Defining property in MSO.

Question

We say that a relationship $R$ has property $S$ when for $R \subset A \times A$ if for every $a \in A$ every path starts in $a$ has finite length.

Prove that there exists the such sentence $\phi$ in Monadic Second Order that $(A, E) \models \phi $ iff $E$ has property $S$.

Let's consider every paths in any graph $G = (A, E)$:

1) Finite paths of the form: $a_1 \to a_2 \to ... \to a_n$ we encode as word: $a_1a_2a_{n-1}$

2) Infinite paths that starts in $a$ we encode as $a^*$.

Let a langugage ( regular) defined by graph's paths be $L(G)$.

We define a regular language: $L_w = \{ a^* | a \in A\}$. We take a complement $L_w'$ of $L_w$. It is also regular. From Buchi-Elgot we have a formula $\phi'$ defining $L_w'$. Now we define a $\phi$:

$$G \models \phi \iff \forall w \in L(G) \psi(w) $$ where $\psi(w) \iff w \models \phi'$.

By path we consider asequence of vertexes which are all distinct from one another.

Right?

Answer 1

Assuming that MSO means monadic second-order logic:

I don't think you're allowed to make $\phi$ depend on $G$. If you were, you could just take $\phi$ to be $(\forall x)x=x$ or $\neg(\forall x)x=x$ depending on whether $G$ has your property.

Also, unless someone promises that $G$ is finite (which it doesn't like anyone does), you have no reason to think the set of all (finite) paths in $G$ make up a regular language. And the Büchi-Elgot-Trakhtenbrot theorem is certainly not directly relevant, because $(A,E)$ is not a word model.

Instead, Hint. It would be easier to find a $\neg\phi$ that is true exactly in graphs that do have an infinite path. The nodes in such a path have the property that each of them has a successor that is also in the path. Conversely if you have a set of nodes without any dead ends, can you always find an infinite path?