The infinite distributive law on a join-complete lattice $L$ is as follows:

$\displaystyle a \wedge\left( \bigvee_{b \in B} b \right) = \bigvee_{b \in B}(a \wedge b) $ for all $a \in L$ and $B \subset L$

Note any collection of sets satisfies the infinite distributive law if the meet and join operations are intersection and union respectively. Hence any concrete lattice $-$ i.e. a lattice of sets taken under union and intersection $-$ satisfies the infinite distributive law if it is closed under arbitrary unions.

In Peter T. Johnstone's Stone Spaces, Exercise 2.5 is to prove every distributive lattice $L$ is isomorphic to a concrete lattice $\Phi$. In this case we identify each $x \in L$ with the set of all morphisms $\phi \colon L \to \{0,1 \}$ for which $\phi (x) = 1$. Some form of Choice is needed to guarantee the existence of sufficiently many morphisms.

If we assume $L$ is finitely distributive and join-complete $-$ has arbitrary suprema $-$ then $\Phi$ will be closed under arbitrary unions. But then $\Phi$ hence $L$ satisfies the infinite distributive law. The conclusion is that join-complete lattices satisfy the infinite distributive law.

Is there a flaw to this reasoning? I cannot see one, but the conclusion is not what the book has led me to expect.

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    $\begingroup$ Yes, a lattice isomorphic to a lattice of sets is distributive. But if a complete lattice isn't distributive, it's not isomorphic to a lattice of sets. I don't really see how you conclude that all complete lattices are distributive. Notice, a lattice of sets is by definition closed under union, so this is not enough. $\endgroup$ Mar 12, 2015 at 11:23
  • $\begingroup$ Sorry, we're assuming L is join-complete AND finitely distributive. $\endgroup$
    – Daron
    Mar 12, 2015 at 11:35
  • $\begingroup$ A concrete lattice that is closed under finite intersection and infinite union is distributive. But a concrete lattice may have infinite joins without them being given by infinite union. $\endgroup$
    – Zhen Lin
    Mar 12, 2015 at 11:54

1 Answer 1


Let $L$ be the lattice of closed subsets of $\Bbb R$. Finite meets and joins in $L$ are just unions and intersections, so $L$ is distributive, and for arbitrary $F\subseteq L$ we have $\bigvee F=\operatorname{cl}\bigcup F$, so $L$ is join-complete. However,


  • $\begingroup$ So the point is that while the isomorphism $\mathscr I \colon L \to A$ will preserve finite meets and arbitrary joins, and finite joins correspond to unions in $A$, we cannot guarantee arbitrary joins correspond to unions in $A$. Thank you. $\endgroup$
    – Daron
    Mar 13, 2015 at 11:18
  • $\begingroup$ @Daron: You’re welcome. $\endgroup$ Mar 13, 2015 at 21:10

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