This is (a translation of) an excerpt from a model theory textbook that shows that $2\mathbb Z$ is not a definable set in the structure $(\mathbb Z, 0, S, <)$, where $S$ is the successor function.
Suppose $2\mathbb Z$ is defined by a formula $\phi(x)$. Then let $\mathcal M \succ (\mathbb Z, 0, S, <)$ be a proper elementary extension and $D$ be the set defined by $\phi(x)$ in $\mathcal M$. In $\mathbb Z$, odd numbers and even numbers appears in turn. A similar property holds in $\mathcal M$, thus (1): $a\in D \Rightarrow S(a) \not \in D$. Let $\sigma : \mathcal M \rightarrow \mathcal M$ be a map defined by $a \mapsto a\ (a \in \mathbb Z)$ and $a \mapsto S(a)\ (a \not \in \mathbb Z)$. Then $\sigma$ is an isomorphism on $\mathcal M$. By (1), $\sigma$ does not preserve $D$. Thus $2\mathbb Z$ is not definable in $\mathbb Z$.
This uses the following proposition:
Suppose $A\subset |\mathcal M|^n$ is definable. Then for every isomorphism $\sigma$ on $\mathcal M$, $\sigma(A) = A $.
What I don't understand is the argument that $\sigma$ does not preserve $D$. I suppose this argument assumes $D\setminus\mathbb Z\neq\emptyset$. This intuitively holds, because the language is not expressive enough to exclude non-integers from $D$ (I'm thinking of $\mathbb R$ as $\mathcal M$). But I am unable to show this.
How can you prove that $D\setminus\mathbb Z\neq\emptyset$?