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In Russell's paradox, if there is a set that has everything as an element, it can be seen that it is a contradiction. I think if there is a set X such that $X \in X$, there is contradiction. I want to prove this by axiom of regularity but i cannot. How to prove this?

Arturo Magidin
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    Perhaps begin by stating the Axiom of Regularity. Then, assuming you have such a set $X$, try to fit it in that statement. – GEdgar Mar 24 '21 at 15:41
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  • Russell's paradox does not arise from a universal set; though there are paradoxes that arise from the existence of a universal set. 2. You do need regularity (there are set theories that do not have the axiom of regularity in which there do exist sets that are elements of themselves). 3. To determine how to prove this using the Axiom of Regularity, it would be helpful to know exactly what version of the Axiom of Regularity you are using.
    • – Arturo Magidin Mar 24 '21 at 15:42
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    The proof is given on Wikipedia. (Look at the first section on elementary implications of regularity.) – symplectomorphic Mar 24 '21 at 15:47