# creating an infinite model

From the syntactic compactness theorem for first order logic, we get this nice (and useful) result:

Let T be a theory of first-order logic. If T has finite models of unboundedly large sizes, then T also has an infinite^{} model.

###### Proof.

Define the propositions^{}

$${\mathrm{\Phi}}_{n}\equiv \underset{\xaf}{\exists {x}_{1}\mathrm{\cdots}\exists {x}_{n}.({x}_{1}\ne {x}_{2})\wedge \mathrm{\cdots}\wedge ({x}_{1}\ne {x}_{n})\wedge ({x}_{2}\ne {x}_{3})\wedge \mathrm{\cdots}\wedge ({x}_{n-1}\ne {x}_{n})}$$ |

(${\mathrm{\Phi}}_{n}$ says “there exist (at least) $n$ *different* elements in the world”). Note that

$$\mathrm{\cdots}\u22a2{\mathrm{\Phi}}_{n}\u22a2\mathrm{\cdots}\u22a2{\mathrm{\Phi}}_{2}\u22a2{\mathrm{\Phi}}_{1}.$$ |

Define a new theory

$${\text{\mathbf{T}}}_{\mathrm{\infty}}=\text{\mathbf{T}}\cup \{{\mathrm{\Phi}}_{1},{\mathrm{\Phi}}_{2},\mathrm{\dots}\}.$$ |

For any *finite* subset ${\text{\mathbf{T}}}^{\prime}\subset {\text{\mathbf{T}}}_{\mathrm{\infty}}$, we claim that ${\text{\mathbf{T}}}^{\prime}$ is consistent: Indeed, ${\text{\mathbf{T}}}^{\prime}$ contains axioms of T, along with finitely many of ${\left\{{\mathrm{\Phi}}_{n}\right\}}_{n\ge 1}$. Let ${\mathrm{\Phi}}_{m}$ correspond to the largest index appearing in ${\text{\mathbf{T}}}^{\prime}$. If ${\mathcal{M}}_{m}\vDash \text{\mathbf{T}}$ is a model of T with at least $m$ elements (and by hypothesis^{}, such as model exists), then ${\mathcal{M}}_{m}\vDash \text{\mathbf{T}}\cup \{{\mathrm{\Phi}}_{m}\}\u22a2{\text{\mathbf{T}}}^{\prime}$.

So every finite subset of ${\text{\mathbf{T}}}_{\mathrm{\infty}}$ is consistent; by the compactness theorem for first-order logic, ${\text{\mathbf{T}}}_{\mathrm{\infty}}$ is consistent, and by Gödel’s completeness theorem for first-order logic it has a model $\mathcal{M}$. Then $\mathcal{M}\vDash {\text{\mathbf{T}}}_{\mathrm{\infty}}\u22a2\text{\mathbf{T}}$, so $\mathcal{M}$ is a model of T with infinitely many elements ($\mathcal{M}\vDash {\mathrm{\Phi}}_{n}$ for any $n$, so $\mathcal{M}$ has at least $\ge n$ elements for all $n$). ∎

Title | creating an infinite model |
---|---|

Canonical name | CreatingAnInfiniteModel |

Date of creation | 2013-03-22 12:44:29 |

Last modified on | 2013-03-22 12:44:29 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 9 |

Author | CWoo (3771) |

Entry type | Example |

Classification | msc 03B10 |

Classification | msc 03C07 |

Related topic | CompactnessTheoremForFirstOrderLogic |

Related topic | GettingModelsIModelsConstructedFromConstants |