Separation and Russell's paradox I just want to be sure that I understand the connection between "Naive Comprehension", the Axiom of Separation, Russell's Paradox, and the existence of a universal set.  Is the following correct?
The Naive Comprehension Axiom has as an instance the following:
$$
\exists y\forall z(z\in y\leftrightarrow z\notin z)
$$
This entails:
y is an element of y just in case y is not an element of y
(where "just in case" means "iff")
which is a contradiction.
By contrast, the Axiom of Separation has as an instance the following:
$$
\forall x\exists y\forall z(z\in y\leftrightarrow (z\in x\land z\notin z)).
$$
The assumption that y is an element of y leads to contradiction.  But the assumption that y is not an element of y just entails that y is not an element of x.  So, we have shown that for every set, y, there is another not in it -- namely, y itself.  That is, there is no universal set.
Is this correct?
 A: Russell's paradox can be seen, in the presence of separation as the following theorem:

For every set $A$, there is a set $B$ such that $B\subseteq A$ and $B\notin A$.

The theorem does not necessarily mean that $A=B$. But it does tell you that there is no universal set.
It is consistent with separation (and most other standard axioms of mathematics) that there is a set $A=\{A\}$. In that situation, the set obtained by Russell's paradox would be $\varnothing$, and indeed $\varnothing\notin A$. It is also possible for a set $X=\{X,\varnothing\}$ to exist, and I will leave it to you to figure out what subset of $X$ is not an element of $X$.
A: Almost correct, but not quite: all is well, until the 3rd to last sentence: 
$$
\text{"So, we have shown that for every set, $y$, there is another not in it -- namely, $y$ itself."}
$$
No — $y$ is constructed from $x$ using Separation. What we can show is:

For every set $x$ there is a set $y$ that is not a member of it: 
  $$\forall x\exists y\, y\notin x.\tag{*}$$ 

We have this instance of Separation:
$$
\forall x\exists y\forall z(z\in y\leftrightarrow (z\in x\land z\notin z)). \tag{$SepInst$}
$$
Given $x$, there is some $y$ such that 
$$\forall z(z\in y\leftrightarrow (z\in x\land z\notin z)).
$$
In particular this holds for $z = y$:
$$y\in y\leftrightarrow (y\in x\land y\notin y)).\tag{$SepInst_{x,y}$}
$$
If $y\in y$, then by ($SepInst_{x,y}$) we have  $y\notin y$. As $(p\to\neg p)\to \neg p$ is a tautology, it follows that $y\notin y$.
Thus, we can conclude now that $y$ is not the universal set. But that doesn't warrant the conclusion that no set contains everything: $y$ is hardly "arbitrary".
Now, if $y\in x$, then the righthand side of ($SepInst_{x,y}$) holds, and the we can infer that the lefthand side holds too — that is, $y\in y$, which we have already shown to be false. So $y\notin x$ after all. $\,\square$
Notice that moving the inner negation in (*) outward, across the quantifiers, results in the logically equivalent sentence $\neg\exists x\forall y\, y\in x$, which clearly expresses "there is no universal set".
