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In "Analysis I", Tao 2006, the Axiom of Regularity is defined on p. 54 as "If A is a non-empty set, then there is at least one element x of A which is either not a set, or is disjoint from A."

Let $A$ be the set (and pair) $\{ 1 , A \}$.

$A \neq \emptyset$ because it contains some element. The Axiom of Regularity tells us we must find an $x \in A$ such that $x$ is not a set or $x \cap A = \emptyset$. $1$ is not a set and $1 \in A$, in consequence of which the first condition is satisfied and the axiom is not violated, making $A = \{1, A\}$ a legal set.

But I know this reasoning must be incorrect because the goal of the Axiom of Regularity is precisely to prevent the existence of sets that contain themselves to avoid Russell's paradox.

I already read multiple threads on this topic on MSE, including:

... but unfortunately I still couldn't understand why my reasoning above is incorrect. I would much appreciate some help.

David Doret
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  • Suppose $x=A$. Then the first condition is violated as $A$, and hence $x$, is a set. Furthermore, since $A=A$, $A\cap A \neq \emptyset$, i.e. $A$ (and hence $x$) and $A$ are not disjoint, which means the second condition is violated and the construction violates the axiom. – H. sapiens rex Apr 01 '24 at 18:39

3 Answers3

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The violation comes from your statement Let $A$ be the set $\{1,A\}$. You assign this set $A$ as you have described, then arrive at a contradiction. Logically, the only thing that could have been wrong here is that $A$ is in fact not a set.

By itself, the Axiom of Regularity does not prohibit the existence of the set $A=\{1,A\}$ as you have described, but the combination of this axiom and other axioms does. Specifically, if we join this axiom with the Axiom Schema of Specification, then we have a problem.

The Axiom Schema tells us that if we have a set $A$, then any rule applied to the elements of $A$ to restrict the elements creates a new set $B$. Formally, if $\varphi$ is a rule of formal logic with a finite number of symbols that ascribes a truth value to sets, then the set

$$ B=\{x\in A|\varphi(x)\} $$

is indeed a set. Let $\varphi(x)$ be the rule $x=A$. Then $B=\{A\}$; by the Axiom of Regularity, we know that since $B$ is nonempty it must contain an element that is either not a set or disjoint from $B$. If $A$ is a set then it must be disjoint from $B$, however,

$$ B\cap A=\{A\}\cap\{1,A\}=\{A\}\neq\emptyset $$

This is our contradiction.

All of this is to say that there is nothing wrong with the axiom itself. It all comes down to the fact that the statement Let $A$ be the set... is a big assumption when dealing with first principles of set theory. You can't just assume something is a set by the definition of its elements. Are those elements sets? How do you know? Did you construct them from first principles? It's a big rabbit hole.

Josh B.
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  • Technically $x = A$ is not a formula unless you mean $x = {1, x}$, which is not about a specific set $A$ but any potential set of that form. You could take $\varphi(x)$ to be $x \in x$ though. – ronno Apr 02 '24 at 09:08
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Let $A$ be the set $\{ 1 , A \}$.

That makes no sense. It's like saying "let $x$ be the number $1 + x$".

What makes sense is:

Let $A$ be a set satisfying $A = \{1,A\}.$

This is analogous to "let $x$ be a number satisfying $x=1+x.$" And just like no such $x$ can exist in the latter, no such $A$ can exist in the former, assuming the axiom of regularity, since, as other have explained, otherwise every element of $\{A\}$ intersects $\{A\}$.

This equivalent statement of regularity is easier to use:

There is no infinite sequence of sets $A_1,A_2,\ldots$ such that $A_{n+1}\in A_n$ for all $n$

(The equivalence of this to the usual statement of regularity requires some nontrivial strength from the other ZFC axioms, including the axiom of (dependent) choice.)

Then we can easily see that we have $A\ni A\ni A\ni A\ldots,$ so we violate that if $A = \{1,A\}$, and also that many other equations and systems of equations like $\{C = \{B\} ,B = \{C\}\}$ can't have any solutions under regularity.

But this doesn't mean that "Let $A$ be the set $\{ 1 , A \}$" makes any more sense when regularity is removed. Even if a set $A$ satisfying $A = \{1,A\}$ existed, there could be more than one, similarly to if we tried to define "the" number $x$ satsifying $x = 1-x^2.$ There is however, one popular antifoundation axiom, the AFA that proves essentially that every system of equations of the above sort has a unique solution. Another popular antifoundation axiom (Boffa's) yields a whole proper class of solutions to every system. And one can also arrange for violations of regularity where say, $A = \{A,1\}$ has no solution, but $\{C\in \{B\}, B\in \{C\}\}$ does.

Finally:

But I know this reasoning must be incorrect because the goal of the Axiom of Regularity is precisely to prevent the existence of sets that contain themselves to avoid Russell's paradox.

This is a common misconception. Adding an axiom can't prevent contradictions... it can only cause them. As the continued study of the aforementioned antifoundation axioms show, there is not anything inherently contradictory about sets containing themselves and so on. Russell's paradox was prevented by weakening the naive comprehension axiom and not anything to do with the axiom of regularity.

spaceisdarkgreen
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The set $B=\{A\}$ violates regularity. Its only element is $A$, and $A\cap B=\{A\}$, so that is not disjoint from $B$.

The strength of the Axiom of Regularity (aka Foundation) comes from its assumption that the condition holds for all sets. It's interesting to note that a universe in which Regularity does not hold, can still have "paradoxical" sets that don't directly violate the axiom.

Michael Weiss
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