that neither $\mu F$ nor $\nu F$ exists.

Show solution

Let $L = \langle \mathbb R, \leq \rangle$ and let $F$ be the identity function $F(x) = x$. $x \leq y \implies F(x) = x \leq y = F(y)$, and so $F$ is monotone. The fixpoints of $F$ are all elements of $\mathbb R$. Because $\mathbb R$ does not have a maximum or minimum element, neither $\mu F$ nor $\nu F$ exist.

If that was too easy, here is a harder exercise: think of a complete lattice $L$ and monotone function $F : L \to L$ for which neither $\mu F$ nor $\nu F$ exist.

Show solution

There are none. :p

In fact, every monotone function on a complete lattice has both least and greatest fixpoints. This is a consequence of the Knaster-Tarski fixpoint theorem.

Theorem (The Knaster-Tarski fixpoint theorem): Let $L$ be a complete lattice and $F : L \to L$ a monotone function on $L$. Then $\mu F$ exists and is equal to $\bigwedge \{x \in L \mid F(x) \leq x\}$. Dually, $\nu F$ exists and is equal to $\bigvee \{x \in L \mid x \leq F(x) \}$.

Proof

We know that both $\bigwedge \{x \in L \mid F(x) \leq x\}$ and $\bigvee \{x \in L \mid F(x) \leq x \}$ exist due to the closure of complete lattices under meets and joins. We therefore only need to prove that $\bigwedge \{x \in L \mid F(x) \leq x\}$ is a fixpoint of $F$ that is less or equal to all other fixpoints of $F$. The rest follows from duality.

Let $U = \{x \in L \mid F(x) \leq x\}$ and $y = \bigwedge U$. We seek to show that $F(y) = y$. Let $V$ be the set of fixpoints of $F$. Clearly, $V \subseteq U$. Because $y \leq u$ for all $u \in U$, $y \leq v$ for all $v \in V$. In other words, $y$ is less than or equal to all fixpoints of $F$.

For $u \in U$, $y \leq u$, and so $F(y) \leq F(u) \leq u$. Since $F(y)$ is a lower bound of $U$, the definition of $y$ gives $F(y) \leq y$. Hence, $y \in U$. Using the monotonicity of $F$ on the inequality $F(y) \leq y$ gives $F(F(y)) \leq F(y)$, and so $F(y) \in U$. By the definition of $y$, we then have $y \leq F(y)$. Since we have established $y \leq F(y)$ and $F(y) \leq y$, we can conclude that $F(y) = y$.