# proof of Vaught’s test

Let $\phi $ be an $L$-sentence^{}, and let $\mathcal{A}$ be the unique model of S of cardinality $\kappa $. Suppose $\mathcal{A}\models \phi $. Then if $\mathcal{B}$ is any model of $S$ then by the upward (http://planetmath.org/UpwardLowenheimSkolemTheorem) and downward Lowenheim-Skolem theorems, there is a model $\mathcal{C}$ of $S$ which is elementarily equivalent to $\mathcal{B}$ such that $|\mathcal{C}|=\kappa $. Then $\mathcal{C}$ is isomorphic^{} to $\mathcal{A}$, and so $\mathcal{C}\models \phi $, and $\mathcal{B}\models \phi $. So $\mathcal{B}\models \phi $ for all models $\mathcal{B}$ of $S$, so $S\models \phi $.

Similarly, if $\mathcal{A}\models \mathrm{\neg}\phi $ then $S\models \mathrm{\neg}\phi $. So $S$ is complete^{} (http://planetmath.org/Complete6).$\mathrm{\square}$

Title | proof of Vaught’s test |
---|---|

Canonical name | ProofOfVaughtsTest |

Date of creation | 2013-03-22 13:00:44 |

Last modified on | 2013-03-22 13:00:44 |

Owner | Evandar (27) |

Last modified by | Evandar (27) |

Numerical id | 4 |

Author | Evandar (27) |

Entry type | Proof |

Classification | msc 03C35 |