Does there exist a set $A$ such that $A \in A$?
The empty set doesn't work. Also, it seems clear that if $A$ exists, then it must be infinite.
If no such set exists, then is there a simple proof of this?
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1This was covered before on the site. – Asaf Karagila Jan 19 '14 at 18:02
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Why would it have to be infinite? It could just be the one-element set $A={A}$. – bof Jan 19 '14 at 18:11
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@bof That doesn't Work. It is not true that ${A} \in {A}$. – Gamma Function Jan 19 '14 at 18:16
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2@JacobMayle Obviously this is impossible with the axiom of foundation. Without that axiom, it is consistent to have sets $A={A}$, have sets $A\ne B$ with $A={B}$ and $B={A}$, have sets $A={A,B}$ with $B\ne A$, etc. – Andrés E. Caicedo Jan 19 '14 at 18:22
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Related: Non-well-founded set theory – Jan 19 '14 at 18:25
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http://math.stackexchange.com/questions/253818/example-of-set-which-contains-itsel and http://math.stackexchange.com/questions/200255/is-the-statement-a-in-a-true-or-false – Asaf Karagila Jan 19 '14 at 18:27
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@AndresCaicedo: My favorite are Boffa atoms--$A={x: A\in x}$--which you can get in NF. – Malice Vidrine Jan 19 '14 at 19:03
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In NF and NFU such sets exist, and in Positive Set Theory as well; most obviously in the case of the universe $V$, $V\in V$. NF can also have objects called Quine atoms which are sets $A$ such that $A=\{A\}$, an object which one also finds when one replaces Foundation in ZF with Aczel's Anti-foundation axiom. But as mentioned by other answerers, ZF typically explicitly precludes such things.
Malice Vidrine
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Such a set does not exist in $\sf ZFC$, with emphasis on $\sf F$.
By the axiom of foundation, if such a set $x$ existed, then since $\{x\}$ isn't empty it would follow that there exists $y\in \{x\}$ such that $y\cap \{x\}=\varnothing$. So $y=x$, $x\in y$ and $x\in \{x\}$ contradicting $y\cap \{x\}=\varnothing$.
Git Gud
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What if $x$ contains itself and also some other members.. a,b,c...?$x={x,a,b,c...}$ – Michal Dvořák May 13 '18 at 23:09
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2@MichalDvořák My proof is independent of that hypothesis. Assume $x={x,a,b,c}$. Now allow me to completely ignore that assumption and allow me to consider the set ${x}$ (which isn't empty). From here read the rest of my proof. Does this help? – Git Gud May 14 '18 at 20:33
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This is the Russell's paradox- right?
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@DavidMitra: I think the ideas are related. These inconsistencies led to the ZFC formulation I think. – voldemort Jan 19 '14 at 18:06
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5There is nothing inconsistent about sets that belong to themselves. It is not related to Russell's paradox. The theory $\mathsf{ZFC}$ explicitly rules out these sets, because they do not belong to the cummulative hierarchy, but this is a separate thing. We also explicitly rule out urelements, and they are not paradoxical either. – Andrés E. Caicedo Jan 19 '14 at 18:15
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Yeah, if ZF minus Foundation were inconsistent with self-membership then it would refute any anti-foundation axiom, which we know isn't the case... – Malice Vidrine Jan 19 '14 at 18:46