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I am currently reading an article from Stanford university related to set theory. In this article, I came across a term that I find intriguing but don't fully understand. Here it is

Consider the property of sets of not being members of themselves. If the property determines a set, call it $A$, then $A$ is a member of itself if and only if $A$ is not a member of itself.

Could you please help me better understand this specific point?

Srijan
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  • If we were to suspend disbelief for a moment that we were allowed to do such a thing... define a "set" $X = {A~:~A\notin A}$, that is the "set" of all sets who are not elements of themselves. Certainly, several sets will be in this... for instance ${1}$ is not itself an *element of ${1}$ and so will be in this "set". We ask the question... will $X$ itself be in this "set"? Well, if it were in the set that should have meant it was one of those sets who were not elements of themselves... That is to say... if it is an element of itself then it is not an element of itself and vice versa. – JMoravitz Oct 05 '23 at 15:21
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    This is the famous Russell's Paradox. The resolution of it is that we should not have been allowed to define this in the first place or at the least if we were to define it, that it should not qualify as a "set" but rather as something more generic... a class... which does not need to follow the same rules that sets do. – JMoravitz Oct 05 '23 at 15:23
  • @JMoravitz Thanks for this explanation. You may write it as an answer. – Srijan Oct 05 '23 at 15:28
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    Does this answer your question? On the class $A={x\mid x\text{ is not an element of }x}$ – JMoravitz Oct 05 '23 at 15:30
  • @JMoravitz I understood now. Thank you so much – Srijan Oct 06 '23 at 03:17

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This is an early result in Naive Set Theory from Bertrand Russell.

It is a refutation of the Unrestricted Axiom of Comprehension, which lets one form a set out of any collection of objects that satisfy some property P.

Intuitively, Unrestricted Comprehension characterizes sets as "collections of objects", Bertrand Russell showed that such a formulation is unsuitable for mathematics with the following argument.

Theorem:¬∃y∀x(x$\in$ y ↔ ¬(x$\in$x))

Proof by Contradiction: Assume ∃y∀x(x$\in$ y ↔ ¬(x$\in$x))

Let $y_0$ be arbitrarily chosen

∀x(x$\in$ $y_0$ ↔ ¬(x$\in$x))

Let x = $y_0$

x$\in$x ↔ ¬(x$\in$x) Produces Our Desired Contradiction.

That may be too formal to be helpful.

Theorem (informal): R = { x | x∉x }, the Russell Class, Does not Exist

Proof by Contradiction: Assume R exists

R$\in$R → R∉R ( by definition of R)

Likewise,

R∉R → R$\in$R ( by definition of R)

Thus, R$\in$R ↔ R∉R is our desired Contradiction.

So R is not a set, we can still refer to it by the Formula: $\phi$(x) = ¬(x$\in$x), and denote R = { x | $\phi$(x) } as a Proper Class. Informally, we can say that "x is in R" if $\phi$(x) is true.

An interesting fact about this theorem is that we didn't need any axioms of Set Theory, just basic logic. This means, that in any Set Theory based on the same logical principles The Russell Class is not a set.

Michael Carey
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