Join-irreducible elements in an infinite lattice.

Question

In Chapter 3.4 of Enumerative Combinatorics, Stanley says that an element of a lattice is join-irreducible if $s \neq \hat0$ and one cannot write $s=t\vee u$ where $t < s$ and $u < s$.

In a finite lattice, an element is join-irreducible if and only if it covers exactly one element.

Is there a simple example of a join-irreducible element in an infinite lattice that covers zero elements or more than one element?

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Some Definitions

The minimal element of a lattice $\hat0$, if it exists, is the unique element such that $\hat0 \leq s$ for all $s$ in the lattice.

The least upper bound (alternatively join) of $s$ and $t$ gives an element $u = s \vee t$ in the lattice such that if $r \geq s$ and $r \geq t$ then $r \geq u$.

Answer 1

Every element in an unbounded linear order is join irreducible. Take a dense linear order, and nothing covers anything, so there's your example. It's not possible for a join irreducible element to cover more than one element, because then it would be the join of any two of them.

Answer 2

As Matt Samuel stated, $ℤ\cup\{-∞,∞\}$ is a lattice that consists only of irreducible elements. It is even a complete lattice with this property. On the other hand in the set of extended reals considered as a lattice every element is irreducible, but as complete lattice every element is supremum reducible as $x=\sup\{ y∈ℝ:y<x\}$ holds.

Given two lattices $L,M$ and a 2-element chain $C$ it is easy to construct a lattice on $L,M,C$ by defining:

• $x\dot \vee y := x\vee y$, if either $x,y∈L$, or $x,y∈M$, or $x,y∈C$.
• $x\dot \vee y := x$, if [$x∈L$ and [ $y∈M$ or $y∈C$]] or [$x∈C$ and $y∈M$]

In that case the upper Element of $C$ is supremum irreducible with respect to $\dot \vee$ and its lower neighbour is infimum irreducible.