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Let $\Sigma$ be a signature: $\Sigma = \{e, P\}$. Let $\mathcal{A}$ be a class of undirected graphs of the form $\mathfrak{A} = (A, e, P) $ where $P$ is single-argument relation. Prove that for every connected component there exists a such $v \in A$ that $P(a)$

Write a Monadic Second Order formula which defines that property:

My approach:

$$\forall x \exists y (\forall X (\forall u \forall v ( u \in X \wedge e(u,v) \implies v \in X) \wedge x \in X ) \implies y \in X) \wedge P(y)$$

The second version:

$$ \forall X X\text{ is connected component } \implies \exists x x \in X \wedge P(x)$$ $$ X \text{ is connected component is defined: } \forall u,v (u \in X \wedge e(u,v)) \implies v \in X$$

Right?

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    A couple of questions: What prevents $P$ from being the empty relation? Why is case used to distinguish $e$ from $E$, but not to distinguish the unary relation symbol $P$ from its interpretation? – 2017-02-21
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    I edited. I cannot see that $P$ can be empty. – 2017-02-21
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    Why not? What's wrong with $A =\{0,1\}$, $e = \{ (0,1) \}$, $P = \emptyset$? – 2017-02-21
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    Your example doesn't satisfies neither the first approach nor the second. – 2017-02-21
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    I'm afraid we are miscommunicating. I'll try to explain. Your post, as I read it, refers to proving that for an arbitrary predicate $P$, there is vertex $x$ in every connected component such that $P(x)$. That cannot be. The example I gave does not satisfy the condition--on purpose. I'm not discussing (yet) the correctness of your MSO rendition, because I'm still trying to understand what the question is about. – 2017-02-21
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    Yes, you right, we miscommunicated. I would like construct a such $\phi$ in MSO that $G \models \phi \iff \text { every connected component has a such vertex v that } P(v)$. – 2017-02-21
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    Very well. Thanks. I'll answer in a bit. – 2017-02-21

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I'll use the convention that the scope of a quantifier is the parenthesized expression following the quantifier. The first formula, slightly adjusted to conform to that convention, is:

$$\forall x \exists y \,((\forall X \,(\forall u \forall v \,( u \in X \wedge e(u,v) \rightarrow v \in X) \wedge x \in X ) \rightarrow y \in X) \wedge P(y)) \enspace. $$

It says that for every vertex $x$ there is a vertex $y$ such that $P(y)$ holds and, for every subset $X$ including $x$ with a certain property $C(X)$ to be discussed, $y$ is also in $X$.

$C(X)$ is not "$X$ is a connected component of the graph." Rather, it is "$X$ is closed under the edge relation $e$." However, every (maximal) connected component is closed under the edge relation, and every set of vertices that is closed under the edge relation consists of an integral number of connected components. Hence $C(x)$ is adequate for the task at hand.

Bottom line: This sentence is OK.

The second sentence is, with minor adjustments,

$$ \forall X \,(\text{ConnectedComponent}(X) \rightarrow \exists x \,(x\in X \wedge P(x))) \enspace. $$

It is obviously OK if we come up with a way to write a formula with one free variable $\text{ConnectedComponent}(X)$ that is true of a set of vertices $X$ if and only if $X$ is a connected component of the graph. One solution is this:

$$ \forall Y \,((Y \neq \emptyset \wedge Y \subseteq X \wedge Y \neq X) \rightarrow \exists u \exists v \,(u \in Y \wedge v \in (X \setminus Y) \wedge e(u,v)))\enspace. $$

This formula says that no matter how you split $X$ into two nonempty subsets, there's an edge of the graph that joins the two subsets.

The subformulae like $Y \neq \emptyset$ and $Y \subseteq X$ are obviously abbreviations of formulae like $\forall v \,(\neg (v \in Y))$ and $\forall v \,(v \in Y \rightarrow v \in X)$.

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    What about "X is connected component is defined: $\forall u,v (u\in X\wedge e(u,v))\implies v\in X", is it ok? – 2017-02-21
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    $\forall x,v \, ((u \in X \wedge e(u,v)) \rightarrow v \in X)$ means that $X$ is disconnected from the rest of the graph, not that it is a connected component. The union of a collection of maximal connected components satisfies this property. – 2017-02-21
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    "The union of a collection of maximal connected components satisfies this property". That is true. But, in our example it satisfies us. We demand, every connected component has a such vertex $v$ that $P(v)$. That formula is satisfied by: (1) every connected component and (2)the union of a collection of maximal connected components satisfies this property Because in (1) a formula enforces that every connected component has a expected vertex ( I mean $P(v)$) it means that also (2) is satisfied. – 2017-02-21
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    I agree. If you read my comment on your first formula, I believe you'll find that we are saying the same thing. It's just that if you insist on characterizing a *connected component*, as in the second formula, you need a different formula. – 2017-02-21
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    Oh, I forget. Let's consider a simple formula MSO: $\forall x \forall y \forall X$. My intention is that we quanitfy firstly by elements, then by set ( x,y represents elements, X represents set). Is it allowed in MSO (SO) to write a such formula? I mean order of quantification. Firstly, I quantify by first order elements, then I quantify by second order elements. Is it ok? – 2017-02-21
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    If I understand the question, the answer is yes. There is no restriction on the order or alternation depth of quantification, unless you deliberately choose to impose one (as in existential SO, which is of interest in descriptive complexity theory). So, it is allowed. – 2017-02-21
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    you understand well. Thanks! – 2017-02-21