---
tags: logic, category theory, stone duality
---

# A Short Introduction to Stone Duality (Topology)

[up](https://hackmd.io/@alexhkurz/S1W8SC0Tc)

(under construction)

## Part 4: Topological duality

Previously, we developed the duality between finite Boolean algebras and finite sets. This duality can be generalised in many different ways.

Here we want to investigate, dropping the finiteness condition, which category is dual to Boolean algebras.

A finite Boolean algebra, as we have seen, is isomorphic to the powerset of its atoms. But infinite Boolean algebras do not need to have atoms.

## Defining Algebras by Generators and Relations

---

Notes from Aug 3.

We want to show that for algebras defined by operations and equations, free algebras always exist.

We want to profit from what we learned about adjunctions to give this is a more abstract formulation.

**Theorem:** Let $\mathcal A$ be a category of algebras defined by operations and equations. Let $U:\mathcal A\to Set$ be the functor that maps an algebra to its underlying set. Then $U$ has a left adjoint $F$.

**Remark:** Why is $U$ a functor? $FX$ is called the free algebra over the set of generators $X$.

We prove the theorem in two steps. First without equations then with equations.

### The case without equations

Special case: only (fundamental, basic) operations $\Sigma$ no equations. In this case we use the letter $T$ instead of $F$ for reasons that will become clear as we go along.

$\Sigma$ is called a signature

$\Sigma: \mathbb N \to Set$ ... $Set$ is the class (category) of all sets

$\Sigma(n)$ is the set of $n$-ary operations

**Example:** For groups we can take $\Sigma(0)=\{1\}$, $\Sigma(1)=\{(-)^{-1}\}$, $\Sigma(2) = \{\cdot\}$, $\Sigma(n)=\{\}$ for $n>2$.

How do we construct the free algebra $T_\Sigma X$ over $X$?

We need to define the elements of the algebra $T_\Sigma X$ and the operations.

(I am going to drop the subscript in $T_\Sigma$ at some point.)

**Idea:** The set $TX$ is the set of all terms, for example, in groups we would have $((1\cdot x)\cdot y)\cdot z^{-1}$ for $X=\{x,y,z\}$.

We define $TX$ by induction, that is, $TX$ is the smallest set closed under the rules:

- base case:

$$\frac {x\in X} {x \in TX}$$

$$\frac {c\in \Sigma(0)} {c \in TX}$$

- inductive case:

$$\frac {t_1,\ldots t_n \in TX \quad \quad f\in\Sigma(n)}
{f(t_1,\ldots t_n) \in TX}$$

**Proposition:** Let $A$ be an algebra for the signature $\Sigma$. There is a 1-1 corresponds between "interpretations of variables" $X\to A$ and algebra morphisms $TX\to A$.

*Proof:* By induction ... fill in the details ...

**Corollary:** Let $\mathcal A$ be the category of $\Sigma$-algebras and $U:\mathcal A\to Set$ the forgetful functor. Then $T$ is left-adjoint to $U$.[^terminology]

[^terminology]: The category of $\Sigma$-algebras is the category of all algebras for the signature $\Sigma$. The forgetful functor maps an algebra to its underlying set.

### The case with equations

**Challenge:**

- What happens in the presence of a set of (fundamental, basic) equations $E\subseteq TX\times TX$?
- We now want to define the free algebra.
- We still have terms $TX$ but now we want $q:TX\to FX = \{ [t]_\equiv \mid t\in TX\}$ such that $q(t)=q(t')$ if $t$ and $t'$ are in the same equivalence class for which equivalence relation $\equiv$?
- Define
$${\equiv} \subseteq TX\times TX$$

by induction on the structure of a proof that $t=t'$ given the equations. (Maybe we should instead of $t=t'$ write $t\equiv t'$ or even $(t,t')\in {\equiv}$.) What are the rules that we use to prove equations of the type $t=t'$?
- for example $$\frac {t\equiv t'} {t'\equiv t}$$

---

Much of the following has been/will be moved to Part 3 (Logic).

>We start with an example:
>
>Let $X=\{p_1,\ldots p_n\}$ be a set of 'generators'. Let $T(X)$ be the smallest set containing $X$ that is closed under the operations join, meet, complement. Let $B(X)$ be the quotient of $T(X)$ wrt the equations of a Boolean algebra. [^equivalencerelation]
>
>[^equivalencerelation]: Formally, this means that there is a function
>$$[-]:T(X)\to B(X)$$
>
> mapping $t\in T(X)$ to its equivalence class $[t]$ wrt the equivalence relation generated by the equations of Boolean algebra. Note that $[-]$ is onto.
>
>**Exercise:** $[-]$ is a homomorphism, that is, it preserves the operations.
>
>**Exercise:** $B(X)$ is a Boolean algebra.

> **Terminology:** $T(X)$ is the absolutely free algebra over $X$ and $B(X)$ is the free Boolean algebra over $X$.
>
> **Proposition:** $B(X)$ is isomorphic to $2^{2^X}$.
>
> *Hint:* First define $T(X)\to 2^{2^X}$. Then show that this factor through $[-]:T(X)\to B(X)$:
>
>
> <iframe class="quiver-embed" src="https://q.uiver.app/?q=WzAsMyxbMCwwLCJUWCJdLFsyLDAsIkJYIl0sWzIsMiwiMl57Ml5YfSJdLFswLDJdLFswLDEsIlstXSJdLFsxLDIsIiIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==&embed" width="432" height="432" style="border-radius: 8px; border: none;"></iframe>
>
> Then show that the dashed arrow is an isomorphism.

## An example of an atomless Boolean algebra

**Proposition:** The Boolean algebra that is generated from the countably infinite set $\{p_0,p_1,\ldots\}$, closed under conjunction, disjunction, negation and quotiented by the equations of Boolean algebra does not have atoms.

## Find the points of infinite Boolean algebras

... to be continued ...