---
tag: KT2GNAY91E
kind: definition
name: Countably generated structure
msc: [08A30, 03C07]
defines: 2JFAZSGB6J
uses: [983GZBQYT4, WDVVEYD6AT, 2TQD8NMG9P, 01TQB88H4C]
sources:
  - url: https://en.wikipedia.org/wiki/Substructure_(mathematics)
---

## Definition

Let $L$ be a [[signature|WDVVEYD6AT]] and let $\mathcal{M}$ be an [[$L$-structure|983GZBQYT4]] with universe $M$.
For $X \subseteq M$, let $\operatorname{Sg}(X)$ denote the [[substructure generated by|2TQD8NMG9P]] $X$.
It is the smallest subset of $M$ that contains $X$ and the constants and is closed under the functions of $\mathcal{M}$.
The structure $\mathcal{M}$ is *countably generated* when there is a [[countable set|01TQB88H4C]] $X \subseteq M$ with $\operatorname{Sg}(X) = M$.

## Remarks

Here countable means finite or countably infinite, so every [[finitely generated algebra|WR4WF1K2KW]] is countably generated.

Example: the group $(\Q, +, -, 0)$ is countably generated.
Take $X = \set{1/n : n \ge 1}$; then $p/q = p \cdot (1/q)$ is a sum of copies of $1/q$ or of its negative, so $\operatorname{Sg}(X) = \Q$.

Non-example: the group $(\R, +, -, 0)$ is not countably generated.
For a countable $X$, each element of $\operatorname{Sg}(X)$ is the value of a term at finitely many elements of $X$.
There are only countably many such values, so $\operatorname{Sg}(X)$ is countable, while $\R$ is not.

The same counting shows that, when $L$ is countable, an $L$-structure is countably generated if and only if its universe is countable.