# Why does the Axiom of Selection solve Russell's Paradox in Set Theory?

I am a beginner in mathematics and I was reading a text on Set Theory that talked about how Zermelo's Axiom of Selection "solves" Russel's Paradox.

I understand that the the axiom does not allow constructions of the form $$\{x \:: \text S(x) \}$$ and only allows$$\{x \in \text A \:: \text S(x) \}$$ but how does this change the outcome of the paradox when we have:
$$S = \{x \in \text A \:: \text x \notin \text x \}$$ where $S$ is still the set of all sets that do not contain themselves.

Won't we still get the paradox?

• We have not constructed $A$, which presumably is supposed to be the "set" of all sets. So we have not constructed $S$. Mar 2, 2016 at 23:53
• Work through the details and you'll see that the paradox doesn't arise. $S$ is now just the set of all members of $A$ that aren't members of themselves. The closest we come is: $S\in S\iff S\in A \land S\notin S$, so $S\notin A$. (Furthermore, by the axiom of Regularity, $x\notin x$ for all $x$, so $S = A$, thus, again, $S\notin A$.) Mar 3, 2016 at 0:07

Axiom of regularity (or Foundation) rules out the case $S\in S$.

Otherwise, we would have $S\notin S$. Now does this imply $S\in S$?

The problem with Russel's paradox is it implicitly implies existence of universal set ("set of all sets").

If $A$ is a set and $S=\{x\in A:x\notin x\}$, $S\notin S$ implies either $S\in S$ or $S\notin A$.

In case of Russell's paradox, it is not possible to have $S\notin A$, because $S$ must belong to set of all sets, if that existed, and leads us to paradox $S \in S$. But with our construction such set cannot exist (exactly because existence of such set leads us to Russell's paradox). So paradox does not arise.

• Regularity is not required to avoid Russell's Paradox. Mar 6, 2016 at 4:33
• Russell's Paradox does not imply the existence of universal set. In fact, it can be used to prove the universal set does not exist. Mar 6, 2016 at 4:54
• @DanChristensen That's not quite true - for instance, in NF there is a universal set. Roughly speaking, all Russell's paradox disproves is the axiom (scheme) of full comprehension. As a corollary, this shows that you can have either a universal set (NF) or separation (ZFC), but not both. Mar 6, 2016 at 5:11
• @NoahSchweber How does NF block the proving of the the non-existence of the universal set? In regular set theory, it can be done using some equivalent of the ZF Specification (Subset) Axiom. Is that not available in NF? Mar 6, 2016 at 5:46
• @DanChristensen Indeed it is not. NF has stratified comprehension instead of separation: the formula "$x\not\in x$" is not stratified, so that blocks Russell's paradox, while "$x=x$" is stratified, leading to the existence of a universal set. (See en.wikipedia.org/wiki/New_Foundations.) Mar 6, 2016 at 5:52

If we assume the existence of a set $$R$$ such that $$R=\{x: x\notin x\}$$then we can obtain the contradiction $$R\in R$$ and $$R\notin R$$. So, $$R$$ cannot exist.

To avoid Russell's Paradox then, all we need to do is not assume that, for every unary predicate $$P$$, there exists a set $$S$$ such that $$S=\{x:P(x)\}$$

If, however, $$A$$ is assumed or proven to be a set, then we can assume without fear of contradiction that there exists a subset $$S\subset A$$ such that $$S=\{x\in A: P(x)\}$$Or equivalently$$S=\{ x: x\in A \text{ and } P(x)\}$$

You can think of $$P(x)$$ as the criterion for selecting elements from the set $$A$$ for the subset $$S$$. The only restriction is that the variable $$S$$ may not occur in the selection criterion. This is the Axiom of Specification (Selection).

If, for example, we have set $$A$$, then we can assume that there exists a subset $$S\subset A$$ such that $$S=\{x\in A: x\notin x\}$$ Or equivalently $$S=\{x:x\in A \text{ and } x\notin x\}$$ Then we would not obtain a contradiction, but we would have $$S\notin S$$ and $$S\notin A$$). (Proof left as an exercise.)

The way that the Axiom of Selection prevent's Russell's Paradox is by preventing you from selecting from all sets. Rather you are selecting from the elements of A. Since $A \in A$ is forbidden by the Axiom of regularity the paradox can't arise.

• The axiom of regularity has nothing to do with it. Without regularity, we may have $A\in A,$ we may have $A\notin A,$ in any case $S\notin A,$ no paradox.
– bof
Mar 3, 2016 at 0:35
• More generally, adding axioms can never resolve an inconsistency. Dec 25, 2022 at 5:23