0
$\begingroup$

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!

  • 0
    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. – 2018-02-10

1 Answers 1

6

It's a consequence of the axiom of foundation (AF): 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 $y \cap \{x\} \neq \varnothing$, which is the negation of AF.

Furhtermore, 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.

  • 0
    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}. – 2017-02-26
  • 0
    Is my above restatement of your argument correct? – 2017-02-26
  • 0
    @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 – 2017-02-26
  • 0
    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 – 2017-02-26
  • 0
    @MauroALLEGRANZA Very clear explanation! Thanks so much. – 2017-02-26
  • 0
    @ButterMath You're welcome! – 2017-02-26