if every set is an element of some set (axiom of pairing), how come if A is any set and B={x∈A:x∉x}, B∉A? I'm reading naïve set theory by Halmos and at page 6, while explaining the axiom of specification (and Russell's paradox) he says:
"whatever the set A may be, if B={x∈A:x∉x} B∈A is impossible"
then at page 10 he says:
The axiom of pairing ensures that every set is an element of some set
so my question is, if every set is an element of some set, how come B={x∈A:x∉x} is not element of A even though A can be any set?
 A: This is a linguistic trap.
If you say $A$ can be any #thing#, once you specify it, it is no longer arbitrary.  You have now labelled it as a specific #thing# and even though you know nothing about it everything you do with it now, is not arbitrary but about $A$
Here's a silly example.  If you say "Let $A$ be any number".  Then we say.  Let $B = A-1$ so $B < A$.  Now suppose we argue "But we said $A$ could be any number.  So $A$ doesn't have to be less than $B$".   So it is perfectly possible, in theory, that there is some number where $A -1 \ge A$.
Do you see the problem there?  
You argument is much the same.  Let $A$ be any set.... but once we label it, it is no longer any set.  It is $A$; a specific set.  We define $B$ based on $A$ but we can't just change what $A$ is.
....
Analogy.  Suppose we claimed: Every number is larger than some number.  But ten pages later we claim.  Let $B = A-1$.  Then we must have $B< A$.
Suppose we argue:  But why is $B < A$?  $B$ is greater than some number and $A$ can be any number, so why can't $A < B$?
Well, because we already defined $A$ and then we defined $B$.  $A$ can't be any number now.  It's too late.  It could be any number when we brought it up but now it's committed to being $A$.
A: The point is that there isn't a single set $B$: each $A$ determines its own $B_A$. So e.g. we'd have $B_A\not\in A$ but $B_A$ might well be in some other set $C$. Of course $B_C\not\in C$, but so what? $B_A$ and $B_C$ can be different sets (and indeed will have to be if $B_A\in C$).
