proof of Vaught’s test

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

Similarly, if $𝒜\models \mathrm{\neg }\phi$ 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