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I was reading through some notes on closure operations, and the example was given that for a poset $(S,\preceq)$, the operation on $(\mathcal{P}(S),\subseteq)$ given by $E\mapsto LU(E)$ is a closure operation. Here $LU(E)$ is the set of lower bounds of the upper bounds of $E$ for some $E\subseteq S$. I verified that indeed $E\subseteq LU(E)$, and if $E\subseteq E'$, then $LU(E)\subseteq LU(E')$.

My question is, if I take a family of sets $\xi\subseteq \mathcal{P}(S)$, such that $\xi$ consists only of $LU$-closed subsets of $S$, is it then true that the union is again $LU$-closed? That is,


where the first equality comes from the fact that $E=LU(E)$. I see that for $e\in\bigcup_{E\in\xi}E$, then $e\in LU(\bigcup_{E\in\xi}E)$, however, the other containment has eluded me. I know that the Kuratowski closure axioms require a closure operator to preserve unions in topology, but is such an equality even the case for partially ordered sets?

EDIT The reason I ask is that there is the following exercise in the notes, and here $\mathcal{L}(S)\subseteq\mathcal{P}(S)$ is the set of all $LU$-closed subsets of $S$. I want to show for any family $\xi\subseteq \mathcal{L}(S)$ of $LU$-closed subsets of $S$, then $\sup\xi$ exists in $\mathcal{L}(S)$ and


I know that $\sup\xi$ in $\mathcal{P}(S)$ is the union of all sets $E\in\xi$. My strategy was if one could show that $\cup_{E\in\xi}E=LU(\cup_{E\in\xi}E)$, then it follows that $\cup_{E\in\xi}E$ does indeed exist in $\mathcal{L}(S)$, and thus $\sup\xi=\cup_{E\in\xi}E=LU(\cup_{E\in\xi}E)$. Perhaps I have misinterpreted the question?

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up vote 2 down vote accepted

This isn't even true in finite posets. Let $S=\{a,b,c\}$ where the partial order is defined by $a < c$, $ b < c $ and $a$ and $b$ are incomparable. Then $\{a\}$ and $\{b\}$ are LU-closed but $\{a,b\}$ isn't.

Added The supremum you seek is not $\bigcup_{E\in\xi}E$ but rather $LU(\bigcup_{E\in\xi}E)$.

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I believe I see it now, I wish I had seen it sooner. Thank you. – yunone Oct 4 '10 at 8:45

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