# 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,

$$\text{xymatrix}\mathrm{\&}1\text{ar}\mathrm{@}-[ld]\text{ar}\mathrm{@}-[rd]\text{ar}\mathrm{@}-[d]a\text{ar}\mathrm{@}-[rd]\mathrm{\&}b\text{ar}\mathrm{@}-[d]\mathrm{\&}c\text{ar}\mathrm{@}-[ld]\mathrm{\&}0\mathit{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.5em}}\text{xymatrix}\mathrm{\&}1\text{ar}\mathrm{@}-[ld]\text{ar}\mathrm{@}-[rd]a\text{ar}\mathrm{@}-[rd]\mathrm{\&}\mathrm{\&}b\text{ar}\mathrm{@}-[ld]\mathrm{\&}0$$ |

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.

Title | irredundant |
---|---|

Canonical name | Irredundant |

Date of creation | 2013-03-22 18:10:08 |

Last modified on | 2013-03-22 18:10:08 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 8 |

Author | CWoo (3771) |

Entry type | Definition |

Classification | msc 06B05 |

Defines | redundant |

Defines | irredundant meet |

Defines | irredundant join |

Defines | meet irredundant |

Defines | join irredundant |