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Write formula in MSO such that for each component in undirected graph there exists some node which satisfy $P$.

My idea is: for each node there exists path to node satisfying $P$.
$\forall_{x} \exists_y [Path(x,y) \wedge P(y)]$

$Path(x,y) = $ there exists finite directed path from $x$ to $y$
$Path(x,y) = \exists_R [x = y \vee [(\text{$R$ is finite directed order}) \wedge (\forall_z R(x,z))\wedge (\forall_z R(z,y))\wedge \phi(x,y)]]$

$\phi(x,y)$ = for each sequent nodes between $x$ and $y$ there exists edge:
$\phi(x,y)=\forall_{z,t}[[R(z,y) \wedge R(x,z)\wedge(\forall_{u}((R(z,u)\wedge R(u,t))) \to (u = z\vee u = t)))] \to \exists E(z,t)]$

$\text{$R$ is finite directed order} = \forall_{u} \neg R(u,u) \wedge (\forall_{u,v} (R(u,v)\vee R(v,u) ) \wedge (\forall_{u,v} (R(u,v)\wedge R(v,u))\to u=v)\wedge ((\forall_{u,v,w} (R(u,v)\wedge R(v,w)))\to R(u,w))\wedge (\text{$R$ is finite relation})$

I believe that finitness of relation can be easily expressed in MSO, but at this moment I would prefer wait for you answer about this stage. I can write such formula using second order: something like: not exists bijection between set and its subset..... I think that it may be difficult without quantyfying over functions..

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I think you meant to ask:

Write a monadic second order formula which is satisfied in an undirected graph if and only if each component of the graph contains a node which satisfies $P.$

$$\forall X\ [\forall_x(\neg x\in X)\lor\exists_x\exists_y\ (x\in X\land E(x,y)\land\neg y\in X)\lor\exists_x(x\in X\land P(x))]$$

In plain English that says that every nonempty set which is a union of components contains a node which satisfies $P,$ but of course that's equivalent to saying that every component contains a node which satisfies $P.$

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    Is it ok for infinite graphs ? – 2017-02-13
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    I can't see that finite or infinite makes a difference. Why do you think there would be a problem for infinite graphs? – 2017-02-13
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    It is diffucult to think about connected nodes, but infinite path. We should that are reachable each other ? Or not ? What about your formula in this case ? (Before I read it I would like to know more). What about approach with relation ? – 2017-02-13
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    I'm sorry, I can't understand your last comment. I can't even parse the sentence "We should that are reachable each other?" – 2017-02-13
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    Yes, it is too blur. If $x$ and $y$ are connected only by infinite path - if they are reachable each other ? – 2017-02-13
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    In my understanding, there is no such thing as "connected by an infinite path." A graph is connected if each pair of points is connected by a ***finite*** path. – 2017-02-13
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    I think that I got it. Since, we quantify over all subset, so in particular we quantify over all single components. Then, the only chance that such subset is accepted by formula is last part of alternative. Yeah >? – 2017-02-13
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    bof, I think that neighter finitness of graph (structure) nor connetivity of graph are not expressible in MSO. Moreover, I invite to this thread: http://math.stackexchange.com/questions/2142276/church-rosser-property-in-unidrected-grap-proof-of-existence-formula-in-mso?noredirect=1#comment4406471_2142276 It the best subject for you :) – 2017-02-14
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    A graph $G=(V,E)$ is connected if and only if, for any nonempty sets $X,Y$ of vertices such that $X\cup Y=V$ and $X\cap V=\emptyset,$ there is an edge $xy$ xith $x\in X$ and $y\in Y.$ – 2017-02-14
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    Do you mean somehting like that ? $\forall_X\forall_Y [X\cap Y \neq \emptyset \wedge X\cup Y = V \wedge \exists_{x\in X}\exists_{y\in Y} E(x,y) ]$ And here, my permanent problem: Maybe, it should be: $\forall_X\forall_Y [X\cap Y \neq \emptyset \wedge X\cup Y = V ]\to\exists_{x\in X}\exists_{y\in Y} E(x,y) $ – 2017-02-15