Set Theory HT25, Foundation

\[\forall x (x \ne \emptyset \to \exists y ( y \in x \land y \cap x = \emptyset)\]

Every non-empty set $x$ has an $\in$-minimal element, i.e. an element $y \in x$ such that no element of $x$ is an element of $y$.

1. If $x \in x$, then $\lbrace x\rbrace$ violates the axiom of foundation, since the only element of $\lbrace x\rbrace$ is $x$, but $x \cap \lbrace x\rbrace = \lbrace x\rbrace \ne \emptyset$.
2. If $x \in y \in x$, then $\lbrace x, y\rbrace$ violates foundation, since $y \in x \cap \lbrace x, y\rbrace$ and $x \in y \cap \lbrace x, y\rbrace$.
3. If $\mathcal P(x) \in x$, then $y = \lbrace x, \mathcal P(x)\rbrace$ violates foundation, since $x \cap y \ne \emptyset$ and $\mathcal P(x) \cap y \ne \emptyset$.
4. If $\langle x, y \rangle \in x$ for some $y$, then $x \in \lbrace x\rbrace \in \langle x, y \rangle \in x$ and hence $Y = \lbrace x, \lbrace x\rbrace , \langle x, y \rangle\rbrace$ has no $\in$-minimal element.

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For reference, the axiom of foundation states:

\[\forall x (x \ne \emptyset \to \exists y \in x \text{ }y \cap x = \emptyset)\]

Every non-empty set $x$ has an $\in$-minimal element, i.e. an element $y \in x$ such that no element of $x$ is an element of $y$.

Suppose $X$ is such a set. Then $X$ has an $\in$-minimal element, say $x$. Suppose that $x \ne \emptyset$, so that $y \in x$. But then by transitivity, $y \in X$, which contradicts $x$ being $\in$-minimal.

@exam~ @example~