properties of Heyting algebras
Let be a Brouwerian lattice. The following properties hold:
The first three equations are proved in this entry (http://planetmath.org/BrouwerianLattice). We prove the last equation here. For any , iff iff and iff and iff . Hence the equation holds. ∎
Let be a lattice with a binary operation on it satisfying the identities above. We want to show that iff for any . First, suppose . Then . Conversely, suppose . Then by the property 6 in this entry (http://planetmath.org/BrouwerianLattice). As a result, . ∎
The class of Brouwerian lattices is equational. The class of Heyting algebras is equational.
The first fact is the result of the two propositions above. The second comes from the fact that is not used in the proofs of the propositions. ∎
Let be a Heyting algebra. Then iff for all .
Suppose . Since in any Heyting algebra, we only need to show that . Since is distributive, we have . The last equation comes from the fact that . As a result, . Conversely, suppose . Now, , and therefore . ∎
First, assume is a Heyting algebra homomorphism, and . Clearly, is a filter. Suppose , then . Now, , so . If is not maximal, let be a proper filter containing and , then , so that , and hence , contradicting the fact that is proper. So is maximal.
Conversely, suppose is an ultrafilter of . Define by iff . Let . We first show that is a lattice homomorphism:
First, iff iff (since is a filter) iff . So respects .
Next, if , then , which means neither nor is in , or that . On the other hand, if , then neither nor is in , since is an ultrafilter. As a result, neither is , which means . So respects .
So is a lattice homomorphism. Next, we show that is a Heyting algebra homomorphism, which means showing that respects : . It suffices to show iff and .
First, if and then and . If , then . Since , , a contradiction. So .
On the other hand, suppose . So . Now, since , , or . If , then , so there is some with . But this means , or . Since , we would have , a contradiction. Hence .
Therefore is a Heyting algebra homomorphism. ∎
In the proof above, we use the fact that, for any ultrafilter in a bounded lattice , if , then there is such that (for otherwise, the filter generated by and would be proper and properly contains , contradicting the maximality of ). If in addition were distributive, then implies that either or . To see this, suppose . Then there is such that . Similarly, if , there is such that . Let . So , and . Furthermore, . If , so would , a contradiction. Hence .
|Title||properties of Heyting algebras|
|Date of creation||2013-03-22 19:31:45|
|Last modified on||2013-03-22 19:31:45|
|Last modified by||CWoo (3771)|