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Write a such senetence $\phi$ in the first order logic that: $$G \models \phi \text{ iff } G \text{ is connected} $$

We consider only finite graphs. When it comes to infinite graphs it is obvious that FO cannot express that fact.

$$\forall p_1 \, \forall p_2 \, p_1 \neq p_2 \implies \exists n \, \exists p_2 \, \cdots \, \exists p_{n-1} \bigwedge_{1 \le i < n} e(p_i, p_{i+1})$$

Is correct? Why, why not?

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    In FOL you cannot quantify on the "lenght" of a formula ($\exists n$)... – 2017-01-23
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    I think that the class of *finite connected graphs* is not axiomatizable in FOL... – 2017-01-23
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    @MauroALLEGRANZA, but why, I was learnt that a formula must be of finite length but the length of my formula is finite because a graph is finite. – 2017-01-23
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    As Mauro stated, the length of your formula "varies with $n$", which shows that it's not a real formula. Moreover, assume you have such a formula $\phi$. Then by an easy compactness argument, it's obvious that there will be infinite graphs that satisfy $\phi$, so again as Mauro stated, the class of finite connected graphs is not axiomatizable in FOL. – 2017-01-23
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    @Max, I agree that the class of finite graphs is not axiomatizable. But, I cannot see how do you get it from the compacntess argument. After all, compactness is expressed for set of sentences, not for models ( graphs are models in fact). – 2017-01-23
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    It's well known that if a theory (necessarily consistant with what follows) has arbitrarily large finite models, then it must have an infinite model. Compactness is enough, as you can just consider constant symbols $\{c_n, n\in \mathbb{N}\}$ and add to your theory the axioms $c_n \neq c_m$ whenever $n\neq m$, which gives you a finitely satisfiable theory, and therefore a satisfiable theory. A model of this theory is a model of the previous one, and must be infinite. – 2017-01-23
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    @Max, I don't understand, please explain: "you can just consider constant symbols $\{c_n,n\in \mathbb{N}\}$ and add to your theory". Why do you take infinitely many constans while you have only finite universum ( after all, constants have to point to elements from universum). What do you mean by "add to **your** theory". I agree that it is finitely satisfable. But, I cannot see how do you know that it implies that infinite model satisfies our sentence ( theory). – 2017-01-24
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    Let's do it generally. Let $T$ be a consistant theory expressed in the language $L$. Assume $T$ has finite models of arbitrarily large size. Then consider the language $L'$ to which you've added different constant symbols $\{c_n, n\in \mathbb{N}\}$, which we can assume weren't in $L$. Then consider the theory $T^*$ expressed in $L'$ that consists of the theory $T$ + the axioms $c_n \neq c_m$ for $n\neq m$. Then, according to our assumption, $T^*$ is finitely satisfiable (is that what you have trouble seeing ?). Therefore, through compactness, $T^*$ is satisfiable. – 2017-01-24
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    I let you see how that's enough to show that $T$ has an infinite model – 2017-01-24
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    "Assume T has finite models of arbitrarily large size." Now, I cannot see why it is necessary. "let you see how that's enough to show that T has an infinite model " Probably it is connected with my the first question. I guess that the first one implies that it is possible to get infinite universum for $T$. – 2017-01-24

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You made some basic errors. The question is whether or not the following holds:

There is a single first-order sentence $φ$ in the language of graphs such that every finite graph $G$ satisfies $φ$ iff $G$ is connected.

You cannot create $φ$ after being given $G$. Rather you must write one $φ$ down that works for every finite graph $G$. Secondly, quantifiers are interpreted (in any structure) to quantify over all objects in that structure, so you cannot quantify over natural numbers. Thirdly, you cannot have "$\cdots$" anywhere in your formula. In general "$\cdots$" is forbidden in any rigorous mathematics; if you cannot express it without ellipsis then it is not rigorous.

After you understand the above, then you can try your best to construct a desired $φ$, to get a feel for why it is impossible. Then carefully study the proof of the compactness theorem to see why. And you may wish to try the four problems in this question and the two answers.