---
tag: YVBHE2SEE6
kind: conjecture
name: Vaught conjecture
msc: [03C15, 03C45]
uses: [655K8Y7VS8, MVCH6N9SNW, 01TQB88H4C, 0V2H5PGJ9B]
about: [655K8Y7VS8]
sources:
  - url: https://en.wikipedia.org/wiki/Vaught_conjecture
same_as:
  wikidata: Q7917378
  wikipedia: Vaught conjecture
history:
  posed_by: "Robert Vaught"
  first_stated: "1961"
---

## Statement

Let $T$ be a [[complete theory|655K8Y7VS8]] in a countable first-order signature.
Let $I(T, \aleph_0)$ be the number of [[countable|01TQB88H4C]] [[models|MVCH6N9SNW]] of $T$ up to isomorphism.
Then $I(T, \aleph_0)$ is finite, or equal to [[$\aleph_0$|0V2H5PGJ9B]], or equal to $2^{\aleph_0}$.

## Remarks

The conjecture is open.
Morley proved in 1970 that $I(T, \aleph_0)$ is at most $\aleph_1$ or equal to $2^{\aleph_0}$.
So the conjecture holds under the continuum hypothesis, and the open question is whether a theory can have exactly $\aleph_1$ countable models when $\aleph_1 < 2^{\aleph_0}$.
Shelah, Harrington and Makkai proved the conjecture for $\omega$-stable theories.
Vaught himself proved that $I(T, \aleph_0)$ is never equal to $2$.

All the allowed values occur.
The theory of dense linear orders without endpoints has exactly one countable model.
The theory of algebraically closed fields of characteristic $0$ has $\aleph_0$ countable models, one for each transcendence degree $0, 1, 2, \dots, \aleph_0$.
The complete theory of $(\Z, <)$ has $2^{\aleph_0}$ countable models.

## Explanations

### L1 human

A complete theory can have many non-isomorphic countable models.
The conjecture says that their number is never strictly between "countably many" and "as many as there are real numbers".
Without the continuum hypothesis, there could be a size in between, and the question is whether some theory has exactly that many countable models.