In the structure $\langle \mathbb{Q}, < \rangle$ which of the ZF axioms hold?

Question

In the structure $\langle \mathbb{Q}, < \rangle$ which of the following axioms hold? How about when we use the weak versions of the axioms (all $\leftrightarrow$ replaced with $\rightarrow$ )?

Extensionality, Empty Set, Power Set, Infinity

Find an instance of Separation that is true in $\langle \mathbb{Q}, < \rangle$, and on that is false.

What I'm struggling with is: What does the $\in$ in the axioms mean in$\langle \mathbb{Q}, < \rangle$? I assume it has to mean

In which case for the strong axioms, I'd say the Extensionality, Power Set and Infinity holds and Empty set doesn't as there's no minimal element, is that right? For the weak versions, just the same ones?

And what is meant by "An instance of Separation"?

Answer 1

You have to interpret $\in$ with $<$. In order to check whether extensionality holds you have to consider the sentence

for all $x$ and $y$, $x=y$ if and only if, for every $z$, $z<x$ if and only if $z<y$

or, in symbolic form, $$ \forall x\forall y (x=y \leftrightarrow (\forall z(z<x\leftrightarrow z<y))) $$ where the universe is, of course, $\mathbb{Q}$.

It's easy to see this sentence is true.

On the other hand, it's also true that $$ \forall x\exists y(y<x) $$ so the interpretation of the empty set axiom is false.

Can you go on?

You can notice that the relation “$x$ is a subset of $y$” is interpreted into $x\le y$. The power set of $x$ is a set $y$ such that, for every $z$, if $z\subseteq x$, then $z\in y$. So, given $x\in\mathbb{Q}$, you should find $y$ such that, for all $z$, if $z\le x$ then $z<y$. Note that such an element exists, but it's not unique: $x+1$ and $x+2$ both work. Can you conclude something about the power set axiom?

For the axiom of separation, which is an axiom scheme, you have to find a case where it holds and one where it doesn't.

Answer 2

One easy way to see this is false is that the $\sf ZF$ axioms prove that $\in$ is not transitive on every set. Namely, there exists $a,b,c$ such that $a\in b$ and $b\in c$, but $a\notin c$.

For example, $a=\varnothing, b=\{\varnothing\}, c=\{\{\varnothing\}\}$.

On the other hand, $<$ is transitive on $\Bbb Q$.

(Another way, using far heavier guns, would be to point that the first order theory of $(\Bbb Q, <)$ is decidable, consistent and complete, whereas $\sf ZF$ has no extension which satisfies all those three properties by Gödel's second incompleteness theorem.)