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<record version="2" id="10730">
 <title>irredundant</title>
 <name>Irredundant</name>
 <created>2008-06-30 03:15:57</created>
 <modified>2008-06-30 05:01:40</modified>
 <type>Definition</type>
 <creator id="3771" name="CWoo"/>
 <author id="3771" name="CWoo"/>
 <classification>
	<category scheme="msc" code="06B05"/>
 </classification>
 <defines>
	<concept>redundant</concept>
 </defines>
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 <content>Let $L$ be a lattice.  A finite join $$a_1\vee a_2\vee \cdots \vee a_n$$ of elements in $L$ is said to be \emph{irredundant} if one can not delete an element from from the join without resulting in a smaller join.  In other words, 
$$ \bigvee \lbrace a_j \mid j\ne i\rbrace &lt; a_1\vee a_2\vee \cdots \vee a_n$$
for all $i=1,\ldots, n$.  This definition can be extended to the case where the join is taken over an infinite number of elements, provided that the join exists.  If the join is not irredundant, it is \emph{redundant}

Irredundant meets are dually defined.

\textbf{Example}.  In the lattice of all subsets (ordered by inclusion) of $\mathbb{Z}$, the set of all integers, the join $$\mathbb{Z}= \bigvee \lbrace p\mathbb{Z} \mid p \mbox{ is prime} \rbrace$$ 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, $$\mathbb{Z}=\bigvee \lbrace n\mathbb{Z} \mid n \mbox{ is any positive integer} \rbrace$$ is redundant, since $n\mathbb{Z}$ can be removed whenever $n$ is a composite number.  The join of all doubletons is also redundant, for $\lbrace a,b\rbrace \le \lbrace a,c\rbrace \vee \lbrace c,b\rbrace$, for any $c\notin \lbrace a,b\rbrace$.</content>
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