Alternative proofs that connectedness is not a first-order property of simple graphs

Question

Are there proofs that connectedness is not a first-order property of simple graphs besides the two proofs (or perhaps one proof) shown below?

For motivation, I'm interested in techniques for showing that properties are not first-order. I know essentially two tricks for doing this: building a theory that can distinguish infinite cardinals and invoking Löwenheim–Skolem, and basically forming two sets of sentences that enforce mutually exclusive conditions but are collectively finitely satisfiable and invoking compactness.

I'm curious whether there are other ways to do this, and am picking a concrete and easy to understand condition as a test case.

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Let's define a simple graph as a model of the $(R)$-theory $\{\forall x (\lnot xRx), \forall x y (xRy \to yRx)\}$.

I proved this earlier today using the following argument:

• Lemma: Any simple graph where every vertex has degree $2$ is a disjoint union of circular paths of length $3$ through $\omega$ inclusive.
  • As proof of the lemma, suppose a graph $G$ where every vertex has degree $2$ is connected and strictly larger than $\omega$. Pick an arbitrary point $c$ and begin building the circular path containing it; this path has size $\omega$. That means that there must exist a point $d$ outside the circular path generated by $c$ and therefore that $G$ is disconnected, contradicting the hypothesis
• Suppose for contradiction there is a first-order $(R)$-theory $\Phi$ that holds of $G$ if and only if $G$ is connected.
• The sentence $\alpha \;\textbf{is}\; \forall x y z w (xRy \land xRz \land xRw \to y=z \lor z=w \lor y =w)$ holds if and only if the degree of every vertex of a simple graph is at most two.
• The sentence $\beta \;\textbf{is}\; \forall x (\exists y z (y \neq z \land xRy \land xRz))$ holds if and only if the degree of a simple graph is at least two.
• The theory $\{\alpha, \beta \} \cup \Phi$ holds of a graph if and only if it is connected and every vertex has degree two.
• The theory $\{\alpha, \beta\} \cup \Phi$ has a countable infinite model.
• The theory $\{\alpha, \beta \} \cup \Phi$ has no uncountable models, by the lemma.
• This contradicts the Löwenheim–Skolem theorem.
• Therefore $\Phi$ does not actually exist and connectedness is not a first-order property.

This argument seems (to me at least) to be slightly different than the one(s) I've been able to find on this site which invoke compactness instead (although upward LS is a corollary of compactness, so it's entirely possible that my proof is just a more complicated version of the following proof):

• Add two points $c$ and $d$ to a connected graph that are not connected by a path of any finite length and invoke compactness
  • This is a proof of a stronger statement that is the same as the above argument when restricting the language to the language of graphs.
  • This answer is also the same proof as the above.

Answer 1

Here's another way to do it.

Let $P$ be the theory of Peano Arithmetic and $T$ be the $(R)$-theory of a single connected graph. Let $\Sigma$ be the combined signature of $(R)$ and the language of Peano Arithmetic.

• Let $\alpha$ be the sentence $\forall x y (s(x) = y \leftrightarrow xRy)$.
• The theory $\{\alpha\} \cup P \cup T$ is categorical; $\mathbb{N_S}$, an extension of $\mathbb{N}$, is its sole model.
• The theory $\text{Th}(\mathbb{N}_S)$ is complete, consistent, recursively axiomatizable, and extends PA, contradicting Gödel's first incompleteness theorem.
• Therefore $T$ does not exist and is there no first-order $(R)$-theory of connected grahps.