First order logic, finite partially ordered structure problem

Question

The exercise below is part of an exam for a Logic course. There were other questions leading up to this, but I've included their "result" in the wording below.

Suppose alphabet $\mathcal{L} = \lbrace <, =\rbrace $, where $<$ is a 2-place relation.

Suppose also, that every structure $\mathcal{A}$ of $\mathcal{L}$ is strictly partially ordered ($<^\mathcal{A}$ is irreflexive and transitive).

1. Find a proposition $\phi$ such that every structure $\mathcal{A}$ of $\mathcal{L}$ that satisfies it, has at least n elements where $n \in \mathbb{N}$. (I've done this)
2. Using the compactness theorem prove that there is no set of propositions $\Sigma$ such that if $\mathcal{A}$ is a structure of $\mathcal{L}$ then

$$\mathcal{A} \text{ model of } \Sigma \Leftrightarrow \mathcal{A} \text{ is finite strictly partially ordered}$$

Any ideas for the second part will be very appreciated :)

Answer 1

Let $\phi_n$ be a proposition satisfied precisely by those structures with at least $n$ elements. Consider $$\Sigma' = \Sigma \cup \bigcup_{n\in \mathbb{N}} \phi_n$$

as a new family of propositions. Any model of $\Sigma'$ is also a model of $\Sigma$. Can you show that any model of $\Sigma'$ is infinite, and also, using the Compactness Theorem, that models of $\Sigma'$ exist?

Answer 2

Suppose for reductio there were such a set $\Sigma$ whose models were exactly the finite strict partial orders. Consider the set:

$$\Gamma = \Sigma \cup \{\psi_n \mid \psi_n \text{ says, "there exist at least $n$-many things" for } n \in \mathbb{N}\}$$

(You should try to make that explicit if you don't know how to do that already). Consider now any finite subset $\Gamma' \subset \Gamma$. There will be a maximum $m \in \mathbb{N}$ such that $\psi_m \in \Gamma'$ (why?). Hence, a model of $\Gamma'$ will be the partial order of size $m$ (since this is a model of $\Sigma$ and has at least $m$ elements). Hence, every finite subset of $\Gamma$ has a model. So by Compactness, $\Gamma$ has a model, which is also a model of $\Sigma$ by construction. But this model of $\Gamma$ has to be infinite, and hence $\Sigma$ has an infinite model, contrary to supposition. Hence, there cannot be such a $\Sigma$.