Prove that the language of only constant symbols is complete

Question

Take the language made up of only constant symbols and no function or relation symbols $L = \{c_1\ne c_2, c_2 \neq c_3, c_1 \neq c_3, ... \}$. I want to show that $L$ is a complete theory.

The way I think this should be done is to first show that $L$ has countably many non-isomorphic countable models and then secondly show that there is a countable model of $L$ into which all countable models can be elementary embedded. And then lastly we can conclude that $L$ is complete. Can somebody just write down how the first part is done?

Answer 1

This should really be a comment, but it's too long:

What you describe is not the easiest way to answer the question. Here are a couple approaches which will go quicker:

• Isomorphisms of countable reducts. Suppose $\mathcal{M,N}$ are countable models of $L$ and $\varphi$ is a sentence in the language of $L$; we'll show that $\mathcal{M}\models\varphi\iff\mathcal{N}\models\varphi$. Since $\varphi$ is a single sentence, it only involves finitely many constant symbols $c_1,...,c_n$. Let $\mathcal{M}',\mathcal{N}'$ be the reducts of $\mathcal{M},\mathcal{N}$ respectively to the language consisting of just these finitely many constant symbols. We have $\mathcal{M}'\models\varphi\iff\mathcal{M}\models\varphi$ and similarly for $\mathcal{N}'$ and $\mathcal{N}$. With a bit of thought, you can show that $\mathcal{M}'\cong\mathcal{N}'$ and this finishes the problem.

  • This works for models of $L$ of the same cardinality, not just countable ones, but it might be helpful to restrict attention to countable models when they suffice.

• Uncountable categoricity. Rather than look at the countable models, think about the models of size (say) $\aleph_1$. It's not hard to show that there is exactly one model up to isomorphism of cardinality $\aleph_1$, and by the Lowenheim-Skolem theorem this means that $L$ is complete.

Note that both approaches above require understanding what models of $L$, or $L$'s finite fragments, look like. So broadly speaking this is still running into the concern you have in the OP. But I think each of the approaches above, being meaningfully simpler, will help demystify things.

On that note, here are a couple hints to get started:

• Suppose $\mathcal{M}\models L$. Then $\mathcal{M}$ has a particular distinguished countable subset, namely the set $\{c_i^\mathcal{M}: i\in\mathbb{N}\}$ of elements named by constant symbols. Can you show that this does not have to be all of $\mathcal{M}$?

• After doing so, you have that any model of $\mathcal{M}$ splits into two pieces, the piece above and everything else; what properties of these pieces determine $\mathcal{M}$ up to isomorphism?