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?


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.

  • $\begingroup$ 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

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