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Let $f$ will be one-argument function and let $A=⋃_{n∈N}Mod(∀_xf^n(x)=x)$,where $f^n(x)$ is $n$-th composition. Prove that:
(a) $A$ can't be axiomatized with set of first order formulas
(b) complent of $A$ can't be defined with one first order formula

$A$ is class of directed finite cycles of each length $1,2,3...$.
(a)

EF-games approach:
(1) Directed cycle of length $2^{n+1}$ belongs to $A$
(2) Doubly infinite directed chain
It is well known example, strategy for duplicator is easy.
Compactness theorem approach
We extends language by two constants $u$ and $v$. We suppose that $\Delta$ axiomatizing this class exists and add some formulas:
$\{\phi_n\equiv \text{there exists path from u to v of length at least $n$} \ |\ n\in\mathbb{N}\}$
And we can see that it is finitely satisfable, but not infinitely.

Other idea it to add infinitely many constants : $c_1,c_2...$ and add only only one formula: $\forall_{i\in\mathbb{N}}E(c_i, c_{i+1})$
Other approach
Your :)

(b)
I think that to show that $C$ is not definable (=only one formula) it is sufficient to show that complement $C$ is not axiomatizable. Because, I above did show that class is not axiomatizable. So (b) is ready. Look, let assume that complement of $A$ is definable (one formula). Then I can negate it and I can define complement of complement so define $A$.

I ask for checking my solutions and some alternative approach to (a) especially using Skolem-Lowenheim is welcome.

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    Your description of $A$ makes no sense. In particular, your use of the universal quantifier inside your formula for $A$ is incomprehensible. Please revise your question to match the exact wording of the textbook or course notes you are working with.2017-02-19
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    It is exactly rewritten from my textbook. Yes, is not formal and precisely, but we know that it is about class funtion being finie directed cycles2017-02-19
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    @RobArthan: As written it actually makes sense, but does not axiomatize the class of finite directed cycles at all. For example a graph comprising a pair of disjoint 2-cycles will also be a member of $A$. So it's still wrong.2017-02-19
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    @user21820: I've made the necessary guesses to decipher it now and I agree with your analysis.2017-02-19

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