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Prove that it is not possible in the FO logic to define that there exists a such connected component $G_0 \in G$ where $G$ undirected graph that for every $v \in G_0 P(v)$ where $P$ is any single-argument relation-symbol.

So we consider structures-graphs: $\mathbb{A} = (V, E, P) $.

  1. From my intuition it's possible for finite graphs, yes?

  2. My solution:

Let's assume that there exists a such set of sentences $\Delta$ that $G \models \Delta \iff \text{ There exists a such connected component } G_0$ that for every $v \in G_0 P(v)$

$V$ is a universum- it is a set of vertexes.

Let's assume that considered graph is countable. We don't lose a generality. So, every vertex can be labeled with natural number.

Let $\Delta' = \Delta \cup \Gamma, \Gamma = \{E(v_i, f) \mid i \in \mathbb{N}\}$ and $f$ is a constant (argumentless function) which we add to signature. $f$ points to a such vertex $v$ that $\neg P(v)$.

Let $\Delta_0 \subset \Delta' \text { and } \Delta_0 \text { is finite }$ Note, that $\Delta_0 $ is satisfable because it is easy to point a model.

From compactness theorem we have that $\Delta'$ is satisfable but it's impossible beacuse of $\Delta$

Is it ok?

What can you say about the larger theory $S=T\cup \{\phi_n : n\in \mathbb{N}\}$?

Why $S$ is finitely satisfable? $S = T \cup N = T \cup \{\phi_n | n \in \mathbb{N} \}$ Let's take a finite subset $\Delta$ of $S$. It is easy for me to show that there is a model for $N \cap \Delta$. But, I have a problem to show that $T \cap \Delta$. Certainly, the crucial fact is that it is finite. However, intuitively I imagine that model looks like:

enter image description here

From compactness theorem we get that $S$ is satisfable.

What can you say about the connected component of c in any model $M$ of $S$?

As we can see from the image.

Now suppose $M\models S$; let $H$ be the reduct of $M$ to $(E,P)$. What can you say about $H$ versus $G$? (HINT: look at their theories)

$H$ is not connected while $G$ is connected. Their theories are equal ( I am not convinced that it is a problem). Perhaps, the grap you've chosen is a special?

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    It is not clear to me if you are asking about "The set of nodes satisfying $P$ is connected" or "there is a connected component for which every node satisfies $P$". The former is a much more common kind of exercise. – 2017-01-21
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    In the solution, could you explain the set $V$? I am not sure how to form $\Gamma$ without knowing the set $V$ first. – 2017-01-21
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    I am asking about: "there is a connected component for which every node satisfies $P$." – 2017-01-21
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    @Carl Mummert, I edited – 2017-01-21
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    Because $\Gamma$ only refers to vertices in $V$, it feels like I can make a model of $\Delta'$ by just making a graph that has one more vertex $w$, not in $V$, not connected to any other vertex, and with $P(w)$. Then $\{w\}$ is a component satisfying $P$, so $\Delta'$ is satisfiable. I am afraid, though, that I don't understand the idea you are trying to use to make the answer. – 2017-01-21
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    @Carl, it is not true. "I can make a model". Let's call your model $\mathbb{M}$ So, you claim that $\mathbb{M} \models \Delta'$. But, it means that $\mathbb{M} \models \Gamma$. So, every vertex in your graph is connected to $f$ because $\Delta'$ says that every vertex in your universum ( $\{ E(v, f) | v \in V\}$) is connected to $f$ where $\neg P(f)$. – 2017-01-21
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    It seems to me that the set of sentences $\Delta'$ was created *before* I made my graph, and I chose $w$ to not be in the set $V$ that was used to create $\Delta'$. If you want to say that every vertex in the graph is connected to $f$, you have to use a universal quantifier: $(\forall n)E(n,f)$. But that is a single sentence, not an infinite set of sentences. Indeed, whatever set $V$ is, I could choose my graph to have a larger cardinality, so there is no way $V$ could refer to every node in my graph. – 2017-01-21
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    You right. I edited. – 2017-01-21
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    THe edit does not resolve the problem, though. It is true that if a graph is countable then the vertices can be labeled with a countable set. However, if you fix the countable set ahead of time (e.g. if the signature has a countable set of constant symbols), there is no way to write a sentence or set of sentences which ensures that every element of the graph is named by one of those constant symbols. We can always make a model with additional elements that are not named by any of the symbols. – 2017-01-21
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    It may help others find a solution if the question mentioned the source of the problem. I suspect the source might have included some examples of similar properties which are not definable. – 2017-01-21
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    Obviously you right. Do you know how to fix it? – 2017-01-21
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    No, I don't see a solution at the moment. It's an interesting problem, because it is not the kind that is usually included in elementary logic texts. – 2017-01-21
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    After some thought, I believe you can show that the property "$G$ has a component for which every node satisfies $P$" is not definable in the signature $(E,P)$ by showing that, if it was definable, then the property "$G$ is connected" would be definable in the signature $(E)$. The second quoted property is known not to be definable in that signature. The main idea is that $G$ is disconnected if and only if there exists a node $w$ and a connected component that does not contain $w$. – 2017-01-21
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    @CarlMummert I don't see how to do that - maybe the predicate $P$ holds on all of the disconnected graph! How will you "see" that it's disconnected? – 2017-01-21
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    @Noah: I posted the method now. The idea is that we can interpret $P$ however we wish - including a syntactic interpretation. – 2017-01-21
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    @CarlMummert Ah, I see. Nice! I'll leave my answer up though, since I think the straight compactness argument is also worth having. – 2017-01-21
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    @Noah: the beautiful fact is that these are really aspects of a single answer, written in very different perspectives (syntactic vs. model theoretic, in some sense) and from different viewpoints. (If you don't see it, note that saying $P = G \setminus \{0\}$ is very similar to interpreting $P(w)$ as $w \not = 0$, and think about how to replace the lemma I used with its proof.) – 2017-01-21
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    @CarlMummert Yes, sorry if my previous comment was unclear - I just meant that the *arguments* (that is, their presentations) are different. Of course it's the same underlying mathematics. – 2017-01-21
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    Yes, I think it is a very pretty example! I did not mean to disparage your answer in any way. – 2017-01-21

2 Answers 2

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Let $G$ be the $(E, P)$-structure consisting of the usual graph structure on $\mathbb{Z}$ (with an edge between $a$ and $b$ iff $\vert a-b\vert=1$ - that is, edges=successivities), where $P$ holds everywhere except at $0$.

Now expand the language by a new constant symbol $c$, and let $\varphi_n$ say "$c$ is at least distance $n$ from any point where $P$ fails." And let $T$ be the $(E, P)$-theory of $G$.

  • What can you say about the larger theory $S=T\cup\{\varphi_n: n\in\mathbb{N}\}$?

  • What can you say about the connected component of $c$ in any model $M$ of $S$?

Now suppose $M\models S$; let $H$ be the reduct of $M$ to $(E, P)$. What can you say about $H$ versus $G$? (HINT: look at their theories . . .)

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    thanks for questions make me to think. But, I have to ask you: 1. What is a reduct? 2. " the connected component of c ". I cannot understand what does it mean connected component of *c*. 3. I have doubts: We have just one constant $c$ and we have a lot of sentences $\phi_n$ saying something else about $c$. For example: $\phi_1$ _says $c$ equals 1_, $\phi_2$ _says $c$ equals 2_ and so on. – 2017-01-21
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    Re: question 1: Suppose I have a structure $S$ in some big language $L$, and I have a sublanguage $L_0\subset L$ that I'm interested in. Then I can also view $S$ as an $L_0$-structure: just "forget" about the symbols in $L$ not in $L_0$! For instance, the *underlying group of a ring* is the reduct of the ring (a $\{+, \times\}$-structure) to the language $\{+\}$. Similarly, if I take one of the structures you're interested in (an $\{E, P\}$-structure) and "forget" the $P$-part, this is the *underlying graph*. When I view $S$ as an $L_0$-structure, I'm taking the reduct of $S$ to $L_0$. (contd) – 2017-01-21
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    Re: question 2: $c$ is a constant symbol. So in any $\{E, P, c\}$-structure, $c$ names an *element* of that structure (that is, a vertex). So the "connected component of $c$" is the connected component of that vertex. Re: question 3, I'm not really sure what your doubts are. $\varphi_n$ says that there is no vertex distance $\le n$ from $c$ which is not in $P$ (remember, the thing named by $c$ is a vertex!); it doesn't say "$c$ equals $n$" (and I'm not sure what that would mean anyways). For example, $\varphi_1$ is the sentence "$\forall x(E(c, x)\implies P(x))$". Does that make sense? – 2017-01-21
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    Yes, it makes sense- thanks. Now- I am trying to answer your questions. – 2017-01-21
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    "What can you say about the larger theory $S=T\cup\{\phi_n:n\in \mathbb{N}\}$?". It is not satisfable because T enforces that there exists an edge between $0$ and $1$ and $\phi_1$ forbids it. – 2017-01-21
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    "What can you say about the connected component of $c$ in any model $M$ of $S$?" Nothing. There is no model because $S is not satisfable. – 2017-01-21
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    @Logic You misunderstand what $\varphi_n$ says. It says that there is no path of length $n$ connecting $0$ and $c$. For instance, consider the expansion of $G$ where $c$ is interpreted as $4$. Then this structure satisfies $\varphi_1, \varphi_2$, and $\varphi_3$. It fails $\varphi_4$, though, since $c$ (that is, $4$) *is in fact* within distance $4$ of a point where $P$ fails. – 2017-01-22
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    Note that $c$ is *some specific point*. $\varphi_1$ (for instance) doesn't say that *every* vertex where $P$ holds is more than $1$ unit away from $0$, it says that *the specific vertex labelled by $c$* is more than $1$ unit away from $0$. – 2017-01-22
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    "What can you say about the larger theory $S=T\cup\{\varphi_n: n\in\mathbb{N}\}$?". There exists only one model: $(E, P, 0)$ "What can you say about the connected component of $c$ in any model $M$ of $S$?" c is 0. And for every vertex it exists path to $0$ because of $T$ ensures it. So, the connected component is just the whole graph. – 2017-01-22
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    "Now suppose $M\models S$; let $H$ be the reduct of $M$ to $(E,P)$. What can you say about $H$ versus $G$?" I see that $H$ and $G$ are same. I see also that theory of $H$ is equal = $T \cup \{ \phi_n | n \in \mathbb{N} \}$ where $\phi_n$ says that there is no vertex distance $\leq n$ from $0$ which is not in $P$. But I don't see what does it mean for us. – 2017-01-22
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    I post that comment because, perhaps, you didn't see my answers. If you just don't want to reply, feel free to ignore it :) – 2017-01-22
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    @Logic Re: your first comment, no that's not correct. I'm not sure how you concluded $c=0$: the whole point of the sentences $\varphi_n$ is that they force $c$ to be *far away from* $0$. I think the following warm-up will make things easier: can you prove that there is a graph which is *not isomorphic to* (the underlying graph of) $G$, but *is* elementarily equivalent to it? This is a standard application of compactness, and it's essentially what is going on in my answer, but with bells and whistles added. – 2017-01-22
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    This is how I concluded that model with $ c = 0 $ satisifies the expanded theory: $\{\phi_n | n \in \mathbb{N} \} = \{\phi_1, ....\} $ $\phi_n $ says that there is no path of length $\le n$ between $0$ and $c$. And it is true for $c = 0$. There is no path between $0$ and $0$ of length $\ge 1$. Therefore, $(E, P, 0)$ satisfies expanded theory – 2017-01-22
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    @Logic That is incorrect, there is such a path: $0E1$ and $1E0$. So there's a path of length $2$. More generally, there's a path of every even length. – 2017-01-22
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    vertices and edges cannot repeat on the path ( definition from Wilson- Introduction to graph theory) – 2017-01-22
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    @Logic Alright, then by "path" I meant a path with repetition allowed (that's the definition I learned btw). Now do you see how to do the problem? Or, if you prefer, add the axiom "$c\not=0$". – 2017-01-22
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    at that moment I am thinking of how to solve a simpler problem you proposed. The second graph is obvious and it is easy to prove it with Ehrenfeucht's game, but using compactness theorem makes me troubles ;) – 2017-01-22
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    I edited my post, please help. – 2017-01-22
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    @Logic There's a reason you're having trouble using compactness alone to show that those two structures are elementarily equivalent - it does take a bit more! The structure I had in mind was the one consisting of (countably) *infinitely many* "$\mathbb{Z}$-graphs". Here Compactness is much easier to apply. – 2017-01-22
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    please give me an advice. – 2017-01-23
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    @Logic Let's go back to the theory in my answer. What does it say about $c$ and $0$ (okay fine, there isn't a symbol "$0$" in the language; by "$0$" I mean the unique element where $P$ fails)? (HINT: let $M$ be a model of that theory; are $c$ and $0$ in the same connected component, in $M$?) – 2017-01-23
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    You mean obviously $S$ theory, right? – 2017-01-23
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    @Logic Yes, I'm talking about $S$. (Note that $S$ includes $T$, and $T$ includes the sentence "$P$ fails at exactly one element," so my using "$0$" as shorthand in my above comment is a benign abuse of notation.) – 2017-01-23
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    on my eye there is no such model: 1. The theory of $G$ enforces connection between neighbouring vertexes. 2. $S$ enforces that there is no path of length $n$ between $0$ and $c$ – 2017-01-23
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    @Logic You seem to think that the theory of $G$ implies the graph of any model is connected, but a standard application of the Compactness theorem is that *no theory can enforce connectedness of the graph of any model!* (Assuming there is no finite upper bound on the diameter of models of that theory, that is.) On that note: can you show that $S$ is finitely consistent? – 2017-01-23
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    It might be a good idea to do the following easier compactness exercise: show that if $R$ is a theory in the language of graphs containing the theory of graphs, and $R$ has an infinite model where every vertex has order $2$ (nothing special about $2$ here, we just need a finite bound), then $R$ has non-connected models. – 2017-01-23
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    Wow, I see that I have a really mess in my head ;) – 2017-01-23
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    "You seem to think that the theory of G implies the graph of any model is connected". Yes, I was convinced that it is so in fact! – 2017-01-23
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    And it was a problem with understanding. – 2017-01-23
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    ok, so, how to know what can say theory of model? How do you know that $S$ theory ( created by expanding of $G$ theory) is not unsatisfable? – 2017-01-23
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    @Logic By showing that every finite subset of it is satisfiable; the consistency of $S$ then follows by the Compactness Theorem. – 2017-01-23
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    Ok, but you have to know what sentences the theory of $G$ contains. – 2017-01-23
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    @Logic No, you don't! HINT: show that any finite subtheory of $S$ has a model whose reduct to $\{E, P\}$ is $G$ itself. Note that to build such a model, you just start with $G$ and then choose where to put $c$. HINT: a finite subtheory can only contain finitely many of the $\varphi_n$s . . . *The moral: showing that a given theory is consistent is often best done by showing how to modify an already-known model of a related theory.* – 2017-01-23
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    thanks for your help. I come back here tomorrow becaue in my country it is almost morning so I have to go sleep. Tomorrow, I rethink everything again and try to do excerise. Good night. – 2017-01-23
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    I've edited my post. – 2017-01-25
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Suppose for a contradiction that $\phi$ is a sentence in the language $(E,P,=)$ such that every graph/predicate pair $(G, E, P,=)$ satisfies $\phi$ if and only if there is a connected component of $G$ for which all the nodes satisfy $P$.

Trivially, a graph $(G,E)$ is disconnected if and only if it has at least two connected components. This is equivalent, in turn, to saying there is some node $v_0 \in G$ such that $(G,E,P,=)$ satisfies $\phi$, when $P(w)$ is interpreted as "$w \not = v_0$".

We will obtain a contradiction using the following result.

Lemma. The property that a graph is connected is not definable in the signature $(E,=)$.

We can use $\phi$ to find a sentence $\phi'$ in the signature $(E,=)$ which is true about a graph $(G,E,=)$ if and only if $G$ is connected.

The sentence $\phi'$ can be taken to be $\lnot (\exists v) \phi[P(w)/(w \not= v)]$ where $\phi[P(w)/(w\not =v)]$ is the formula obtained from $\phi$ by replacing $P(w)$ with "$w \not = v$" everywhere. (We assume that $v$ is chosen to be a variable that does not occur in $\phi$).

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    Ok, I grasp it! :). I understand that $v_0$ is a constant which points to any vertex, yeah? – 2017-01-21