Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 26 / 2 / ii / Solution

Suppose a nonempty set $S$ had no minimal member for the intersection-power-set predecessor relation. Put $b=\bigcap S$, which is a set by separation. For every $y\in S$ there is an $x\in S$ with $\mathcal{P}(x\cap y)\subseteq y$. Since $b\subseteq x\cap y$, this gives $\mathcal{P}(b)\subseteq y$. Intersecting over every $y\in S$ yields $\mathcal{P}(b)\subseteq b$.

Now form the diagonal subset $d=\{u\in b:u\notin u\}$. It belongs to $\mathcal{P}(b)$, hence to $b$, so its defining condition implies $d\in d⟺d\notin d$, a contradiction. Thus $\boxed{R\text{ is well-founded}.}$ This is a Cantor diagonal argument; it uses neither the Axiom of foundation nor the axiom of choice. In particular, no selection of an infinite descending sequence is needed.