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I am having troubles with proving the following statement:

if $x$ is a set, then $x \notin x$.

I am only allowed to use the axioms from the Zermelo-Fraenkel Set Theory. Thanks so much if someone could give me a hand!

Asaf Karagila
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  • Because then $x$ would be an element, and not a set. Contradiction. It follows, then, that $x\subseteq x$ and $x\notin x$, as desired. – Mr Pie Feb 10 '18 at 10:28

1 Answers1

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It's a consequence of the axiom of foundation (AF, aka axiom of regularity).

Specifically, if $x \in x$, then you cannot find an element $y$ of the non-empty set $\{x\}$ (which actually exists by the ZF axiom of pairing!) such that $y \cap \{x\} = \varnothing$; this contradicts AF.

Furthermore, it can be shown that neither AF nor it's negation are consequences of the other ZF axioms, which means the above argument is essentially the only possible proof.

kjo
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Tig la Pomme
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    Let me restate your argument so that you can check whether I understand this correctly. The negation of the axiom of foundation is ∃x( (x≠∅) ∧ (∀y∈x (y∩x ≠ ∅)) ). This statement is true because that set x does exist, which is the set {x}. –  Feb 26 '17 at 12:37
  • Is my above restatement of your argument correct? –  Feb 26 '17 at 12:44
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    @ButterMath - yes, your negation is correct. Consider it : $∃z( (z≠∅) ∧ (∀y∈z (y∩z ≠ ∅))$ and consider ${ x }$ as $z$. We have that $x \in { x }$ and thus ${ x }$ it is not empty. What about the second conjunct ? We have that $∀y ∈ { x } (y∩ { x } ≠ ∅)$; but there is only one $x \in { x }$ : it is $x$ itself. 1/2 – Mauro ALLEGRANZA Feb 26 '17 at 12:54
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    Thus : $x∈ { x }$ and if $x \in x$ we have that $x \in (x \cap { x })$ and thus also the second conjunct is true. Conclusion: if $x \in x$ then the negation of (AF) is true and this is a contradiciton. So, we have to discard the assumption that $x \in x$. 2/2 – Mauro ALLEGRANZA Feb 26 '17 at 12:55
  • @MauroALLEGRANZA Very clear explanation! Thanks so much. –  Feb 26 '17 at 13:04
  • @ButterMath You're welcome! – Tig la Pomme Feb 26 '17 at 15:39
  • @TiglaPomme: I made several edits to your answer, including a pretty crucial correction (namely, I replaced a $\neq$ with $=$, which is cearly a pretty big deal, as far as the post's argument goes). I hope you agree with them. – kjo Jan 26 '22 at 14:58