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I have a doubt regarding Russell's paradox.

We know that the collection of sets not containing themselves is not a set. Hence, let that collection be a class $\mathcal{A}$.

Does this imply that $\mathcal{A}$ cannot contain itself? Because if it does, then it will be a set not containing itself (as it is inside $\mathcal{A}$), which is a contradiction as we just stated that $\mathcal{A}$ is not a set.

Thank you.

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    $\begingroup$ As $\mathcal{A}$ is a proper class, it cannot be an element of any class. Only sets can be elements of classes. $\endgroup$ – Daniel Fischer May 23 '14 at 10:33
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    $\begingroup$ $\in\ne\subset$. $\endgroup$ – Martín-Blas Pérez Pinilla May 23 '14 at 10:41
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By your definition, $\mathcal{A}$ is the collection of sets that do not contain themselves. Also, you have acknowledged that $\mathcal{A}$ is not a set.

Since $\mathcal{A}$ contains only sets and $\mathcal{A}$ is not a set, then no, it cannot contain itself.

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Assuming you're working in NBG or ML or something, $\mathcal{A}$ is a proper class that doesn't contain itself. Theories with proper classes only let you form collections of those classes which are sets. Here, Russel's paradox proves that $\mathcal{A}$ is not a set, so it doesn't even match the defining condition of $\mathcal{A}$ in the first place.

If you're working in ZFC, there's simply no such thing as $\mathcal{A}$.

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  • $\begingroup$ So what $\mathcal{A}$ is in reality is the class of all sets not containing themselves? In ZFC, I assume. $\endgroup$ – algebraically_speaking May 23 '14 at 10:44
  • $\begingroup$ Yes, it is the class of sets that don't contain themselves. Though there's an important point in $ZFC$ that $\mathcal{A}$ doesn't exist as a set or class internally to the theory. More precisely, none of the quantifiers of ZFC range over any such thing as $\{x: \exists y(x\in y)\wedge x\notin x\}$. $\endgroup$ – Malice Vidrine May 23 '14 at 10:59
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In the context of modern set theory, a proper class is not an element of another class. In other words, if $X$ is a class, and there is some $Y$ such that $X\in Y$, then $X$ is a set.

Since we already know that $\{A\mid A\notin A\}$ is not a set, it cannot be an element of another class, let alone of itself.

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