The class $B$ is well-founded, if every descending chain of subclasses is finite. The class $B$ is not well-founded if there exists an infinite chain of classes $B_n$ such that $\quad \dots \in B_n \in \dots \in B_2 \in B_1 \in B.$

Paradox: The collection $\mathbf{WF}$ of well-founded sets (on a given set of atoms) is not itself a set.

Let me try to elaborate on the formulation of the paradox. Let $\mathbf{WF}$ be the set of well-founded sets; then $\mathbf{WF}$ itself is well-founded, for were $\mathbf{WF} \ni x_0 \ni x_1 \ni x_2 \ni\dots\,$, then $x_0$ would be a not well-founded member of $\mathbf{WF}$, which is preposterous. Therefore $\mathbf{WF} \in \mathbf{WF}$, so $\mathbf{WF}$ is not well-founded, i.e. contradiction.

Is this paradox only present in the naive set theory? If the answer is no, then how is it resolved?

I think this paradox is similar to Burali-Forti antinomy, but honestly I do not understand (the resolution of) the latter one as well.

• How is that a paradox? Is there any particular reason that collection ought to be a set? Commented Jun 6, 2015 at 1:53
• The statement boils down to something like "a certain collection of sets is not a set itself". How is it not a paradox?
Commented Jun 6, 2015 at 9:47
• There is a mistake in your very first line. When you say "subclass" you are talking about $\subseteq$ chains, not $\in$, and those do not have to be finite at all. Not to mention that classes which are elements of other classes are called sets. Commented Jun 6, 2015 at 9:47

Indeed if $\bf WF$ is a set then it is well-founded, then $\bf WF\in WF$, which is a contradiction to the well-foundedness.
Some modern set theories (e.g. $\sf ZFC$) resolve this by requiring that properties establish subsets of previously constructed sets; other set theories (e.g. $\sf NF$) resolve this by requiring that the formulas themselves will have certain syntactic properties which prevent them from defining these classes.
And when all this has been said and done, I'd like to point out that the axiom of choice is necessary for the proof that well-foundedness is equivalent to the lack of infinite decreasing chains. The correct formulation should be that every non-empty set has an $\in$-minimal element (namely, if $x$ is not empty, then for some $y\in x$, $x\cap y=\varnothing$).
On the other hand, if you prove (without appealing to choice) that if $x$ is a well-founded set, then $\{x\}$ is a well-founded set, and that for every $x\in\bf WF$ satisfies $x\notin x$, from which we can conclude that $\bf WF\notin WF$.