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I understand that the set of all sets isn't a set intuitively. But the argument in the textbook I'm reading says that the set of all sets (labeled $V$) is not a set, because if it were, we could define a subset of V in the following way:

$\{ S \in V \mid S \notin S \}$

which is Russell's class. I'm confused as to how can the Russell's class be defined as a subset, if it's not a set? Isn't the Russell's class not part of the set of all sets, since it isn't a set? Where am I going wrong?

Mixoftwo
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    It is a hypothetical statement: if the class of all sets $V$ were a set then defining the subset ${ S \in V \mid S \notin S }$ would be meaningful, but that would lead to Russell's paradox and so $V$ is not a set. – Henry Nov 14 '24 at 19:31
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    The existence of a universal set is not inherently paradoxical (NF set theory does have a universal set), but existence of a universal set is incompatible with the ZFC axiom schema of separation (Russell’s Paradox would result). See this answer – NikS Nov 14 '24 at 21:53
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    Roughly, If X is a set then {y$\in$X | $\phi$(y) } is a Set by the Axiom Schema of Separation. ( There is a slight restriction on $\phi$(y), but that is not the focus here. So if V = {x | x = x } is a set, then {y$\in$V| y∉y} = R is a Set. But, you can then produce a contradiction by considering what happens when R$\in$R – Michael Carey Nov 15 '24 at 13:16

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The point is that every sub-class of a set is a set, then if we assume that $V$ is a set, the sub-class $\{S\in V: S\not\in V\}$ would be set which leads to Russell's paradox.

On the other hand, if $V$ is a proper class, i.e., is not a set, then there is no paradox inside set-theory.

Francesco Peña-Garcia
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