Bipartite graph is axiomatizable in FO.

The class of bipartite graph in logic is axiomatizable. That's a fact.

So, it means that the following reasoning is incorrect:

We'll show that there exists a winning strategy for player 2 in Ehrenfeucht-Fraise's game.

Let's take a two structures:

$k$ is a number of runds. The first one is not bipartite- it has a odd number of vertices. The second one is an infinite graph- it is bipartie because of it easy to point 2-coloring function. It is easy to point strategy when in every $k-$rund EF game II player wins.

What is wrong?

Obviously you are right:

there's no reason to believe that any one of these Gs is a model of the whole set of axioms describing the bipartite graphs.

Let $K$ be a class of structures. Let $A$ be a independent structure on $k$ and $A \not \in K$. Let $B \in K$ depends on $k$ where $k$ is number of runds. Let's assume that we have always a such strategy that a duplicator wins. But, we can also conclude that it doesn't mean that $K$ is not axiomatizable, right? Because, for any $B_k$ it is truth that $B_k \equiv_k A$ but, perhaps, there exists $(k+1)-$rund game in which spoiler wins. Right?

there's no reason to believe that any one of these Gs is a model of the whole set of axioms describing the bipartite graphs.

When we can belive by using E-F games?

If for each $M \in \mathbb{K}$ there is some $N \not \in \mathbb{K}$ such that Duplicator wins the length-n(*) game, then $\mathbb{K}$ is not axiomatizable by $k$-quantifier formulas; in particular, if this is true for every $k$ (as in the case of $\mathbb{K}=${bipartite graphs}), then $\mathbb{K}$ is not finitely axiomatizable. Note that $k$ may still be axiomatizable - e.g. {bipartite graphs} is axiomatizable via the axioms

1. If for each $M \in \mathbb{K}$ there is some $N \not \in \mathbb{K}$ such that Duplicator wins the...

Why each? It is enough we show that one structure from $\mathbb{K}$ cannot be distinguished from structure $S \not \in \mathbb{K}$ by the $k$-quanitifer formula for every $k$.

1. is not finitely axiomatizable

Why finitely? You show that for **every $k \in \mathbb{N}$. So, for every formulas. Probably, the reason is the fact that an axiomatizaion can be $> \aleph_0$

1. So how to show with E-F games that the class is not axiomatizable ( not finitely axiomatizable?)

You did a mess in my head, again... Thanks! :)

• Quite right, the $n$ should be a $k$; I fixed it. I think you didn't finish your edit, though . . . – Noah Schweber Feb 18 '17 at 22:42
• Yes, I didn't finish, I will in the moment. – user376326 Feb 18 '17 at 22:44

Nothing's wrong; but I think you are missing the distinction between axiomatizable and finitely axiomatizable.

Let $Z$ be the graph on the right. What you've shown is that, for each $k$, there is a non-bipartite graph $G$ with $G\equiv_kZ$ (where "$\equiv_k$" means that Duplicator has a winning strategy in the $k$-round EF-game). But as $k$ changes, we have to change $G$ too! So there's no contradiction - there's no reason to believe that any one of these $G$s is a model of the whole set of axioms describing the bipartite graphs.

What's going on here is that the class of bipartite graphs, though axiomatizable, is not finitely axiomatizable - for any finite set $\Delta$ of sentences true of every bipartite graph, there is a non-bipartite graph $G$ with $G\models \Delta$. Interestingly, there's a compactness result here:

If a class $\mathcal{C}$ of structures is axiomatizable but not finitely axiomatizable, then its complement is not axiomatizable.

Put another way:

If $\mathcal{C}$ and $\mathcal{C}^c$ are each axiomatizable, then they are each finitely axiomatizable.

Proof: Suppose $\Gamma_0$ axiomatizes $\mathcal{C}$ and $\Gamma_1$ axiomatizes $\mathcal{C}^c$. Consider the theory $T=\Gamma_0\cup\Gamma_1$. $T$ is inconsistent, so there are finite $\Delta_i\subseteq\Gamma_i$ such that $\Delta_0\cup\Delta_1$ is inconsistent. But then - since $\Delta_0$ is true in every structure in $\mathcal{C}$ - every model of $\Delta_1$ is in $\mathcal{C}^c$! And since $\Delta_1\subseteq\Gamma_1$ and $\Gamma_1$ axiomatizes $\mathcal{C}^c$, every structure in $\mathcal{C}^c$ is a model of $\Delta_1$. So $\Delta_1$ axiomatizes $\mathcal{C}^c$. Similarly, $\Delta_0$ axiomatizes $\mathcal{C}$.

EDIT: The following is a response to the OP's edit:

• If Duplicator wins the $k$-round EF game between $A$ and $B$, then $A\equiv_k B$ - they agree on $k$-quantifer formulas. (Strictly speaking this only works for relational languages, like graphs - a bit of work needs to be brought into things to make it true on the nose for functional languages, like rings.)

• In particular, if Duplicator wins each finite-length game, then $A\equiv B$. Note that this gives us a single game characterization of elementary equivalence: Spoiler plays some $n\in\mathbb{N}$, and then Spoiler and Duplicator play the $n$-length game as usual.

• If for each $M\in\mathbb{K}$ there is some $N\not\in\mathbb{K}$ such that Duplicator wins the length-$k$ game, then $\mathbb{K}$ is not axiomatizable by $k$-quantifier formulas; in particular, if this is true for every $k$ (as in the case of $\mathbb{K}=\{$bipartite graphs$\}$), then $\mathbb{K}$ is not finitely axiomatizable. Note that $k$ may still be axiomatizable - e.g. $\{$bipartite graphs$\}$ is axiomatizable via the axioms $$\psi_k\equiv\mbox{"There is no cycle of length 2k+1},$$ but $\varphi_k$ is a $(2k+1)$-quantifier sentence, so it's no surprise that $\{$bipartite graphs$\}$ has no finite "depth" axiomatization. And in fact this is what your observation proves.

• Finally, it's worth pointing out that there are EF-style games which capture much different properties - specifically, elementary (or bounded elementary) equivalence for logics other than first-order. For example, consider the version of the EF-game where the game continues on for infinitely many moves; this captures elementary equivalence for the infinitary logic $\mathcal{L}_{\infty\omega}$, and when restricted to countable structures captures isomorphism.