4

I came across Los-Vaught's test:

Let $\mathcal{L}$ be countable and $\Sigma$ a $\mathcal{L}$-theory. $\Sigma$ is complete, if

  1. $\Sigma$ only has infinite models.

  2. There is a cardinal number $\kappa$ for which any two models of cardinality $\kappa$ are isomorphic.

In the proof we use 2. just to get elementary equivalence:

Let $A,B \in Mod (\Sigma)$, then $A \equiv A' \cong B' \equiv B$ hence $A \equiv B$.

Wouldn't it be enough to just ask for elementary equivalence in the second statement?

user1868607
  • 5,791
Eriien
  • 211
  • 1
  • 9
  • 1
    You are correct, it would be enough. But in practice, the notion of $\kappa$-categoricity comes more often. I think there was a question on this site a while ago about whether your condition implied $\kappa$-categoricity. – Maxime Ramzi Sep 26 '19 at 14:36
  • @Max The question raised in your comment wouldn't make much sense: Any complete theory with infinite models satisfies the property that all models of size $\kappa$ are elementarily equiavlent (for every infinite $\kappa$), and of course this doesn't imply $\kappa$-categoricity. Is it possible you were thinking of this question? – Alex Kruckman Sep 26 '19 at 16:34
  • @AlexKruckman : you are of course right - I must misremember... but I don't think it was that question you linked to – Maxime Ramzi Sep 26 '19 at 17:33

1 Answers1

5

Yes, you are correct. We can replace "all models of size $\kappa$ are isomorphic" with "all models of size $\kappa$ are elementarily equivalent" and everything goes through fine.

spaceisdarkgreen
  • 58,508