Brouwerian lattice (original) (raw)
Let L be a lattice, and a,b∈L. Then a is said to be pseudocomplemented relative to b if the set
has a maximal element. 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:
- b≤a→b (since b∧a≤b)
- (Birkhoff-Von Neumann condition) a≤b iff a→b=1 (since 1∧a=a≤b)
- a∧(a→b)=a∧b
- a=1→a (consequence of 4)
- if a≤b, then (c→a)≤(c→b) (use 4, c∧(c→a)=c∧a≤a≤b)
- if a≤b, then (b→c)≤(a→c) (use 4, a∧(b→c)≤b∧(b→c)=b∧c≤c)
- 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 proposition 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 formula: 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 space. 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 complement
of A.
References
- 1 G. Birkhoff, Lattice Theory, AMS Colloquium Publications, Vol. XXV, 3rd Ed. (1967).
- 2 R. Goldblatt, Topoi, The Categorial Analysis of Logic, Dover Publications (2006).
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 |