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I was reading, https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory#2._Axiom_of_regularity_.28also_called_the_Axiom_of_foundation.29

And it claims the Axiom of regularity implies that no set can be a member of itself. I don't follow whats being conveyed here. Consider this example:

Let $L = \lbrace{1 , L} \rbrace$

or to do this in ZFC

Let $L = \lbrace \lbrace \emptyset \rbrace , L \rbrace $

Then consider $L \cap \lbrace \emptyset \rbrace $

Obviously this is empty. So we meet the disjointedness condition, yet we can still have self membership. What am I missing here?

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The Axiom of Regularity states that if $A$ is a non-empty set, then there is $B\in A$ with $A\cap B=\emptyset$.

In your example, consider $A=\{L\}$. Is there $B\in A$ such that $A\cap B$ is empty.

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  • $\begingroup$ I see. So by wrapping L into a set we can find a contradiction. $\endgroup$ – frogeyedpeas Oct 15 '17 at 6:43

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