I'm doing some research on naive set theory and was a little confused over the statement of the axiom of unrestricted comprehension, $\exists$B$\forall$x(x$\in$B$\iff$$\phi$(x)). I am curious as to why this is an iff statement. I was under the assumption that, in naive set theory, a set only exists if it is determined by some property, but this seems to say that a property exists for each set as well. Why doesn't the implication $\exists$B$\forall$x(x$\in$B$\implies$$\phi$(x)) suffice?
[Math] Axiom of unrestricted comprehension
axiomselementary-set-theorylogic
Related Solutions
Here is a model: $\{1,2\}$ with $$ 1\in 1 \qquad 1\notin 2 \qquad 2\in 2 \qquad 2\notin 1 $$
That is, two Quine atoms side by side in an otherwise empty world.
In this model $\in$ and $=$ behave identically, so $\phi^=\leftrightarrow \phi$. Thus, the antecedent of your comprehension axiom demands that there is a $y$ that satisfies $\phi$ and another $y$ that doesn't satisfy $\phi$, which means that the collection you're looking for is either $\{1\}$ or $\{2\}$, which are exactly the options you have.
(Note that this model does not have a universal set).
That is the only model, though. If you strengthen your multiplicity axiom such that it promises three distinct sets $a,b,c$, then the following loophole will give unrestricted comprehension, and therefore an inconsistent theory:
The restricted comprehension axiom allows you to form $\{a,b\}$ -- because there is now a $c$ such that $\neg(c=a\lor c=b)$ -- and this unordered pair cannot equal both $a$ and $b$ at the same time, so there are some sets $p,q$ such that $p\in q$ but $p\ne q$.
Now to form unrestricted $\{y\mid \psi(y)\}$, use your comprehension axiom with a parameter $a$ and $$ \phi \equiv (\forall p,q(p\in q\leftrightarrow p=q)\land y=a) \lor (\neg \forall p,q(p\in q\leftrightarrow p=q) \land \psi(y)) $$
No, this form of comprehension is inconsistent with even the ability to form singletons and unions of two sets.
Suppose $A=\{x: x\not\in x\wedge x\not=A\}$. Then I claim $A\cup\{A\}$ is the usual Russell set.
If $x\in A\cup \{A\}$ then $x\not\in x$. This is because for such an $x$, either $x\in A$ or $x=A$, and no element of $A$ contains itself by the "$x\not\in x$"-clause of the definition of $A$ and $A\not\in A$ by the "$x\not=A$"-clause of the definition of $A$.
If $x\not\in x$ then $x\in A\cup\{A\}$. Again, we reason by cases. If $x\not\in x$ then either $x\not=A$ (in which case $x\in A$) or $x=A$ (in which case $x\in\{A\}$), and either way $x\in A\cup\{A\}$.
And now we ask whether $A\cup\{A\}$ is an element of itself.
Best Answer
Recall, by definition of the empty set, $\forall x\;x \notin \emptyset$.
So trivially, for any $\varphi$, $\forall x(x \in \emptyset \to \varphi(x))$.
So, still trivially, for any $\varphi$, $\exists B\forall x(x \in B \to \varphi(x))$, just so long as the empty set exists. Not enough, then, to give you a set as the extension of $\varphi$, so not what you want!