---
tag: YFT50501A0
kind: statement
role: theorem
name: Birkhoff–Frink representation of algebraic closure systems
msc: [08A30, 06A15, 06B15]
uses: [2HJGRABSPZ, QJGPT2W0FT, 3J0BTXSCVN, T0TNG7AGXT]
about: [2HJGRABSPZ, T0TNG7AGXT]
sources:
  - url: https://en.wikipedia.org/wiki/Closure_operator
---

## Statement

Let $A$ be a set and let $C$ be a [[closure operator|2HJGRABSPZ]] on $A$.
Assume that $C$ is algebraic: $C(X) = \bigcup \set{C(Y) : Y \subseteq X,\ Y \text{ finite}}$ for every $X \subseteq A$.
Then there is an [[algebra|QJGPT2W0FT]] $\mathbf{A}$ with universe $A$ and finitary operations with the following property.
A set $X \subseteq A$ is a subuniverse of $\mathbf{A}$ if and only if $C(X) = X$.

## Remarks

Proof idea.
For each finite set $B = \set{b_1, \dots, b_n} \subseteq A$ with $n$ distinct elements and each $c \in C(B)$, add an $n$-ary operation.
It sends $(b_1, \dots, b_n)$ to $c$ and sends every other tuple to its first entry; for $n = 0$ it is the constant $c$.
A closed set is closed under these operations because $C(B) \subseteq C(X)$ when $B \subseteq X$.
A set closed under them contains $C(B)$ for each finite $B$ inside it, so it is closed because $C$ is algebraic.

Consequence: every [[algebraic lattice|3J0BTXSCVN]] is isomorphic to the [[subalgebra lattice|T0TNG7AGXT]] of some algebra.
Indeed, an algebraic lattice is isomorphic to the lattice of closed sets of some algebraic closure operator.

The hypothesis that $C$ is algebraic is necessary.
For any algebra with finitary operations, the union of a chain of subuniverses is a subuniverse.
On $A = \N$, let $C(X) = X$ for finite $X$ and $C(X) = \N$ for infinite $X$.
This is a closure operator, and it is not algebraic.
The closed sets $\set{0, 2, \dots, 2k}$ form a chain whose union, the set of even numbers, is not closed.
So no algebra on $\N$ has exactly the closed sets of $C$ as its subuniverses.

## Explanations

### L1 human

A closure operator takes a set of elements and adds everything that the set "forces".
It is algebraic when whatever a set forces is already forced by some finite part of it.
The theorem says that every such operator comes from an algebra: the closed sets are exactly the subsets closed under some operations.
So the abstract idea of "generating" is no more general than generating with operations.

### L2 human

The theorem is the converse of a basic fact: in any algebra, the subuniverse operator $\operatorname{Sg}$ is an algebraic closure operator.
Together they show that algebraic closure systems, subuniverse lattices and algebraic lattices describe the same objects.
A similar but harder result for congruences is the theorem of Grätzer and Schmidt, which says that every algebraic lattice is a [[congruence lattice|REE07CSXHR]].