# properties of Heyting algebras

###### Proposition 1.

Let $H$ be a Brouwerian lattice. The following properties hold:

1. 1.

$a\to a=1$

2. 2.

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

3. 3.

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

4. 4.

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

###### Proof.

The first three equations are proved in this entry (http://planetmath.org/BrouwerianLattice). We prove the last equation here. For any $x\in H$, $x\leq a\to(b\wedge c)$ iff $x\wedge a\leq b\wedge c$ iff $x\wedge a\leq b$ and $x\wedge a\leq c$ iff $x\leq a\to b$ and $x\leq a\to c$ iff $x\leq(a\to b)\wedge(a\to c)$. Hence the equation holds. ∎

###### Proposition 2.

Conversely, a lattice with a binary operation $\to$ satisfying the four conditions above is a Brouwerian lattice.

###### Proof.

Let $H$ be a lattice with a binary operation $\to$ on it satisfying the identities above. We want to show that $x\leq a\to b$ iff $x\wedge a\leq b$ for any $x\in H$. First, suppose $x\leq a\to b$. Then $x\wedge a\leq a\wedge(a\to b)=a\wedge b\leq b$. Conversely, suppose $x\wedge a\leq b$. Then $a\to(x\wedge a)\leq a\to b$ by the property 6 in this entry (http://planetmath.org/BrouwerianLattice). As a result, $x=x\wedge(a\to x)\leq(a\to a)\wedge(a\to x)=a\to(a\wedge x)\leq a\to b$. ∎

###### Corollary 1.

The class of Brouwerian lattices is equational. The class of Heyting algebras is equational.

###### Proof.

The first fact is the result of the two propositions above. The second comes from the fact that $0$ is not used in the proofs of the propositions. ∎

###### Proposition 3.

Let $H$ be a Heyting algebra. Then $a\vee a^{*}=1$ iff $a^{**}=a$ for all $a\in H$.

###### Proof.

Suppose $a\vee a^{*}=1$. Since $a\leq a^{**}$ in any Heyting algebra, we only need to show that $a^{**}\leq a$. Since $H$ is distributive, we have $a^{**}=a^{**}\wedge(a\vee a^{*})=(a^{**}\wedge a)\vee(a^{**}\wedge a^{*})=a^{*% *}\wedge a$. The last equation comes from the fact that $a^{**}\wedge a^{*}=0$. As a result, $a^{**}\leq a$. Conversely, suppose $a^{**}=a$. Now, $(a\vee a^{*})^{*}\leq a^{*}\wedge a^{**}=0$, and therefore $a\vee a^{*}=(a\vee a^{*})^{**}=0^{*}=1$. ∎

Note, the last inequality in the proof above comes from the inequality $(a\vee b)^{*}\leq a^{*}\wedge b^{*}$, which is a direct consequence of the fact that pseudocomplementation is order-reversing: $x\leq y$ implies that $y^{*}\leq x^{*}$.

###### Corollary 2.

A Heyting algebra where psuedocomplentation $*$ satisfies the equivalent conditions above is a Boolean algebra. Conversely, a Boolean algebra with $a\to b:=a^{*}\vee b$ is a Heyting algebra.

###### Proof.

Since $a\wedge a^{*}=0$ and $a\vee a^{*}=1$, the pseudocomplementation operation $*$ is the complementation operation. And because any Heyting algebra is distributive, it is Boolean as a result. Conversely, assume $B$ is Boolean. Then $c\leq a\to b=a^{*}\vee b$, so that $c\wedge a\leq a\wedge(a^{*}\vee b)=a\wedge b\leq b$. On the other hand, if $c\wedge a\leq b$, then $c\leq c\vee a^{*}=(c\wedge a)\vee a^{*}\leq a^{*}\vee b=a\to b$. ∎

###### Proposition 4.

A subset $F$ of a Heyting algebra $H$ is an ultrafilter iff there is a Heyting algebra homomorphism $f:H\to\{0,1\}$ with $F=f^{-1}(1)$.

###### Proof.

First, assume $f:H\to\{0,1\}$ is a Heyting algebra homomorphism, and $F=f^{-1}(1)$. Clearly, $F$ is a filter. Suppose $0\neq a\notin F$, then $f(a)=0$. Now, $f(a^{*})=f(a)^{*}=0^{*}=1$, so $a^{*}\in F$. If $F$ is not maximal, let $G$ be a proper filter containing $F$ and $a$, then $a^{*}\in G$, so that $0\in a\wedge a^{*}\in G$, and hence $G=H$, contradicting the fact that $G$ is proper. So $F$ is maximal.

Conversely, suppose $F$ is an ultrafilter of $H$. Define $f:H\to\{0,1\}$ by $f(x)=1$ iff $x\in F$. Let $a,b\in H$. We first show that $f$ is a lattice homomorphism:

• First, $f(a\wedge b)=1$ iff $a\wedge b\in F$ iff $a,b\in F$ (since $F$ is a filter) iff $f(a)=f(b)=1$. So $f$ respects $\wedge$.

• Next, if $f(a\vee b)=0$, then $a\vee b\notin F$, which means neither $a$ nor $b$ is in $F$, or that $f(a)=f(b)=0$. On the other hand, if $f(a)=f(b)=0$, then neither $a$ nor $b$ is in $F$, since $F$ is an ultrafilter. As a result, neither is $a\vee b\in F$, which means $f(a\vee b)=0$. So $f$ respects $\vee$.

So $f$ is a lattice homomorphism. Next, we show that $f$ is a Heyting algebra homomorphism, which means showing that $f$ respects $\to$: $f(a\to b)=f(a)\to f(b)$. It suffices to show $f(a\to b)=0$ iff $f(a)=1$ and $f(b)=0$.

• First, if $f(a)=1$ and $f(b)=0$ then $a\in F$ and $b\notin F$. If $a\to b\in F$, then $(a\to b)\wedge a\in F$. Since $(a\to b)\wedge a\leq b$, $b\in F$, a contradiction. So $a\to b\notin F$.

• On the other hand, suppose $f(a\to b)=0$. So $a\to b\notin F$. Now, since $b\leq a\to b$, $b\notin F$, or $f(b)=0$. If $f(a)=0$, then $a\notin F$, so there is some $c\in F$ with $0=a\wedge c$. But this means $c\leq a^{*}$, or $a^{*}\in F$. Since $a^{*}\leq a\to b$, we would have $a\to b\in F$, a contradiction. Hence $f(a)=1$.

Therefore $f$ is a Heyting algebra homomorphism. ∎

In the proof above, we use the fact that, for any ultrafilter $F$ in a bounded lattice $L$, if $x\notin F$, then there is $y\in F$ such that $0=x\wedge y$ (for otherwise, the filter generated by $x$ and $F$ would be proper and properly contains $F$, contradicting the maximality of $F$). If in addition $L$ were distributive, then $a\vee b\in F$ implies that either $a\in F$ or $b\in F$. To see this, suppose $a\notin F$. Then there is $c\in F$ such that $0=a\wedge c$. Similarly, if $b\notin F$, there is $d\in F$ such that $0=b\wedge d$. Let $e=c\wedge d\in F$. So $e\neq 0$, and $a\wedge e=0=b\wedge e$. Furthermore, $0=(a\wedge e)\vee(b\wedge e)=(a\vee b)\wedge e$. If $a\vee b\in F$, so would $0\in F$, a contradiction. Hence $a\vee b\notin F$.

Title properties of Heyting algebras PropertiesOfHeytingAlgebras 2013-03-22 19:31:45 2013-03-22 19:31:45 CWoo (3771) CWoo (3771) 19 CWoo (3771) Definition msc 03G10 msc 06D20