irredundant

Definition. Let $L$ be a lattice. A finite join

$$a_1\vee a_2\vee \mathrm{\cdots }\vee a_n$$

of elements in $L$ is said to be irredundant if one can not delete an element from from the join without resulting in a smaller join. In other words,

for all $i=1,\mathrm{\dots },n$.

If the join is not irredundant, it is redundant

Irredundant meets are dually defined.

Remark. The definitions above can be extended to the case where the join (or meet) is taken over an infinite number of elements, provided that the join (or meet) exists.

Example. In the lattice of all subsets (ordered by inclusion) of $\mathbb{Z}$, the set of all integers, the join

$$\mathbb{Z}=\bigvee \{p\mathbb{Z}\mid p\text{ is prime}\}$$

is irredudant. Another irredundant join representation of $\mathbb{Z}$ is just the join of all atoms, the singletons consisting of the individual elements of $\mathbb{Z}$. However,

$$\mathbb{Z}=\bigvee \{n\mathbb{Z}\mid n\text{ is any positive integer}\}$$

is redundant, since $n\mathbb{Z}$ can be removed whenever $n$ is a composite number. The join of all doubletons is also redundant, for $\{a,b\}\le \{a,c\}\vee \{c,b\}$, for any $c\notin \{a,b\}$.

Definition. An element in a lattice is join irredundant if it can not be written as a redundant join of elements. Dually, an element is meet irredundant if each of its representation as a meet of elements is irredundant.

Example. In the two lattice diagrams (Hasse diagram) below,

The $1$ on the left diagram is not join irredundant, since $1=a\vee b\vee c=a\vee b$. On the other hand, the $1$ on the right is join irredundant. Similarly, the $0$ on the right is not meet irredundant, while the corresponding one on the right is.