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Let in sentence $\phi$ there are only one-argument relations. Prove that $\text{Spec}(\phi)$ or $\overline{\text{Spec}(\phi)}$ is finite, $\overline{A}$ means complement of $A$.

I thnk think that if sentence $\phi$ is not contradictory then $\text{Spec} = \{0,1,2,3,4,....\}$. Simply, we always can interpret each unary symbol. For example, lets look at example:
$X = \{a_1, a_2, a_3\}$, $|X|=3$.
Now, we have a full control on each unary relation - I mean that if we could make $\exists_x (x)$ to be true then we can interpret $p$ as $p(x) \leftrightarrow x\in\{a_1, a_2, a_3\}$. Of course we can also interpret $\forall_x p(x) $ as false: $p(x) \leftrightarrow x\notin\{a_1, a_2, a_3\}$.
As you can see we can guarantee that $3\in\text{Spec($\phi$)}$;
However, if $\phi$ is contradictory then $\text{Spec($\phi$)}=\emptyset \neq \mathbb{N} = $$\overline{\text{Spec}(\phi)}$

What do you think about this solution ?

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That solution is not correct: consider a sentence like $$\forall x\forall y\forall z\exists w(w\not=x\wedge w\not=y\wedge w\not=z);$$ the spectrum of this statement is exactly $\{4, 5, 6, ...\}$, and this is a sentence in the empty language.

More generally, for any finite set $F\subseteq\mathbb{N}$, we can find a sentence $\varphi_F$ whose spectrum is exactly $\mathbb{N}\setminus F$; and similarly we can get a sentence $\psi_F$ whose spectrum is exactly $F$.

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    But your sentence uses variable, it can use only one-argument relation. And the problem is that I don't understand sense of this assumption. – 2017-02-13
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    @HaskellFun "One-argument relation" means that the language consists entirely of unary predicates. (Note that "$=$" is not part of the language; it's considered a logical symbol.) – 2017-02-13
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    Moreover, I have no idea how to solve it – 2017-02-13
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    @HaskellFun Are you familiar with quantifier elimination? – 2017-02-13
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    No, I don't - only EF-games and compactness. – 2017-02-13
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    @HaskellFun Well, you can solve this by using a modification of EF-games - have you seen the EF-like game associated to a *single* first-order sentence? (See [this](https://en.wikipedia.org/wiki/Game_semantics#Classical_logic).) Incidentally, what text are you using? I'm curious about what order it introduces various topics in . . . – 2017-02-13
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    Simply, I follow my lecturer :) – 2017-02-13
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    Is it difficult to solve it with this modification ? Could you hide in your post? – 2017-02-13
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    As I said, I don't understand why this assumption is helping here – 2017-02-13
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    The general fact *More generally, for any finite set $F⊆N$, we can find a sentence $φ_F$ whose spectrum is exactly$ \mathbb{N}∖F$; and similarly we can get a sentence $ψF$ whose spectrum is exactly $F$* it is obious because we can express in FO: the set has at least/at most $n$ elements, however the exercise is hard, – 2017-02-15
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    Can you try make it easy :D ? – 2017-02-16
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    @HaskellFun Not really - I don't think it *is* very easy, especially without quantifier elimination. I think this is definitely one where talking to your lecturer would be helpful - they can give some suggestions as to how you should attack it (all I can suggest are tedious kludges, like the game semantics thing above). – 2017-02-16
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    ok, thanks! :-) – 2017-02-16
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    here you are welcome :) http://math.stackexchange.com/questions/2147662/prove-that-formula-in-monadic-second-order-logic-exists-for-each-node-path-is – 2017-02-16