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I encountered idea of considering Sample space $\Omega$ as collection of function so far as I saw.

Can you show the proof of thesis $x \neq \{x\}$? In a book I saw this thesis is included in context of Russell's paradox.

Best regards,

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In general, if $x$ is any non-set object, the fact that $x \ne \{x\}$ is trivial.

Most certainly, if $x$ is a set with $|x| \ne 1$, you also have $x \ne \{x\}$ since the cardinalities differ.

The final case is $x$ is a set with cardinality $1$, i.e. $x = \{a\}$. Can you prove that $\{a\} \ne \{\{a\}\}$?

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    $\begingroup$ Can you? You delegate the one actual part of the proof, where one needs to appeal to the axiom of foundation as if it's that simple. Try to prove it yourself. You will quickly find that the proof works only if you assume that $x\neq\{x\}$. That's circular. $\endgroup$ – Asaf Karagila Sep 17 '18 at 14:39
  • $\begingroup$ @AsafKaragila i do not have experience with the axiom of foundation, but i thought one could use the induction on the levels of indirection of the set. In other words, clearly $a \ne \{a\}$ if $a$ is not a set, and then inductively assuming this works for $n$ levels, prove that it works for $n+1$ -- why is this problematic? Sorry to bother you with a beginner's question, but the topic interests me $\endgroup$ – gt6989b Sep 17 '18 at 15:01
  • $\begingroup$ @AsafKaragila I upvoted your answer in the linked question, but I still don't understand why my primitive proof in the above comment does not work $\endgroup$ – gt6989b Sep 17 '18 at 15:04
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    $\begingroup$ The problem is that without the axiom of foundation you can't prove this by induction on rank. There is no notion of rank. So pathologies like $x\in x$ become possible. $\endgroup$ – Asaf Karagila Sep 17 '18 at 15:16

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