4
$\begingroup$

I read a result of Vaught(a little down the page) that says that there cannot be any first order theory which has exactly two countable models upto isomorphism. Is this not a counter example:

The theory of dense linear orders(DLO) has 4 countable models upto isomorphism. They are the intervals $[0,1]$, $(0,1)$, $[0,1)$, and $(0,1]$ (restricted to the rationals). If to DLO we add the statement that there is a lower bound, do we not end up with exactly two of these 4 theories?

Am I misunderstanding Vaught's result or making a mistake with the DLO part?

$\endgroup$
5
$\begingroup$

Vaught's result says that a complete theory cannot have precisely two countable models (up to isomorphism). The theory you suggest does have precisely two models, but it is not complete.

$\endgroup$
  • $\begingroup$ Oh yes! I completely missed that. Thank you very much $\endgroup$ – Asvin Jun 9 '14 at 15:38
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Jun 9 '14 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.