TITLE: Models and Sets
QUESTION [0 upvotes]: I'm asked to find models whose universes are subsets of the reals, where the statement is true in the first model and false in the second.
$$(1) \ \forall x \exists y(x<y)$$
An example answer I have is $\mathbb N \models \forall x \exists y(x<y)$ and $\mathbb Z^- \not\models \forall x \exists y (x<y)$
But I'm struggling to understand the reasoning fundamentally. I think I Intuitively understand it. I acknowledge $\mathbb N \subseteq \mathbb R$, and when I think of $\{0, 1, 2, ..., x\}$ as my subset I see that it is always possible for $x<y$ since $y$ can be found between elements of $\mathbb N$. For example, at $x=3$ $y$ may be $3.5$, so $3<3.5$ checks. But I see there are also examples where some $y$ is not greater than $x$, such as $y=2.2$, then $x=3$ and therefore $2.2\not\ge3$. So $\exists y(y>x)$ holds.
And with the example of the false model, my thinking is that if $x\in \mathbb Z^-$ then $x$ is always less than $y$ for ALL $y$.
Is there a better way I can notate a justification for these answers? Are my interpretations correct?

REPLY [3 votes]: I'm afraid your understanding is not quite right; for instance, in the $<$-structure $\mathbb{N}$, there is no element $3.5$; even though $\mathbb{N}$ is a substructure of $\mathbb{R}$, which does contain $3.5$, by restricting our attention to the (sub)structure $\mathbb{N}$ we are essentially choosing to ignore all the elements of $\mathbb{R}\setminus\mathbb{N}$. In particular, we are choosing to ignore the element $3.5\in\mathbb{R}\setminus\mathbb{N}$.
Also, for example, this sentence

And with the example of the false model, my thinking is that if $x\in\mathbb{Z}^-$ then $x$ is always less than $y$ for ALL $y$.

is not right either; for example, take the element $x=-5\in\mathbb{Z}^-$, and let $y=-7\in\mathbb{Z}^-$. Is $x<y$?
However, your examples are still correct. Let's flesh the question out; what does it mean for the sentence $\forall x\exists y (x<y)$ to hold in an $<$-structure $M$? It means that, for any $x\in M$, we can give a choice of $y\in M$ such that $x<y$.
Let's check whether or not this is true in your examples. First, for $\mathbb{N}$. Let $x\in\mathbb{N}$ be any natural number. Then choose $y=x+1$. Is $y\in\mathbb{N}$? Indeed, for the natural numbers are closed under the successor operation. (For instance, if $x=2$, we can choose $y=3$.) Also, $x<x+1=y$ for any natural number $x$, so this choice of $y$ is a "witness" to the truth of the sentence $\forall x\exists y (x<y)$ in $\mathbb{N}$, and we are done.
Now, what about $\mathbb{Z}^-=\{-1, -2, -3, ...\}$? Well, we wish to show that $\mathbb{Z}^-$ does not satisfy the sentence $\forall x\exists y (x<y)$, and for this it suffices to construct a single counterexample – ie, to give a choice of $x\in\mathbb{Z}^-$ such that, for any $y\in\mathbb{Z}^-$, $x$ is not strictly less than $y$. (Or, restated, a choice of $x\in\mathbb{Z}^-$ such that, for any $y\in\mathbb{Z}^-$, $x\leqslant y$.) I claim $x=-1$ gives such a choice, and is hence a witness to the falseness of the sentence $\forall x\exists y (x<y)$ in $\mathbb{Z}^-$. Can you see why this $x$ is a suitable choice?
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