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Question I want to ask for some more hints for point 2. since I'm stuck. And whether or not the reasoning I made for point 3. makes sense since I'm not totally convinced. I know there is a similar question in There exists no zero-order or first-order theory for connected graphs but I did it in a different ways.

Let $ L = \{ R \} $ be language of the first order containing the symbol $=$ and a binary relation $R$, et $ \mathcal{M} = \left< M, R^{\mathcal{M} } \right>$ an $L$-structure. We say that a graph is non directed if $R^{\mathcal{M}} $ is symmetric and irreflexive. Moreover we say that a graph is connexe if for all $x,y \in \mathcal{M} $ we have that there is a chain connecting $x$ to $y$, i.e. there exist a positive integer $n \in \mathbb{N} $ such that $s_0 =x$ and $s_n=y$ and for all $i < n$ we have $(s_i,s_{i+1}) \in R^{\mathcal{M}} $.

  1. Prove that $ Z = \left< \mathbb{Z}, R^{Z} \right> $ with $R^{Z} = \{ (a,b) : \left| a-b \right| = 1 \} $ is a connected graph.
  2. Use a compactness argument to prove that there exists a countable unconnected graph that is elementarily equivalent to $Z$.
  3. The class of connected graphs is axiomatizable? The class of nonconnected graphs is axiomatizable?

For 3.: By definition the class $C$ of the connected unidrected graph is axiomatizable by a theory $T$ if and only if for any models $ \mathcal{M} \in C$ we have that $\mathcal{M} \models T$, let $ \mathcal{M}_2 $ the model of point 2, we have that is elementarily equivalent to $Z$ hence $ \mathcal{M}_2 \models \varphi \Leftrightarrow Z \models \varphi $, in particular for all $\varphi \in T$ hence $ \mathcal{M}_2 \models T $, resulting that is a connected graph, a contradiction. Similarly for the class of unconnected graphs.

3m0o
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    For 2, you should add constant symbols $a$ and $b$ to your language and add, for each $n$, an axiom saying there’s no path between $a$ and $b$ of length $n$. Apply compactness and downward Lowenheim-Skolem. – Mark Saving Jan 25 '22 at 23:00
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    Also for (3), your definition of a class being axiomatized is incorrect. A class of structures $C$ is axiomatised by a theory $T$ if and only if $C = {\mathcal{M} \mid \mathcal{M} \models T}$. In other words, for all structures $\mathcal{M}$, we have $\mathcal{M} \in C$ if and only if $\mathcal{M} \models T$. But your logic of using models $Z$ and $\mathcal{M}_2$ which are elementarily equivalent, but where one is connected while the other isn't, is sound. – Mark Saving Jan 26 '22 at 00:30

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