Suppes on definition by abstraction I'm in the section of Axiomatic Set Theory on "definition by abstraction," which is just about the introduction of standard set notation of this sort: {x:Gx}. For example, {x:x=1 v x=2}. We've just proved, as an exercise, that if two properties are equivalent they are extensionally identical--that is, if two properties apply to exactly the same things, then the sets of things having those properties have the same members. Now he asks us to give a counterexample to this claim: that if everything that has one property also has a second property, then the set of things having the first property is a subset (not necessarily proper) of the set of things having the second property. That would mean finding two properties such that the first implies the second, but the set of things having the first property is not included in the set of things having the second. I've worked on this for a while and I have no idea what he is after. Any help appreciated.
 A: I struggled with this a lot as well, but I think I have finally found a counterexample! In particular, I thought that $ \{x:\phi(x)\}$ would allow Russell's paradox, but I managed to prove that it doesn't and thus use that as a counterexample. Below is the way I constructed it.

Take $\psi(x) = \lnot (x\in x)$. Then, the definition $\{ x : \psi(x) \}$ becomes:
$$
\{ x : \psi(x) \} = y \iff [ (\forall z)(z \in y \iff \lnot (z \in z))\wedge y\text{ is a set} ] \lor [y = 0\wedge\lnot (\exists B)(\forall z)(z\in B \iff \lnot(z\in z))]
$$
Here, similarly to $\S$1.3, taking $z = y$, the first part of the right side becomes false. Namely, it becomes
$$
y \in y \iff \lnot(y\in y)
 \text{ & } y \text{ is a set}
$$
If $y$ is not a set, then it is trivially false. If $y$ is a set, then the whole formula is logically contradictory, hence false. On the other hand, in the right part we have:
$$
[y = 0 \wedge\lnot(\exists B)(\forall z)(z\in B \iff \lnot(z\in z))]
$$
Again, we take $z = B$ and show that $B\in B \iff \lnot(B\in B)$ is contradictory and thus $\lnot[B\in B \iff \lnot(B\in B)]$ is a truth. Therefore, we can assert that $y = \{x : \psi(x) \} = 0$, i.e. $y$ is the empty set.
Now, all we need is a property $\phi(x)$ such that $\phi(x) \implies \psi(x)$ and $\{ x : \phi(x) \}$ has elements, since a set that has elements is not a subset of the empty set. Here you can pick anything you like, e.g. $\phi(x) \iff x = {2}$ would imply $\phi(x) \implies \psi(x)$, since $\{2\} \not\in \{2\}$. However, $\{ x : \phi(x) \} = \{\{2\}\}$, therefore:
$$
\{\{2\}\} \not\subseteq 0
$$
Q.E.D.
A: So, my first post, which answered the question as posed was rejected, so I looked up the exercise which asks for a counter example for:$$\forall x(\varphi(x)\implies\psi(x))\implies\{x:\varphi(x)\}\subseteq\{x:\psi(x)\}$$
Your counter example is$$\varphi(x)\iff x\in\{\emptyset\}
$$ $$\psi(x)\iff x\notin x$$
Which gives you $$\forall x(x\in\{\emptyset\}\implies x\notin x) $$
But then $\{x:x\in\{\emptyset\}\}$ is not empty (it contains $\emptyset$) and cannot be a subset of $\{x:x\notin x\} $ which is empty.
Stealthmate's answer works as well substituting $\{2\}$ for $\emptyset$, but $\{2\}\notin \{2\}$ requires Regularity which has not yet been introduced into the theory at this point.
