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Prove that class of all structures $\mathbb{A}=\langle A, r^{\mathbb{A}}\rangle$, where $r\in \Sigma_1^{r}$ and $|r^{\mathbb{A}}|=|\mathbb{A}\setminus r^{\mathbb{A}}|$ is not axiomatizable.

I must show two structures:
$A_n\in \mathbb{A}$
and
$B\notin \mathbb{A}$, in $n$ rounds these two strucutres shouldn't be distinguishable. Let $A_n=\{1,2,3,...,2n\}$ and $B=\{1,2,3,4,....\}.$
$r^{A_n} = \{1,2,...,n\}$
$r^{B} = \{2,4,6,8,10,...\}$

Strategy for duplicator is easy, Duplicator always copies moves of spoiler. It is possible because here we must only prevent relation $r$. So if spoiler chooses $x$ such that $r(x)$ then duplicator also chooses $y$ such that $r(y)$.

What do you think ? Maybe compacntess or Skolem-Lowenheim is also ok here ?

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Consider the two structures $({\bf Z}, \chi_{2{\bf Z}})$ and $({\bf R},\chi_{\bf Q})$, where $\chi$ indicates the indicator function. The first has $|r| = |A \setminus r|$ and the second does not.

Duplicator chooses b satisfying r if spoiler's choice a satisfies r, b not satisfying r if spoiler's choice a does not satisfy r.

Thus the two structures are elementarily equivalent.

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    +1. Even easier: any countable elementary substructure of the second structure is isomorphic to the first structure, so use Lowenheim-Skolem . . . – 2017-02-17
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    Ok, but what about my solution ? – 2017-02-17
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    @NoahSchweber I am not sure if I correctly understand your approach. These two structures proposed by author of this answer: here isomorphism must only prevent relation $r$ what is very simple. But I don't know how Skolem-Lowenheim help you here. – 2017-02-18
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    Consider $({\bf R}, \chi_{\bf Q})$. By downward Lowenheim Skolem, there is a countable model elementarily equivalent to it. This countable model has $|r| = |A\setminus r|$ but the original structure did not. – 2017-02-20