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I am slightly confused by the notation used in Russell's paradox. I am following this text.

I understand that $\phi (x)$ is this boolean function, which outputs either True or False. I understand that $R$ is the set of all $x$ which has the image True when subjected to $\phi$. This is denoted by $$R = \{x:\phi (x)\}$$

Great. I am good so far.

Now, the author defines $\phi(x)$ such that $x \in x$. I don't understand this notation. How can an object belong to itself? Shouldn't it be '$=$' instead of '$\in$'?

Then the author goes ahead and defines $R$ is a set that contains all $x$ such that $x$ does not belong in $x$, or using symbols, $$R = \{ x: \, \sim\phi(x)\}$$ I don't understand this $x\in x$ notation used. I believe this very argument is used to explain the paradox. Could someone explain this?

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    $\begingroup$ Intuitively, think of this: if I make a list of all of the things in my room, and the list itself is in my room, then the list itself will be on the list. Therefore, the list belongs on the list of things in my room. This is how something can belong to itself. $\endgroup$ – David Lui Jan 18 at 5:09
  • $\begingroup$ This is great, lol. Thanks @DavidLui $\endgroup$ – megamence Jan 18 at 5:21
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For the $x\in x$ part, you can just think $\phi(x)$ is false for every $x$. This will not affect the paradox. So now $R$ is literally the set of "everything," and the paradox is asking "does $R$ belong to $R$ itself?" If $R\in R$, i.e., the set of "everything" is one of "everything," then $\phi(R)$ is true, contradicting the construction of $R$; if $R\notin R$, then $\phi(R)$ is true, so $R\in R$, contradiction.

See this. ZFC says that the construction of $R$ is not allowed, for we have to specify a "bound" for $x$.

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How can an object belong to itself?

Well, it depends on what "belong" means, doesn't it? I'm inclined to agree with you, but this is a somewhat controversial point; some people have found uses for set theories where a set can belong to itself.

In a sane world, where things can't belong to themselves, Russell's paradox would take the simpler form: $$R=\{x:x=x\}$$ Now $R=R$, so $R\in R$, which is absurd. Paradox!

But what if someone objects: "I'm not so sure. Normal sets can't belong to themselves, but maybe there are some abnormal sets that do belong to themselves? Maybe your $R$ is one of those?"

To satisfy those people, instead of defining $R$ to be the set of all sets, we define it to be the set of all normal sets: $$R=\{x:x\not\in x\}$$ And now, as you know, we get a paradox, regardless of whether abnormal sets exist or don't.

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It's not a typo: although the idea of a set containing itself as an element may be weird, and we may even want to include an axiom in our set theory forbidding such shenanigans (such as the axiom of regularity in the setting of $\mathsf{ZFC}$), it makes sense to ask whether or not a set is an element of itself, and so we can ask whether there is in fact a set consisting exactly of those sets which are not elements of themselves.

Whether or not self-elementhood actually happens doesn't affect the paradox. This may be made clearer by distinguishing between sets and classes (= a collection of sets which may or may not be a set itself). In a theory like $\mathsf{ZFC}$, which prohibits self-elementhood, the Russell class $\{x: x\not\in x\}$ is the whole universe of sets: every set satisfies $x\not\in x$. In theories like $\mathsf{NFU}$ or $\mathsf{GPK^+_\infty}$ which do allow self-elementhood, the Russell class is more complicated. But either way, the Russell class can never be a set, as Russell's argument shows.

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  • $\begingroup$ thank you for your response @NoahSchweber. I have a quick question about the axiom of regularity. The axiom states that for every set $A$, there exists an element $a\in A$ such that $A \cap a = \{ \}$. How is this possible? for any $a\in A$, shouldn't $A \cap a = a$? $\endgroup$ – megamence Jan 18 at 5:04
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    $\begingroup$ @megamence No. We have $A\cap a=a$ if $a$ is a subset of $A$, but here $a$ need only be an element of $A$. Take for example $A=\{\{\emptyset\}\}$ and $a=\{\emptyset\}$. We have $a\in A$ and $A\cap a=\emptyset\not=a$. $\endgroup$ – Noah Schweber Jan 18 at 5:08
  • $\begingroup$ that is a little wacky... but valid nonetheless $\endgroup$ – megamence Jan 18 at 5:22

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