TITLE: Why is "the set of all sets" a paradox, in layman's terms?
QUESTION [102 upvotes]: I've heard of some other paradoxes involving sets (i.e., "the set of all sets that do not contain themselves") and I understand how paradoxes arise from them.  But this one I do not understand.
Why is "the set of all sets" a paradox?  It seems like it would be fine, to me.  There is nothing paradoxical about a set containing itself.
Is it something that arises from the "rules of sets" that are involved in more rigorous set theory?

REPLY [5 votes]: Some of the comments in the previous answers make a subtle mistake, and I think it may be worth clarifying some issues. I am assuming the standard sort of set theory in what follows. 
Cantor's diagonal theorem (mentioned in some of the answers) gives us that for any set $X$, $|X|<|\mathcal P(X)|$. Unlike what some comments claim, this really has nothing to do with cardinalities. All it says is that no map $f\!:X\to\mathcal P(X)$ is onto. The usual proof proceeds by noting that $A=\{x\in X:x\notin f(x)\}$ is not in the range of $f$, because if $A=f(a)$, then $a\in A$ if and only if $a\notin f(a)=A$.
The usual argument for Russell's paradox (also mentioned in some of the answers) proceeds by considering $A=\{a:a\notin a\}$. If $V$ is a set, $A$ would be a set as well (by comprehension, if you wish), and  we reach a contradiction by noting that $A\in A$ if and only if $A\notin A$.
I think it is misleading to think (as some of the comments suggest) that the two proofs are (fundamentally) different. They are essentially the same. 
The point is that if there is a set of all sets (let's call it $V$), then $\mathcal P(V)$ is a set as well (by comprehension) and in fact $\mathcal P(V)=V$ because, on the one hand, any subset of $V$ is a set, and therefore a member of $V$, and on the other hand, any member of $V$ is itself a subset of $V$ (since the members of any set are sets themselves), and therefore a member of $\mathcal P(V)$. 
Now, the identity function is a map from $V$ to $\mathcal P(V)$. Let's call it $f$. The set not in the range of $f$ given to us by the standard proof of Cantor's diagonal theorem recalled above is $A=\{x\in V:x\notin f(x)\}$, which in this case reduces to $\{x:x\notin x\}$, which in turn is precisely the set given by the standard proof of Russell's paradox.
[What to make of this result from a foundational point of view is another matter. In $\mathsf{ZF}$ the conclusion is just that there is no set of all sets, although it is perhaps more accurate to say that the result is used to justify dismissing unrestricted comprehension and adopting instead the version of bounded comprehension used in $\mathsf{ZF}$. In $\mathsf{NF}$ the solution is instead to limit unbounded comprehension by only allowing stratified instances. Other foundational solutions may and have been adopted as well.]