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Let $X$ be any arbitrary set. Set $X$ exists. Now assume on the contrary that $\mathscr{P}(X) \subseteq X$. Since $X \in \mathscr{P}(X) $, we have $X \in X$. But this is not allowed by the axioms. So we reach contradiction and hence, $\mathscr{P}(X) \nsubseteq X$ for any set $X$. Is this a valid proof ?

user9026
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  • Yes, it is a valid proof by contradiction. – Angelo Aug 30 '20 at 08:50
  • Thanks Angelo. I see that others have posted a similar question. Since I got the answer, should I delete the question ? – user9026 Aug 30 '20 at 08:51
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    No, you should not delete your question because it could be useful for other people even though others have already posted similar questions. I have upvoted you. – Angelo Aug 30 '20 at 08:53
  • It is only valid in a purely formal sense. I'm sorry, but it is absurd to prove this statement this way. – Andrés E. Caicedo Aug 30 '20 at 12:42
  • @andrés-e-caicedo Rick and Angelo have said that its a valid proof. What is the error here ? – user9026 Aug 31 '20 at 05:28
  • There is no mathematical error. Just a pedagogical one. There is a point to the axiom of regularity, namely, that sets are ranked. This is a technical feature. It is irrelevant here. Proving things is not just about checking, but more about understanding. I think that this proof gives you no insight into how regularity is actually used, or in which situations it is useful, and also gives you no indication of why the result you are after is true. That $\mathcal P(X)$ cannot be contained in $X$ is much more fundamental, and it holds even if regularity does not. – Andrés E. Caicedo Sep 03 '20 at 21:19
  • This proof makes the result look technical and mysterious, not something that a high school could understand, for example. Why is $\mathcal P({a,b})\not\subset{a,b}$, for instance? Surely it is not because of some axiomatization or other including regularity as an axiom. – Andrés E. Caicedo Sep 03 '20 at 21:21

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This is a perfectly valid proof. Maybe you could also add a clarification on what axioms are used to prove that for any set $X$, $X \notin X$; i.e. the axioms of pairing and regularity.

Rick
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