In this installment, I will introduce the concept of Boolean algebra, one of the main stars of this series, and relate it to concepts introduced in previous lectures (distributive lattice, Heyting algebra, and so on). Boolean algebra is the algebra of classical propositional calculus, and so has an abstract logical provenance; but one of our eventual goals is to show how any Boolean algebra can also be represented in concrete set-theoretic (or topological) terms, as part of a powerful categorical duality due to Stone.

There are lots of ways to define Boolean algebras. Some definitions were for a long time difficult conjectures (like the Robbins conjecture, established only in the last ten years or so with the help of computers) — testament to the richness of the concept. Here we’ll discuss just a few definitions. The first is a traditional one, and one which is pretty snappy:

A Boolean algebra is a distributive lattice in which every element has a complement.

(If $X$ is a lattice and $x \in X$, a complement of $x$ is an element $y$ such that $x \wedge y = 0$ and $x \vee y = 1$. A lattice is said to be complemented if every element has a complement. Observe that the notions of complement and complemented lattice are manifestly self-dual. Since the notion of distributive lattice is self-dual, so therefore is the notion of Boolean algebra.)

• Example: Probably almost everyone reading this knows the archetypal example of a Boolean algebra: a power set $PX$, ordered by subset inclusion. As we know, this is a distributive lattice, and the complement $S^c$ of a subset $S \subseteq X$ satisfies $S \cap S^c = \emptyset$ and $S \cup S^c = X$.
• Example: Also well known is that the Boolean algebra axioms mirror the usual interactions between conjunction $\wedge$, disjunction $\vee$, and negation $\neg$ in ordinary classical logic. In particular, given a theory $\mathbf{T}$, there is a preorder whose elements are sentences (closed formulas) $p$ of $\mathbf{T}$, ordered by $p \leq q$ if the entailment $p \to q$ is provable in $\mathbf{T}$ using classical logic. By passing to logical equivalence classes ($p \equiv q$ iff $p \leftrightarrow q$ in $\mathbf{T}$), we get a poset with meets, joins, and complements satisfying the Boolean algebra axioms. This is called the Lindenbaum algebra of the theory $\mathbf{T}$.

Exercise: Give an example of a complemented lattice which is not distributive.

As a possible leading hint for the previous exercise, here is a first order of business:

Proposition: In a distributive lattice, complements of elements are unique when they exist.

Proof: If both $b$ and $c$ are complementary to $a$, then $b = b \wedge 1 = b \wedge (a \vee c) = (b \wedge a) \vee (b \wedge c) = 0 \vee (b \wedge c) = b \wedge c$. Since $b = b \wedge c$, we have $b \leq c$. Similarly $c = b \wedge c$, so $b = c. \Box$

The definition of Boolean algebra we have just given underscores its self-dual nature, but we gain more insight by packaging it in a way which stresses adjoint relationships — Boolean algebras are the same things as special types of Heyting algebras (recall that a Heyting algebra is a lattice which admits an implication operator satisfying an adjoint relationship with the meet operator).

Theorem: A lattice is a Boolean algebra if and only if it is a Heyting algebra in which either of the following properties holds:

1. $(a \wedge x \leq y)$ if and only if $(a \leq \neg x \vee y)$
2. $\neg \neg x = x$ for all elements $x$

Proof: First let $X$ be a Boolean algebra, and let $x^c$ denote the complement of an element $x \in X$. Then I claim that $a \wedge x \leq y$ if and only if $a \leq x^c \vee y$, proving that $X$ admits an implication $x \Rightarrow y = x^c \vee y$. Then, taking $y = 0$, it follows that $\neg x := (x \Rightarrow 0) = x^c \vee 0 = x^c$, whence 1. follows. Also, since (by definition of complement) $x$ is the complement of $y$ if and only if $y$ is the complement of $x$, we have $x^{c c} = x$, whence 2. follows.

[Proof of claim: if $a \leq x^c \vee y$, then $x \wedge a \leq x \wedge (x^c \vee y) = (x \wedge x^c) \vee (x \wedge y) \leq 0 \vee y = y$. On the other hand, if $x \wedge a \leq y$, then $a = 1 \wedge a \leq (x^c \vee x) \wedge (x^c \vee a) = x^c \vee (x \wedge a) \leq x^c \vee y$. This completes the proof of the claim and of the forward implication.]

In the other direction, given a lattice which satisfies 1., it is automatically a Heyting algebra (with implication $\neg x \vee y$). In particular, it is distributive. From $\neg x \leq \neg x \vee 0$, we have (from 1.) $x \wedge \neg x \leq 0$; since $0 \leq x \wedge \neg x$ is automatic by definition of $0 = \bot$, we get $0 = x \wedge \neg x$. From $1 \wedge x \leq x$, we have also (from 1.) that $1 \leq \neg x \vee x$; since $\neg x \vee x \leq 1$ is automatic by definition of $1$, we have $\neg x \vee x = 1$. Thus under 1., every element $x$ has a complement $\neg x$.

On the other hand, suppose $X$ is a Heyting algebra satisfying 2.: $\neg \neg x = x$. As above, we know $x \wedge \neg x = 0$. By the corollary below, we also know the function $\neg: X \to X$ takes 0 to 1 and joins to meets (De Morgan law); since condition 2. is that $\neg$ is its own inverse, it follows that $\neg$ also takes meets to joins. Hence $\neg x \vee x = \neg x \vee \neg \neg x = \neg(x \wedge \neg x) = \neg 0 = 1$. Thus for a Heyting algebra which satisfies 2., every element $x$ has a complement $\neg x$. This completes the proof. $\Box$

• Exercise: Show that Boolean algebras can also be characterized as meet-semilattices $X$ equipped with an operation $\neg: X \to X$ for which $a \wedge x \leq y$ if and only if $a \leq \neg(x \wedge \neg y)$.

The proof above invoked the De Morgan law $\neg(x \vee y) = \neg x \wedge \neg y$. The claim is that this De Morgan law (not the other $\neg(x \wedge y) = \neg x \vee \neg y$!) holds in a general Heyting algebra — the relevant result was actually posed as an exercise from the previous lecture:

Lemma: For any element $c$ of a Heyting algebra $X$, the function $- \Rightarrow c: X \to X$ is an order-reversing map (equivalently, an order-preserving map $X^{op} \to X$, or an order-preserving map $X \to X^{op}$). It is adjoint to itself, in the sense that $- \Rightarrow c: X^{op} \to X$ is right adjoint to $- \Rightarrow c: X \to X^{op}$.

Proof: First, we show that $a \leq b$ in $X$ (equivalently, $b \leq a$ in $X^{op}$) implies $(b \Rightarrow c) \leq (a \Rightarrow c)$. But this conclusion holds iff $(b \Rightarrow c) \wedge a \leq c$, which is clear from $(b \Rightarrow c) \wedge a \leq (b \Rightarrow c) \wedge b \leq c$. Second, the adjunction holds because

$(b \Rightarrow c) \leq a$ in $X^{op}$ if and only if

$a \leq (b \Rightarrow c)$ in $X$ if and only if

$a \wedge b \leq c$ in $X$ if and only if

$b \wedge a \leq c$ in $X$ if and only if

$b \leq (a \Rightarrow c)$ in $X. \Box$

Corollary: $- \Rightarrow c: X^{op} \to X$ takes any inf which exists in $X^{op}$ to the corresponding inf in $X$. Equivalently, it takes any sup in $X$ to the corresponding inf in $X$, i.e., $(\bigvee_{s \in S} s) \Rightarrow c = \bigwedge_{s \in S} (s \Rightarrow c)$. (In particular, this applies to finite joins in $X$, and in particular, it applies to the case $c = 0$, where we conclude, e.g., the De Morgan law $\neg(x \vee y) = \neg x \wedge \neg y$.)

• Remark: If we think of sups as sums and infs as products, then we can think of implications $x \Rightarrow y$ as behaving like exponentials $y^x$. Indeed, our earlier result that $x \Rightarrow (-)$ preserves infs $\bigwedge_{s \in S} y_s$ can then be recast in exponential notation as saying $(\prod_{s \in S} y_s)^x = \prod_{s \in S} (y_s)^x$, and our present corollary that $(- \Rightarrow y)$ takes sups to infs can then be recast as saying $y^{\sum_{s \in S} x_s} = \prod_{s \in S} y^{x_s}$. Later we will state another exponential law for implication. It is correct to assume that this is no notational accident!

Let me reprise part of the lemma (in the case $c = 0$), because it illustrates a situation which comes up over and over again in mathematics. In part it asserts that $\neg = (-)\Rightarrow 0: X \to X$ is order-reversing, and that there is a three-way equivalence:

$a \leq \neg b$ if and only if $a \wedge b = 0$ if and only if $b \leq \neg a$.

This situation is an instance of what is called a “Galois connection” in mathematics. If $X$ and $Y$ are posets (or even preorders), a Galois connection between them consists of two order-reversing functions $f: X \to Y$, $g: Y \to X$ such that for all $x \in X, y \in Y$, we have $y \leq f(x)$ if and only if $x \leq g(y)$. (It’s actually an instance of an adjoint pair: if we consider $f$ as an order-preserving map $X \to Y^{op}$ and $g$ an order-preserving map $Y^{op} \to X$, then $f(x) \leq y$ in $Y^{op}$ if and only if $x \leq g(y)$ in $X$.)

Here are some examples:

1. The original example arises of course in Galois theory. If $k$ is a field and $k \subseteq E$ is a finite Galois extension with Galois group $G = Gal(E/k)$ (of field automorphisms $g: E \to E$ which fix the elements belonging to $k$), then there is a Galois connection consisting of maps $Aut_{(-)}(E): PE \to PG$ and $Fix: PG \to PE$. This works as follows: to each subset $S \subseteq E$, define $Aut_S(E)$ to be $\{g \in G: g(s) = s \mbox{ for all } s \in S \}$. In the other direction, to each subset $T \subseteq G$, define $Fix(T)$ to be $\{x \in E: g(x) = x \mbox{ for all } g \in T\}$. Both $Aut_{(-)}(E)$ and $Fix(-)$ are order-reversing (for example, the larger the subset $T \subseteq G$, the more stringent the conditions for an element $x \in E$ to belong to $Fix(T)$). Moreover, we have

$S \subseteq Fix(T)$ iff ($g(x) = x$ for all $x \in S, g \in T$) iff $T \subseteq Aut_S(E)$

so we do get a Galois connection. It is moreover clear that for any $T \subseteq G$, $Fix(T)$ is an intermediate subfield between $k$ and $E$, and for any $S \subseteq E$, $Aut_S(E)$ is a subgroup of $G$. A principal result of Galois theory is that $Fix(-)$ and $Aut_{(-)}(E)$ are inverse to one another when restricted to the lattice of subgroups of $G$ and the lattice of fields intermediate between $k$ and $E$. Such a bijective correspondence induced by a Galois connection is called a Galois correspondence.

2. Another basic Galois connection arises in algebraic geometry, between subsets $J \subseteq k[x_1, \ldots, x_n]$ (of a polynomial algebra over a field $k$) and subsets $V \subseteq k^n$. Given $J$, define $Z(J)$ (the zero locus of $J$) to be $\{(a_1, \ldots, a_n): f(a_1, \ldots, a_n) = 0 \mbox{ for each polynomial } f \in J\}$. On the other hand, define $I(V)$ (the ideal of $V$) to be $\{f \in k[x_1, \ldots, x_n]: f(a) = 0 \mbox{ for all } a = (a_1, \ldots, a_n) \in V\}$. As in the case of Galois theory above, we clearly have a three-way equivalence

$V \subseteq Z(J)$ iff ($f(a) = 0$ for all $a \in V, f \in J$) iff $J \subseteq I(V)$

so that $Z(-)$, $I(-)$ define a Galois connection between power sets (of the $n$-variable polynomial algebra and of $n$-dimensional affine space $k^n$). One defines an (affine algebraic) variety $V \subseteq k^n$ to be a zero locus of some set. Then, on very general grounds (see below), any variety is the zero locus of its ideal. On the other hand, notice that $I(V)$ is an ideal of the polynomial algebra. Not every ideal $I$ of the polynomial algebra is the ideal of its zero locus, but according to the famous Hilbert Nullstellensatz, those ideals $I$ equal to their radical $rad(I) = \{f \in k[x_1, \ldots, x_n]: f^n \in I \mbox{ for some } n \geq 1\}$ are. Thus, $Z(-)$ and $I(-)$ become inverse to one another when restricted to the lattice of varieties and the lattice of radical ideals, by the Nullstellensatz: there is a Galois correspondence between these objects.

3. Both of the examples above are particular cases of a very general construction. Let $X, Y$ be sets and let $R \subseteq X \times Y$ be any relation between them. Then set up a Galois connection which in one direction takes a subset $S \subseteq X$ to $S \backslash R := \{y \in Y: (x, y) \in R \mbox{ for all } x \in S\}$, and in the other takes $T \subseteq Y$ to $R/T := \{x \in X: (x, y) \in R \mbox{ for all } y \in T\}$. Once again we have a three-way equivalence

$S \subseteq R/T$ iff $S \times T \subseteq R$ iff $T \subseteq S \backslash R$.

There are tons of examples of this flavor.

As indicated above, a Galois connection between posets $X, Y$ is essentially the same thing as an adjoint pair between the posets $X, Y^{op}$ (or between $X^{op}, Y$ if you prefer; Galois connections are after all symmetric in $X, Y$). I would like to record a few basic results about Galois connections/adjoint pairs.

Proposition:

1. Given order-reversing maps $f: X \to Y$, $g: Y \to X$ which form a Galois connection, we have $x \leq g f(x)$ for all $x \in X$ and $y \leq f g(y)$ for all $y \in Y$. (Given poset maps $f, g$ which form an adjoint pair $f \dashv g$, we have $x \leq g f(x)$ for all $x \in X$ and $f g(y) \leq y$ for all $y \in Y$.)
2. Given a Galois connection as above, $f(x) = f g f(x)$ for all $x \in X$ and $g(y) = g f g(y)$ for all $y \in Y$. (Given an adjoint pair $f \dashv g$ as above, the same equations hold.) Therefore a Galois connection $(f, g)$ induces a Galois correspondence between the elements of the form $f(x)$ and the elements of the form $g(y)$.

Proof: (1.) It suffices to prove the statements for adjoint pairs. But under the assumption $f \dashv g$, $x \leq g f(x)$ if and only if $f(x) \leq f(x)$, which is certainly true. The other statement is dual.

(2.) Again it suffices to prove the equations for the adjoint pair. Applying the order-preserving map $f$
to $x \leq g f(x)$ from 1. gives $f(x) \leq f g f(x)$. Applying $f g(y) \leq y$ from 1. to $y = f(x)$ gives $f g f(x) \leq f(x)$. Hence $f(x) = f g f(x)$. The other equation is dual. $\Box$

Incidentally, the equations of 2. show why an algebraic variety $V$ is the zero locus of its ideal (see example 2. above): if $V = Z(S)$ for some set of polynomials $S$, then $V = Z(S) = Z I Z(S) = Z I(V)$. They also show that for any element $x$ in a Heyting algebra, we have $\neg \neg \neg x = \neg x$, even though $\neg \neg y = y$ is in general false.

Let $(f, g)$ be a Galois connection (or $f \dashv g$ an adjoint pair). By the proposition, $c = gf: X \to X$ is an order-preserving map with the following properties:

$x \leq c(x)$ for all $x \in X$

$c c(x) = c(x)$ for all $x \in X$.

Poset maps $c: X \to X$ with these properties are called closure operators. We have earlier discussed examples of closure operators: if for instance $G$ is a group, then the operator $c: PG \to PG$ which takes a subset $S \subseteq G$ to the subgroup generated by $S$ is a closure operator. Or, if $X$ is a topological space, then the operator $c: PX \to PX$ which takes a subset $S \subset X$ to its topological closure $\bar{S}$ is a closure operator. Or, if $X$ is a poset, then the operator $c: PX \to PX$ which takes $S \subseteq X$ to $\{a \in X: a \leq s \mbox{ for some } s \in S\}$ is a closure operator. Examples like these can be multiplied at will.

One virtue of closure operators is that they give a useful means of constructing new posets from old. Specifically, if $c: X \to X$ is a closure operator, then a fixed point of $c$ (or a $c$-closed element of $X$) is an element $x$ such that $c(x) = x$. The collection $Fix(c)$ of fixed points is partially ordered by the order in $X$. For example, the lattice of fixed points of the operator $c: PG \to PG$ above is the lattice of subgroups of $G$. For any closure operator $c$, notice that $Fix(c)$ is the same as the image $c(X)$ of $c$.

One particular use is that the fixed points of the double negation closure $\neg \neg: X \to X$ on a Heyting algebra $X$ form a Boolean algebra $Fix(\neg\neg)$, and the map $\neg \neg: X \to Fix(\neg \neg)$ is a Heyting algebra map. This is not trivial! And it gives a means of constructing some rather exotic Boolean algebras (“atomless Boolean algebras”) which may not be so familiar to many readers.

The following exercises are in view of proving these results. If no one else does, I will probably give solutions next time or sometime soon.

Exercise: If $X$ is a Heyting algebra and $x, y, z \in X$, prove the “exponential law” $((x \wedge y) \Rightarrow z) = (x \Rightarrow (y \Rightarrow z))$. Conclude that $\neg(x \wedge y) = (y \Rightarrow \neg x) = (x \Rightarrow \neg y)$.

Exercise: We have seen that $(x \Rightarrow y) \wedge x \leq y$ in a Heyting algebra. Use this to prove $(x \Rightarrow y) \wedge (y \Rightarrow z) \leq (x \Rightarrow z)$.

Exercise: Show that double negation $\neg \neg: X \to X$ on a Heyting algebra preserves finite meets. (The inequality $\neg \neg(x \wedge y) \leq \neg \neg x \wedge \neg \neg y$ is easy. The reverse inequality takes more work; try using the previous two exercises.)

Exercise: If $c: X \to X$ is a closure operator, show that the inclusion map $i: Fix(c) \hookrightarrow X$ is right adjoint to the projection $c: X \to Fix(c)$ to the image of $c$. Conclude that meets of elements in $Fix(\neg \neg)$ are calculated as they would be as elements in $X$, and also that $\neg \neg: X \to Fix(\neg \neg)$ preserves joins.

Exercise: Show that the fixed points of the double negation operator on a topology (as Heyting algebra) are the regular open sets, i.e., those open sets equal to the interior of their closure. Give some examples of non-regular open sets. Incidentally, is the lattice you get by taking the opposite of a topology also a Heyting algebra?