This question already has an answer here:

I think I understand the way in which Russell's paradox shows that the following principle is wrong:

" for every predicate, there is a set having as elements all the objects that satisfy this predicate"

Russell's picks up a predicate ( namely the predicate " x is not an element of itself" ) and shows that the corresponding " set" would have contradictory properties, which means that " the set of all x such that x is not an element of x" does not exist.

This counts as a counter example to the alledged " principle".

My understanding of Russell's paradox does not go further than this.

Now, if I am correct, it is often said that Russell showed, with this paradox, that the " set of all sets" does not exist.

What is actually the relation between Russell's paradox and the non-existence of the set of all sets?

It seems difficult to me to answer that the relation consists in the fact that precisely a set has the property of not belonging to itself. For it seems to me that Russell's paradox forbids to define a set in this way.


marked as duplicate by Asaf Karagila logic May 20 at 15:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ There is a "more general fact" behind it. See the post about two simple theorems in the language of $\mathsf {ZFC}$. It is a logic law that $¬(∃y)(∀x)(E(x,y) ↔ ¬ E(x,x))$ for a binary predicate $E(x,y)$ whatever. The proof does not rely on mathematical axioms. $\endgroup$ – Mauro ALLEGRANZA May 20 at 13:13
  • $\begingroup$ @MauAllegranza.Thanks for the link. $\endgroup$ – Eleonore Saint James May 20 at 13:15
  • $\begingroup$ Indeed Russell's paradox did not at all show that there is no universal set. See this post for how one can elegantly resolve Russell's paradox based on the right view of the fundamental concept of "set". In short, if you have a philosophically clear notion in mind when you say "set" then you will not even face any paradox to begin with. And it is obvious to everyone that the notion of "everything" (universal type) is well-defined; it simply accepts every thing as a member. $\endgroup$ – user21820 May 25 at 15:50

Russell's paradox does not in itself prevent a set of all sets from existing. There are set theories that do contain a universal set and are not known to be inconsistent, such as Quine's NF.

It is only together with Zermelo's subset selection axiom (the core idea behind ZFC, which claims that $\{x\in A\mid \varphi(x)\}$ is a set whenever $A$ is), that it has this effect. If a universal set existed, then the subset selection axiom would effectively provide a universal comprehension principle, and then Russell's paradox would produce a contradiction.

it is often said that Russell showed, with this paradox, that the "set of all sets" does not exist.

While it is probably not difficult to find a claim such as this in print, it is an ahistoric oversimplification. Russell published the paradox several years before Zermelo proposed his axiom system, so the ingredients for making the jump from "unrestricted set comprehension doesn't work" to the specific claim "there is no universal set" was not present at that time.

  • $\begingroup$ @HenningMakholm.I really thought that the Axiom Schema of separation had been adopted to prevent such a thing as Russell's paradox to happen! I'd really like to understand how (1) selection axiom + (2) sets of all sets brings Frege's universal comprehension axiom. Apparently, it is rather obvious, but I sincerely cannot see how it happens. $\endgroup$ – Eleonore Saint James May 20 at 13:36
  • 1
    $\begingroup$ @EleonoreSaintJames: If you have a universal set $U$, then the unrestricted $\{x\mid\varphi(x)\}$ can be made using separation as $\{x\in U\mid\varphi(x)\}$. $\endgroup$ – Henning Makholm May 20 at 13:39
  • $\begingroup$ With " phi(x) " = " x does not belong to x"? Right? $\endgroup$ – Eleonore Saint James May 20 at 13:43
  • 2
    $\begingroup$ @EleonoreSaintJames: Yes, or anything. Here my point is that if you have separation and a universal set then you effectively have universal comprehension for every formula $\varphi$. $\endgroup$ – Henning Makholm May 20 at 13:46

The thing is that the principle is wrong, but we'd like it to be true : you want to be able to consider the set of things satisfying a certain property, it's one of the bases of constructing new sets.

To make that possible without introducing (or so we hope) contradictions, the ZF axioms have the so-called comprehension schema, which says that if you have a formula $\varphi (x)$ and a set $A$ then you can consider the set of all sets satisfying $\varphi$ which are also in $A$; which is usually denoted $\{x\in A\mid \varphi(x)\}$

Now if $A$ is a set containing all sets, then $\{x\in A\mid \varphi(x)\}$ is precisely the set of all sets that satisfy $\varphi$ , and so having a set of all sets would imply having the naive principle which we know is contradictory : therefore there is no set containing all sets.

This is how Russell's paradox relates to the existence of a set of all sets; although at first sight it just implies that there is no set of all $x$ such that $x\notin x$ (which is a way more general fact and is very little related to set theory, as Maura Allegranza points out in the comments)


Russell's Paradox arose from an inconsistency in Frege's early attempt to axiomatize Cantor's set theory. As you point out, by using his axiom of unrestricted comprehension, it was possible to prove:

$$\exists r: \forall x: [x\in r \iff x\notin x]$$

Using only the rules of logic in Frege's system, however, it was also possible to prove the negation:

$$\neg\exists r: \forall x: [x\in r \iff x\notin x]$$

We kept his rules of logic, but got rid of unrestricted comprehension and are still able to prove the latter.

It wasn't necessary to ban set self-membership to resolve the Paradox. You can substitute any binary relation E to obtain:

$$\neg\exists r: \forall x: [E(x,r) \iff \neg E(x,x)$$


Not the answer you're looking for? Browse other questions tagged or ask your own question.