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Similar questions have already been posted, but they require an understanding of Set theory. I'm not a mathematician and was hoping the site could provide background for laymen in plain English.

In Frege's attempt to lay a foundation for mathematics, a fundamental axiom in his work was that every concept had a corresponding extension. However, Russell's paradox proved this wasn't always the case. The set of all sets that are not members of themselves cannot be a member of itself, and it also can't not be a member itself. No such set exists. From reading other answers, this somehow implies that the set of all sets does not exist. How is that so?

Several ad hoc attempts were made to reconcile this contradiction (set hierarchy) until Zermelo–Fraenkel set theory came along. How does Zermelo–Fraenkel overcome Russell's paradox?

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In Naive set theory (the kind where Russell's paradox can be made in), it is assumed that any given property provides us with the existence of a set of exactly those things that satisfy the property. This is why Russell's set can exist in Naive set theory.

To solve this, in Zermelo-Fraenkel set theory, we work with two notions: that of a class and that of a set. A class is basically what we considered to be sets in Naive set theory, in other words, a collection of things satisfying a property.

A set however, is a more refined concept than a class. Every set is still a class (and thus still can be seen intuitively as a collection of objects satisfying a property), but instead of defining sets through the properties their elements have, we define sets by constructing them from the ground up. We carefully lay down the rules that these constructions have to obey and try to make them not too general, so that it is impossible to create sets like Russell's set.

Zermelo-Fraenkel set theory (to be precise, their Axioms) describes exactly which rules of construction are allowed. Some rules are very simple, such as that you can combine two sets together to form a new set containing all the elements of the both of the previous sets, or that you can create a set containing exactly all the subsets of another set. Other rules are very delicate and counterintuitive, even to many mathematicians.

One particular rule that is reminiscent of Naive set theory, is the rule that if we start with a set and any property, then we can take the collection of all the elements inside that set that satisfy said property. So once we have a set, we can get a smaller set with elements satisfying a property. This gives us much of the power that we had in Naive set theory, but without the trouble of sets that are "too general" in a certain sense.

So this is how Zermelo-Fraenkel set theory solves the problem of Russell's paradox. But where lies the problem in having a set of all sets? In Zermelo-Fraenkel set theory the answer is quite easy to see: we have the rule that we can construct from any set and any property, a new set with elements of the original set that exactly satisfy the property. But if there is a set of all sets, let's call it $V$, then we can get Russell's paradoxical set from taking those elements from $V$ that are not contained in themselves.

Thus, if the set of all sets were a set in Zermelo-Fraenkel's set theory, then Russell's paradox would also give us a set of sets that do not contain themselves.

This is not the only solution to Russell's paradox. Another logical option, would be to not restrict what sets are by laying out rules for constructing sets, but by laying out bounds for which properties we consider appropriate enough to define sets. In particular, we can consider only properties that avoid a certain type of bad self-reference. This is the basis behind Quine's New Foundations set theory. In this theory it is possible to have a set of all sets, and not have problems with Russell's paradox.

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  • $\begingroup$ Thank you. Just to confirm, "This is why Russell's set can exist in Naive set theory." Is that correct? I thought Russell's set can't exist (The set of all sets that are not members of themselves cannot be a member of itself, and it also can't not be a member itself. No such set exists). $\endgroup$ – user27343 Jun 20 at 22:56
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    $\begingroup$ @user27343 The fact that the Russell set can be proven to exist in naive set theory, and that this leads to a contradiction means that naive set theory is inconsistent. $\endgroup$ – spaceisdarkgreen Jun 20 at 23:19
  • $\begingroup$ Got it. Thanks. $\endgroup$ – user27343 Jun 21 at 2:11
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ZF does not allow you to say “the set of all objects that...”. Instead it only allows you say “the set of all objects in $A$ that...”, where $A$ is some set. This avoids the paradox because “the set of all sets in A that do not contain themselves” is not paradoxically forced to contain itself... it just won’t be in $A$.

But even allowing this restricted form of comprehension has consequences. One such consequence is that there can be no set $U$ of all sets. If there were, we get back the paradox, since “the set of all sets in $U$ that do not contain themselves” is exactly the same thing as “the set of all sets that do not contain themselves”.

As others have already noted, New Foundations is an alternative approach to ZF that both avoids Russell’s paradox (and all known paradoxes) and allows a universal set. However it puts a lot more restrictions on what kind of properties can be the “...” in “the set of all objects that ...”. For instance, “is not an element of itself” must not be an allowed property. The nature of these restrictions are somewhat obscure at first pass, which at least partially accounts for the NF approach being much less popular than ZF despite whatever advantages might be perceived in having a universal set.

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If the set of all sets exists, then you can take the subset of the sets not containing themselves and once again get the contradiction. Thus also the set of all sets cannot exist (or rather cannot exist as a set).

ZF is a list of axioms for sets. The set of all sets is what one calls a class and classes have axioms which are less restrictive.

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    $\begingroup$ This assumes that you can make a set by taking some of the elements in an existing set, defined by a property such as "does not contain itself". ZF has an axiom to claim that this is possible, but the cost of that is that a set of all sets cannot exist. In some other set theories different trade-offs are made. $\endgroup$ – Henning Makholm Jun 19 at 19:06

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