No theory such that models are just the ones who have finite amount of left-dense points (approach validation)

Question

The problem

Let $(X, \leq_X)$ be a linear ordening. We call $x \in X$ a left-dense point if for every $y \in X$ met $y <_X x$ there exists a $z \in X$ sucht that $y <_X z <_X x$.

Let L be a language with just a binary relationsymbol $<$. Show there is no L-theory such that the models of this theory are just the linear ordenings with a finite number of left-dense points.

My attempt

Suppose there is such a theory T. Then call the language L' = L $\cup$ {$c_n$ for $n \in \mathbb{N}$}. Now call the theory T' = T $\cup (\cap_{i \in \mathbb{N}} \forall y : y < c_i: \exists z : y < z < c_i$). This theory has to be inconsistent because of the assumption that is made. Now we get a contradiction: every subtheory of T' has a model (since it only requires a finite amount of left-dense points). In other words, say the subtheory requires $N \in \mathbb{N}$ amount of left-dense points, then this theory is consistent because of the assumption.

Is this a correct approach? Feedback is appreciated.

Answer 1

Your approach starts well, but quickly deteriorates. The reason is that $\bigcap_{i\in\Bbb N}...$ is not really something that makes sense in first-order logic.

Here's a more general theorem, with a very specific example, and from that I hope that you can see what's the right approach:

Let $\varphi(x)$ be some definable property in some FO language. Then there is no theory $T$ whose models are exactly the models which have a finite number of points satisfying $\varphi(x)$.

The issue here is the combination between "exactly" and "finitely many".

Now let's consider the case where $\varphi(x)$ is just $x=x$. Then this states "There is no theory whose models are exactly the finite sets".

And how would you prove that? Well, with compactness. If $T$ is a theory whose models are exactly the finite models, take $\{c_n\mid n\in\Bbb N\}$ to be new constant symbols and the theory $T\cup\{c_n\neq c_m\mid n<m\}$. Then any finite fragment has a finite model of a suitable size, because $T$ has models of arbitrarily large, but finite size, where we can interpret the finitely many constants as some subset of the model. Therefore by compactness, the whole theory has a model, which has to be infinite. Contradiction.

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So the general idea, yes, you seem to have got it, but you need to be more precise about the theory.