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Can someone help me out? I don't really get what "x does not belong to x" means in this: Russel's paradox

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  • $\begingroup$ Related: math.stackexchange.com/questions/253818/… $\endgroup$ – Asydot Sep 21 '15 at 14:14
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    $\begingroup$ It means that $X$ is not element of $X$. $\endgroup$ – drhab Sep 21 '15 at 14:17
  • $\begingroup$ If you are wondering how a set can be element of itself, it is a construct that you will never see in, say, number theory or analysis. It is one, however, that created chaos in set theory just over a century ago, and inspired various "fixes" to set theory. There was no formally stated rule that you could not have $x\in x$ or $x\notin x$. So why not have $S=\{x:x\notin x\}$? This led to Russell's famous paradox. You could both prove and disprove the existence of $S$ using the set theory of the day. $\endgroup$ – Dan Christensen Sep 21 '15 at 15:26
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In this scheme, the elements $X$ of $A$ may themselves be sets. In fact, the set $X$ might contain itself as an element, so that $X \in X$. For example, a set $Y$ satisfying $Y = \{1,2,Y\}$ contains itself.

What we are defining, then, is the set of elements $X$ (of $A$) that do not contain themselves. In particular, if we take $Y$ as defined above, then $Y \notin B$.

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  • $\begingroup$ I should rather say: a set $Y$ that satisfies $Y=\{1,2,Y\}$ contains itself (as element). $\endgroup$ – drhab Sep 21 '15 at 14:32
  • $\begingroup$ @drhab is the set not uniquely determined by this equality? $\endgroup$ – Omnomnomnom Sep 21 '15 at 14:36
  • $\begingroup$ I don't think so. Am I wrong then? My comment was purely based on intuition. Btw, got to go now. I will come back later. $\endgroup$ – drhab Sep 21 '15 at 14:43
  • $\begingroup$ I say that $Y$ is uniquely described since we have given all of its elements. $\endgroup$ – Omnomnomnom Sep 21 '15 at 14:45
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    $\begingroup$ I don't see where you get that. Suppose the property $Y = \{1,2,Y\}$ holds for two different entities, say $y_1$ and $y_2$. You want to claim based on extensionality that $y_1=y_2$ because they have the same elements, but to do that you need to know $y_1=y_2$ already. $\endgroup$ – MJD Sep 21 '15 at 15:49

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