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pseudocomplement (Definition)

Given an element $ a$ in a bounded lattice $ L$, a complement of $ a$ is defined to be an element $ b\in L$, if such an element exists, such that

$\displaystyle a\wedge b=0,\qquad{ and }\qquad a\vee b =1.$
If a complement of an element exists, it may not be unique. For example, in the middle row of the following diagram (called the diamond)
$\displaystyle \xymatrix{ & 1 \ar@{-}[ld] \ar@{-}[d] \ar@{-}[rd] & \\ a \ar@{-}[rd] & b \ar@{-}[d] & c \ar@{-}[ld] \\ & 0 & }$    

any two of the three elements are complements of the third.

To get around the non-uniqueness issue, an alternative to a complement, called the pseudocomplement of an element, is defined. However, the cost of having the uniqueness is the lost of one of the equations above (in fact, the second one). The weakening of the second equation is not an arbitrary choice, but historical, when propositional logic was being generalized and the law of the excluded middle was dropped in order to develop non-classical logics.

An element $ b$ in a lattice $ L$ with 0 is a pseudocomplement of $ a\in L$ if
  1. $ b\wedge a=0$
  2. for any $ c$ such that $ c\wedge a=0$ then $ c\le b$.
In other words, $ b$ is the maximal element in the set $ \lbrace c\in L\mid c\wedge a=0\rbrace$.

It is easy to see that given an element $ a\in L$, the pseudocomplement of $ a$, if it exists, is unique. If this is the case, then the psedocomplement of $ a$ is written as $ a^*$.

The next natural question to ask is: if $ a^*$ is the pseudocomplement of $ a$, is $ a$ the pseudocomplement of $ a^*$? The answer is no, as the following diagram illustrates (called the benzene)

$\displaystyle \xymatrix{ & 1 \ar@{-}[rd] \ar@{-}[ld] \\ x \ar@{-}[d] & & y \ar@{-}[d] \\ a \ar@{-}[rd] & & b \ar@{-}[ld] \\ & 0 & }$    

The pseudocomplement of $ a$ is $ y$, but the pseudocomplement of $ y$, however, is $ x$. In fact, it is possible that $ a^{**}$ may not even exist! A lattice $ L$ in which every element has a pseudocomplement is called a pseudocomplemented lattice. Necessarily $ L$ must be a bounded lattice.

From the above little discussion, it is not hard to deduce some of the basic properties of pseudocomplementation in a pseudocomplemented lattice:

  1. $ 1^*=0$ and $ 0^*=1$ (if $ c\wedge 1=0$, then $ c=0$, and the largest $ c$ such that $ c\wedge 0=0$ is $ 1$)
  2. $ a\le a^{**}$ (since $ a^*\wedge a=0$ and $ a^*\wedge a^{**}=0$, $ a\le a^{**}$)
  3. $ a\le b$, then $ b^*\le a^*$ (since $ a\wedge b^* \le b\wedge b^*=0$, and $ a\wedge a^*=0$, $ b^*\le a^*$)
  4. $ a^*=a^{***}$ ( $ a\le a^{**}$ by $ 2$ above, so $ a^{***}\le a^*$ by $ 3$, but $ a^*\le a^{***}$ by $ 2$, so $ a^*=a^{***}$)

Furthermore, it can be shown that in a pseudocomplemented lattice, the subset of all pseudocomplements has the structure of a Boolean lattice.

Example. The most common example is the lattice $ L(X)$ of open sets in a topological space $ X$. $ L(X)$ is usually not complemented, because the set complement of an open set is closed. However, $ L(X)$ is pseudocomplemented, and if $ U$ is an open set in $ X$, then its pseudocomplement is $ (U^c)^{\circ}$, the interior of the complement of $ U$.

Remarks.

  • A closely related concept to a pseudocomplemented lattice is that of a pseudocomplemented algebra, or p-algebra for short, which is a pseudocomplemented lattice such that $ ^*$ is considered as an operator. In other words, a morphism between two pseudocomplemented lattices is just a lattice homomorphism, where as a morphism between two p-algebras is a lattice homomorphism $ f$ preserving $ ^*$: $ f(a^*)=f(a)^*$. In the category of p-algebras, the morphism between any pair of objects is a $ \lbrace 0,1\rbrace$-lattice homomorphism, since $ f(1)=f(0^*)=f(0)^*=0^*=1$.
  • The notion of a pseudocomplement can be generalized. Notice first that the definition of a pseudocomplement of an element does not involve the join operation. In fact, all we need is a poset with the least element. A poset $ P$ with the least element 0 is called a pseudocomplemented poset if , for each element $ a\in P$, there is an element $ a^*\in P$ such that their greatest lower bound is 0 and is the largest such element with this property. By definition, $ a^*$ is unique for each $ a\in P$, and that $ P$ itself is bounded, with the greatest element $ 1$, as it is the pseudocomplement of 0. A pseudocomplemented poset that is also a lattice is a clearly a pseudocomplemented lattice. Examples of pseudocomplemented posets that are not lattices are found in the third reference below.

Bibliography

1
T.S. Blyth, Lattices and Ordered Algebraic Structures, Springer, New York (2005).
2
G. Grätzer, General Lattice Theory, 2nd Edition, Birkhäuser (1998).
3
R. Halas, http://www.emis.de/journals/AM/93-34/halas.ps, Archivum Mathematicum (BRNO) 1993.



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See Also: Brouwerian lattice, complemented lattice, pseudodifference

Other names:  pseudocomplemented algebra
Also defines:  pseudocomplemented lattice, benzene, p-algebra, pseudocomplemented poset
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Cross-references: greatest element, bounded, greatest lower bound, least element, poset, operation, join, objects, category, lattice homomorphism, morphism, operator, interior, closed, complemented, topological space, open sets, Boolean lattice, structure, subset, properties, even, easy to see, maximal element, lattice, logics, order, law of the excluded middle, propositional logic, equations, Diamond, row, complement, bounded lattice
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This is version 15 of pseudocomplement, born on 2006-03-20, modified 2008-04-08.
Object id is 7750, canonical name is Pseudocomplement.
Accessed 1848 times total.

Classification:
AMS MSC: 06D15 (Order, lattices, ordered algebraic structures :: Distributive lattices :: Pseudocomplemented lattices)
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