# Brouwerian lattice

Let $L$ be a lattice  , and $a,b\in L$. Then $a$ is said to be pseudocomplemented relative to $b$ if the set

 $T(a,b):=\{c\in L\mid c\wedge a\leq b\}$

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 $a\to b$. So, $a\to b$, if exists, has the following property

 $c\wedge a\leq b\mbox{ iff }c\leq a\to b.$

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

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

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. 1.

$b\leq a\to b$ (since $b\wedge a\leq b$)

2. 2.

$1=a\to 1$ (consequence of 1)

3. 3.

(Birkhoff-Von Neumann condition) $a\leq b$ iff $a\to b=1$ (since $1\wedge a=a\leq b$)

4. 4.

$a\wedge(a\to b)=a\wedge b$

###### Proof.

On the one hand, by 1, $b\leq a\to b$, so $a\wedge b\leq a\wedge(a\to b)$. On the other hand, by definition, $a\wedge(a\to b)\leq b$. Since $a\wedge(a\to b)\leq a$ as well, $a\wedge(a\to b)\leq a\wedge b$, and the proof is complete      . ∎

5. 5.

$a=1\to a$ (consequence of 4)

6. 6.

if $a\leq b$, then $(c\to a)\leq(c\to b)$ (use 4, $c\wedge(c\to a)=c\wedge a\leq a\leq b$)

7. 7.

if $a\leq b$, then $(b\to c)\leq(a\to c)$ (use 4, $a\wedge(b\to c)\leq b\wedge(b\to c)=b\wedge c\leq c$)

8. 8.

$a\to(b\to c)=(a\wedge b)\to c=(a\to b)\to(a\to c)$

###### Proof.

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

 $\displaystyle\big{(}a\to(b\to c)\big{)}\wedge(a\wedge b)$ $\displaystyle=$ $\displaystyle\big{(}a\wedge(b\to c)\big{)}\wedge b$ $\displaystyle=$ $\displaystyle\big{(}b\wedge(b\to c)\big{)}\wedge a$ $\displaystyle=$ $\displaystyle(b\wedge c)\wedge a\leq c.$

So $a\to(b\to c)\leq(a\wedge b)\to c$.

On the other hand, $\big{(}(a\wedge b)\to c\big{)}\wedge a\wedge b=a\wedge b\wedge c\leq c$, so $\big{(}(a\wedge b)\to c\big{)}\wedge a\leq b\to c$, and consequently $(a\wedge b)\to c\leq a\to(b\to c)$.

Second equality: $\big{(}(a\wedge b)\to c\big{)}\wedge(a\to b)\wedge a=\big{(}(a\wedge b)\to c% \big{)}\wedge(a\wedge b)=(a\wedge b)\wedge c\leq c$, so $\big{(}(a\wedge b)\to c\big{)}\wedge(a\to b)\leq a\to c$ and consequently $(a\wedge b)\to c\leq(a\to b)\to(a\to c)$.

On the other hand,

 $\displaystyle\big{(}(a\to b)\to(a\to c)\big{)}\wedge(a\wedge b)$ $\displaystyle=$ $\displaystyle\big{(}(a\to b)\to(a\to c)\big{)}\wedge\big{(}a\wedge(a\to b)\big% {)}$ $\displaystyle=$ $\displaystyle\big{(}(a\to b)\wedge(a\to c)\big{)}\wedge a$ $\displaystyle=$ $\displaystyle(a\wedge b)\wedge(a\to c)$ $\displaystyle=$ $\displaystyle b\wedge(a\wedge c)\leq c,$

so $(a\to b)\to(a\to c)\leq(a\wedge b)\to c$. ∎

9. 9.
###### Proof.
 $a\wedge(b\vee c)\leq(a\wedge b)\vee(a\wedge c).$

To see this: note that $a\wedge b\leq(a\wedge b)\vee(a\wedge c)$, so $b\leq a\to\big{(}(a\wedge b)\vee(a\wedge c)\big{)}$. Similarly, $c\leq a\to\big{(}(a\wedge b)\vee(a\wedge c)\big{)}$. So $b\vee c\leq a\to\big{(}(a\wedge b)\vee(a\wedge c)\big{)}$, or $a\wedge(b\vee c)\leq(a\wedge b)\vee(a\wedge c)$. ∎

If a Brouwerian lattice were a chain, then relative pseudocomplentation can be given by the formula   : $a\to b=1$ if $a\leq b$, and $a\to b=b$ otherwise. From this, we see that the real interval $(\infty,r]$ is a Brouwerian lattice if $x\to y$ is defined according to the formula just mentioned (with $\vee$ and $\wedge$ defined in the obvious way). Incidentally, this lattice has no bottom, and is therefore not a Heyting algebra.

Remarks.

Example. Let $L(X)$ be the lattice of open sets of a topological space  . Then $L(X)$ is Brouwerian. For any open sets $A,B\in X$, $A\to B=(A^{c}\cup B)^{\circ}$, 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