Define a structure $M\neq N$ such that $M \vDash \forall x(y < x \to \exists z(z < y))$ Let $N$ be the natural numbers with the standard order: $N = (N, <)$.
Define a structure $M \neq N$ such that $M \vDash \forall x(y < x \to \exists z(z < y))$
Not entirely sure what the above exercise is asking for due to the "Define a structure $M \neq N$".
Does it want me to define a structure where $M$ where $M$ (meaning $x, y$ and $z$) is not a natural number? From what I can tell the statement is not true for natural numbers regardless since if $x =2$, $y =1$ and $0$ is not a natural number.
 A: Notice that $\forall x(y < x \rightarrow\exists z(z < y))$ is still an open formula with the variable $y$ free. Thus, $y$ assumes a parametric presence. For instance,  if $y$ is given a fixed interpretation as $2$, we get
$$\forall x(2 < x \rightarrow\exists z(z < 2))$$
We consider the equivalent formula $\exists x(2 < x) \rightarrow\exists z(z < 2))$ and immediately see that the consequent is always satisfied. Hence, if $y$ could be assigned uniformly any value other than $0$ (most, if not all, of the widely known books in model theory count $0$ as a natural number), the structure $N$ would fulfil the task. For each such value of $y$, we would have a set of formulas satisfied by $N$. To make the matter more solid, I adapt the following definitions from Chang and Keisler's Model Theory (3rd edition, p. 77) to the present case:
Let $\Phi$ be a set of formulas of a first-order language $\mathcal{L}$ in the free individual variable $y$ such that every formula $\phi(y)$ in $\Phi$ contains at most the variable $y$ free. Then, we denote the statement that $a\in$ M (the domain of $M$) satisfies $\phi$ in $\Phi$ by
$$M\models\phi(a)$$
and the statement that $a\in$ M satisfies every $\phi$ in $\Phi$ by
$$M\models\Phi(a)$$
The exercise asks us to set out a new structure distinct than $N$ to accomplish this idea. A straightforward method is to expand the signature of the structure $N$ with a new constant symbol $y$ (for the sake of simplicity; also delete it from the list of variables against collisions) and fix its interpretation to the constant, say, $1$ in the domain. Hence, we obtain
$$M = \langle\mathbb{N}, <, y\rangle$$
which fulfils the requirements of the question. Sure, other methods are possible as stated in the comments.
A: I don't see any requirement that the universe of $M$ has to be all natural numbers, and we're assuming you mean that $M$ is supposed to be a substructure of $(\mathbb{N}, <)$, not just any structure distinct from it.
Note that the formula is equivalent to: $(\exists x)\,y < x \to (\exists z)\,z < y$, which says "if there are elements greater than $y$, then there are elements less than $y$".
Given these assumptions and observations, you can just take $M = (\{1\}, <)$, which satisfies the formula.
Aside: $0$ is a natural number. Some authors define $\mathbb{N}$ to begin with $1$, but this is wrong-headed :) For one thing, that structure is not a model of Peano arithmetic.
