Brouwerian lattice (original) (raw)

Let L be a latticeMathworldPlanetmath, and a,b∈L. Then a is said to be pseudocomplemented relative to b if the set

has a maximal elementMathworldPlanetmath. The maximal element (necessarily unique) of T⁢(a,b) is called the pseudocomplement of a relative to b, and is denoted by a→b. So, a→b, if exists, has the following property

If L has 0, then the pseudocomplement of a relative to 0 is the pseudocomplement of a.

An element a∈L is said to be relatively pseudocomplemented if a→b exists for every b∈L. In particular a→a exists. Since T⁢(a,a)=L, so L has a maximal element, or 1∈L.

A lattice L is said to be relatively pseudocomplemented, or Brouwerian, if every element in L is relatively pseudocomplemented. Evidently, as we have just shown, every Brouwerian lattice contains 1. A Brouwerian lattice is also called an implicative lattice.

Here are some other properties of a Brouwerian lattice L:

    1. b≤a→b (since b∧a≤b)
    1. (Birkhoff-Von Neumann condition) a≤b iff a→b=1 (since 1∧a=a≤b)
    1. a∧(a→b)=a∧b
    1. a=1→a (consequence of 4)
    1. if a≤b, then (c→a)≤(c→b) (use 4, c∧(c→a)=c∧a≤a≤b)
    1. if a≤b, then (b→c)≤(a→c) (use 4, a∧(b→c)≤b∧(b→c)=b∧c≤c)
    1. a→(b→c)=(a∧b)→c=(a→b)→(a→c)
Proof.

We shall use property 4 above a number of times, and the fact that x=y iff x≤y and y≤x. First equality:

(a→(b→c))∧(a∧b) = (a∧(b→c))∧b
= (b∧(b→c))∧a
= (b∧c)∧a≤c.
So a→(b→c)≤(a∧b)→c.
On the other hand, ((a∧b)→c)∧a∧b=a∧b∧c≤c, so ((a∧b)→c)∧a≤b→c, and consequently (a∧b)→c≤a→(b→c).
Second equality: ((a∧b)→c)∧(a→b)∧a=((a∧b)→c)∧(a∧b)=(a∧b)∧c≤c, so ((a∧b)→c)∧(a→b)≤a→c and consequently (a∧b)→c≤(a→b)→(a→c).
On the other hand,
((a→b)→(a→c))∧(a∧b) = ((a→b)→(a→c))∧(a∧(a→b))
------------------- --------------- -----------------------
= ((a→b)∧(a→c))∧a
= (a∧b)∧(a→c)
= b∧(a∧c)≤c,
so (a→b)→(a→c)≤(a∧b)→c. ∎
9. 9.
Proof.

By the propositionPlanetmathPlanetmath found in entry distributive inequalities, it is enough to show that
To see this: note that a∧b≤(a∧b)∨(a∧c), so b≤a→((a∧b)∨(a∧c)). Similarly, c≤a→((a∧b)∨(a∧c)). So b∨c≤a→((a∧b)∨(a∧c)), or a∧(b∨c)≤(a∧b)∨(a∧c). ∎

If a Brouwerian lattice were a chain, then relative pseudocomplentation can be given by the formulaMathworldPlanetmathPlanetmath: a→b=1 if a≤b, and a→b=b otherwise. From this, we see that the real interval (∞,r] is a Brouwerian lattice ifx→y is defined according to the formula just mentioned (with ∨ and ∧ defined in the obvious way). Incidentally, this lattice has no bottom, and is therefore not a Heyting algebra.

Example. Let L⁢(X) be the lattice of open sets of a topological spaceMathworldPlanetmath. Then L⁢(X) is Brouwerian. For any open sets A,B∈X, A→B=(Ac∪B)∘, the interior of the union of B and the complementPlanetmathPlanetmath of A.

References

Title Brouwerian lattice
Canonical name BrouwerianLattice
Date of creation 2013-03-22 16:32:59
Last modified on 2013-03-22 16:32:59
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 21
Author CWoo (3771)
Entry type Definition
Classification msc 06D20
Classification msc 06D15
Synonym relatively pseudocomplemented
Synonym pseudocomplemented relative to
Synonym Brouwerian algebra
Synonym implicative lattice
Related topic PseudocomplementedLattice
Related topic Pseudocomplement
Related topic RelativeComplement
Defines relative pseudocomplement