complete distributivity

A latticeMathworldPlanetmath L is said to be completely distributive if it is a complete latticeMathworldPlanetmath such that, given any sets KI×J such that K projects onto I, and any subset {xij(i,j)K} of L,

iI(jK(i)xij)=fA(iIxif(i)), (1)

where K(i):={jJ(i,j)K}, and A={f:IJf(i)K(i) for all iI}.

By setting I=J={1,2} and K={(1,1),(2,1),(2,2)}, then K(1)={1}, K(2)={1,2}, and A consists of two functions {(1,1),(2,1)} and {(1,1),(2,2)}. Then, the equation above reads:


which is one of the distributive laws, so that complete distributivity implies distributivity.

More generally, setting I={1,2} and J containing 1 but otherwise arbitrary, and K={(2,j)jJ}{(1,1)}. Then K(1)={1}, K(2)=J, and A is the set of functions from I to J fixing 1, and the equation (1) above now looks like


which shows that completely distributivity implies join infiniteMathworldPlanetmath distributivity (


  1. 1.

    Dualizing the above equation results in the same lattice. In other words, a completely distributive lattice may be equivalently defined using the dual of Equation (1). As a result, a completely distributive lattice also satisfies MID, and hence is infinite distributive.

  2. 2.

    However, a completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath distributive latticeMathworldPlanetmath does not have to be completely distributive. Here’s an example: let be the set of natural numbers with the usual orderingMathworldPlanetmath, and be an identical copy of such that each natural numberMathworldPlanetmath n corresponds to n. Then has a natural ordering induced by the usual ordering on . Take the union N of these two sets. Then N becomes a lattice if we extend the meets and joins on and by additionally setting

    ab:={aif ba,botherwise.


    ab:={bif ba,aotherwise.

    Finally, let L be the lattice formed from N by adjoining an extra element to be its top element. It is not hard to see that L is complete and distributive. However, L is not completely distributive, for 0({aa})=0=0, whereas {0aa}={0}=00.

  3. 3.

    In some literature, completeness assumptionPlanetmathPlanetmath is not required, so that the equation (1) above is conditionally defined. In other words, the equation is defined only when each side of the equation exists first.

  4. 4.

    Another generalizationPlanetmathPlanetmath is the so-called (𝔪,𝔫)-distributivity, where 𝔪 and 𝔫 are cardinal numbersMathworldPlanetmath. Specifically, a lattice L is (𝔪,𝔫)-distributive if it is complete and equation (1) is true whenever I has cardinality 𝔪 and each K(i) has cardinality 𝔫 for each iI.


  • 1 G. Gierz, K. H. Hofmann, K. Keimel, J. D. Lawson, M. W. Mislove, D. S. Scott, Continuous Lattices and Domains, Cambridge University Press, Cambridge (2003).
Title complete distributivity
Canonical name CompleteDistributivity
Date of creation 2013-03-22 15:41:34
Last modified on 2013-03-22 15:41:34
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 22
Author CWoo (3771)
Entry type Definition
Classification msc 06D10
Related topic JoinInfiniteDistributive
Defines completely distributive
Defines n)-distributive