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Let $\mathcal L$ be a sub-lattice of $\mathcal P(X)$, where $X$ is a finite set.

Denote by $\mathcal I(\mathcal L)$ the set of union-irriducible elements of $\mathcal L$ (i.e. $A\in \mathcal I(\mathcal L)$ iff $A\in\mathcal L$ and it is not possible to write $A$ as a union of other elements of $\mathcal L$).

Is it true that $card(\mathcal I(\mathcal L))\leq card(X)$ ?

Edit. An important point is that $\mathcal I(\mathcal P (X))=X$. Therefore it is possible to restate the question in general terms as follows: let $\mathcal L'$ be a finite distributive lattice and let $\mathcal L$ be a sub-lattice of $\mathcal L'$, is it true that $card (\mathcal I(\mathcal L)) \leq card (\mathcal I(\mathcal L'))$ ?

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4 Answers 4

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Yes, $\operatorname{card}(\mathcal I(\mathcal L))\le\operatorname{card}(X)$. For each $x\in\bigcup\mathcal L$, let $M_x=\bigcap\{A\in\mathcal L:x\in A\}$, the least element of $\mathcal L$ containing $x$. Clearly $M_x$ is join-irreducible. These are the only join-irreducible elements: if $A\in\mathcal L$ then $A=\bigcup\{M_x:x\in A\}$, so $A$ is not join-irreducible unless $A=M_x$ for some $x\in A$.

More generally, suppose $\mathcal L$ is a sub-lattice of a finite distributive lattice $\mathcal L'$, and let $1_\mathcal L$ be the greatest element of $\mathcal L$. For each $a\in\mathcal I(\mathcal L')$ with $a\le1_\mathcal L$, define $\hat a=\bigwedge\{x\in\mathcal L:a\le x\}$; then $a\le\hat a\in\mathcal I(\mathcal L)$.

To see that $\hat a$ is join-irreducible in $\mathcal L$, suppose $\hat a=b\vee c$ for some $b,c\in\mathcal L$. By distributivity, $a=ab\vee ac$; since $a$ is join-irreducible in $\mathcal L'$, either $a=ab$ or else $a=ac$. Say $a=ab$; then $b\in\{x\in\mathcal L:a\le x\}$, so $\hat a\le b\le\hat a$, so $\hat a=b$.

For any $b\in\mathcal L$ we have $b=\bigvee\{a:a\in\mathcal I(\mathcal L'),\ a\le b\}=\bigvee\{\hat a:a\in\mathcal I(\mathcal L'),\ a\le b\}$; if $b$ is join-irreducible in $\mathcal L$, then $b=\hat a$ for some $a\in\mathcal I(\mathcal L')$. Hence $$\mathcal I(\mathcal L)=\{\hat a:a\in\mathcal I(\mathcal L'),\ a\le1_\mathcal L\}$$ and so $$\operatorname{card}(\mathcal I(\mathcal L))\le\operatorname{card}(\{a\in\mathcal I(\mathcal L')):a\le1_\mathcal L\})\le\operatorname{card}(\mathcal I(\mathcal L')).$$

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  • $\begingroup$ Thank you for your answer! Can you extend your proof to the second, more general question? $\endgroup$
    – user118866
    Commented May 20, 2014 at 21:58
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I will answer the second question. The answer is "Yes".

Proof:

Write $\mathbf{FinPos}$ for the category of finite posets and isotone maps and $\mathbf{FinDL}$ for the category of finite distributive lattices with top and bottom preserving lattice homomorphisms.

The categories $\mathbf{FinPos}$ and $\mathbf{FinDL}$ are dually equivalent, in other words $\mathbf{FinPos}^{op}\simeq \mathbf{FinDL}$. One of the functors is $\mathcal I:\mathbf{FinDL}\to\mathbf{FinPos}^{op}$. The embedding $j:\mathcal{L}\to \mathcal L'$ is injective, hence a monomorphism in $\mathbf{FinDL}$. Thus $\mathcal I(j):\mathcal I(\mathcal L)\to \mathcal I(\mathcal L')$ is a monomorphism in $\mathbf{FinPos}^{op}$. But that just means that $\mathcal I(j):\mathcal I(\mathcal L')\to \mathcal I(\mathcal L)$ is an epimorphism in $\mathbf{FinPos}$. Every epimorphism in $\mathbf{FinPos}$ is surjective.

Therefore, $card (\mathcal I(\mathcal L)) \leq card (\mathcal I(\mathcal L'))$.

Clearly, this implies that the answer to the first question is "Yes" as well.

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  • $\begingroup$ Thank you very much for your answer! $\endgroup$
    – user118866
    Commented May 23, 2014 at 11:41
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I will outline another proof that the answer to the second question is "Yes". This one is much simpler and gives another insight.

Fix a maximal chain $C$ in a finite distributive lattice $\mathcal L'$. Let $m:\mathcal I(\mathcal L')\to C$ be a mapping given by the rule $$ m(a)=\wedge\{x\in C:x\geq a\} $$ Then $m$ is a bijection (for the proof of this statement see corollary 112 in Grätzer's book).

Clearly, maximal chains of $\mathcal L$ are subsets of maximal chains of $\mathcal L'$.

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Another proof is based on the fact that a join irreducible element has a unique lower neighbour.

Let $A⊆X$ and $B⊆X$ with $A≠B$ be two irreducibles in $\mathcal L$. Let further for any element $A∈\mathcal L$ the operator $'$ be defined by $$ A'= A\setminus ⋃ \{Y∈\mathcal L\mid Y⊂A\} $$

From $\mathcal L⊆\mathcal P(X)$ we know that $$ ⋃\{Y'\mid Y∈\mathcal I(\mathcal L)\} ⊆ X $$

Then for two irreducibles $A,B∈\mathcal I(\mathcal L)$ we get the equations $A'∩B' = A'∩B = A∩ B'=∅$ as also the intersection $A∩B∈\mathcal L$ has been subtracted both from $A'$ and $B'$. On the other hand $A'≠∅$ and $B'≠∅$ hold, because they are irreducibles. Thus the set $$ \{Y'\mid Y∈\mathcal I(\mathcal L)\} ∪ \bigl\{ X\setminus ⋃\{Y'\mid Y∈\mathcal I(\mathcal L)\} \bigr\} $$ is a partition of $X$. As the operator $'$ is injective for irreducibles, we get the inequality $$ card\bigl(\mathcal I(\mathcal L)\bigr) = card\bigl(\{Y'\mid Y∈\mathcal I(\mathcal L)\}\bigr)≤card(X). $$ As each lattice is isomorphic to the lattice of its primary ideals this condition is true for any finite lattice.

If I'm not mistaken this fact is true for any lattice that can be completely described by its join irreducibles and all of its sublattices, not only finite ones.

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