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From my understanding, the Russell class can't be a member of other classes because by definition it contains all the members of other classes and hence any class is contained in the Russell class. But what is confusing me is that why the Russell class can't be a member of itself? What contradiction would that lead to? Thank you!

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2 Answers 2

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I think you've completely misunderstood its definition. The Russell class $R$ satisfies this: for any $x$, $x\in R$ iff $x$ is a set and $x\notin x$. To avoid a paradox, $R$ cannot be a set, since then we'd have $R\in R\iff R\notin R$. Since $R$ isn't a set, $R\notin R$.

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    $\begingroup$ Thank you very much for your time! $\endgroup$
    – Lili Tong
    Nov 6, 2019 at 10:44
  • $\begingroup$ @LiliTong Is there anything else you'd need clarified for me to get the green tick back? $\endgroup$
    – J.G.
    Jan 9, 2020 at 17:08
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If you're talking NBG, where the domain of discourse is classes and any formula specifies a class, then $R=\{x:x\notin x\}$ is a perfectly valid class, but one has to be careful: in the class builder operator $x$ only refers to sets. In NBG a class $a$ is a set if and only if there exists a class $b$ such that $a\in b$. If we denote by $M$ the unary predicate “being a set”, the class $R$ is more precisely defined as $$ R=\{x:M(x)\land x\notin x\} $$ We can have $R\in R$ only if $M(R)$. However, this leads to a contradiction:

If $R\in R$, then $R$ is a set and satisfies $R\notin R$.

Thus we conclude that $R$ is not a set (hence a proper class). In particular $R\notin R$, because otherwise $R$ would be a set.

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  • $\begingroup$ Sorry if it's a stupid question, for the reason why $R$ cannot be a member of some other classes, is it because in {x|(x is a set) and (p(x))} where p(x)="x=$R$", we are presuming $R$ is a set? Thank you $\endgroup$
    – Lili Tong
    Jan 9, 2020 at 10:09
  • $\begingroup$ @LiliTong The predicate $x=R$ is false if $x$ is a set, because $R$ isn't a set. $\endgroup$
    – egreg
    Jan 9, 2020 at 12:14

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