Axiom of Foundation (Strong Form)/Proof 2

From ProofWiki
Jump to navigation Jump to search


Let $B$ be a class.

Suppose $B$ is non-empty.

Then $B$ has a strictly minimal element under $\in$.



This page is beyond the scope of ZFC, and should not be used in anything other than the theory in which it resides.

If you see any proofs that link to this page, please insert this template at the top.

If you believe that the contents of this page can be reworked to allow ZFC, then you can discuss it at the talk page.

Let $x \in B$.

Let $x'$ be the transitive closure of $x$.

Let $L = x' \cap B$.

Then $x \in L$, so $L$ is not empty.

Since $x'$ is a set, so is $L$, by the axiom of subset.

Thus by the Axiom of Foundation, $L$ has an $\in$-minimal element $m$.

By the definition of intersection, $m \in B$.

Aiming for a contradiction, suppose that there is an element $b \in B$ such that $b \in m$.

Then since $m \in x'$ and $x'$ is transitive, $b \in x'$.

Thus $b \in L$, contradicting the minimality of $m$.

So $m$ is an $\in$-minimal element of $B$.