TITLE: Can a founded relation find minimal elements in proper classes?
QUESTION [5 upvotes]: My definition of a founded relation $R$ on a (possibly proper) class $A$ is
$$R\mbox{ Fr }A\iff \forall x\subseteq A\,(x\neq\emptyset\rightarrow\exists y\in x\ \forall z\in x\ \neg zRy),$$
or equivalently,
$$R\mbox{ Fr }A\iff \forall x\subseteq A\,(x\neq\emptyset\rightarrow\exists y\in x\ x\cap R^{-1}\{y\}=\emptyset).$$
I am working in ZF, so obviously $x$ must be a set so that I can quantify over it. But I would like to conclude from this definition that 
$$X\subseteq A\rightarrow(X\neq\emptyset\rightarrow\exists y\in X\ \forall z\in X\ \neg zRy),$$
where $X$ is now an arbitrary class, which we can assume is a proper class. Since $X$ is then not included in the quantifier, I cannot immediately conclude this theorem, but my question is if it is possible for me to derive this by other means. If it is not true, are there definable counterexamples?

REPLY [4 votes]: $\newcommand{\rank}{\operatorname{rank}}$It can be proved using the notion of the rank of a set: $\rank(x)=\min\{\alpha\in\mathbf{ON}:x\in V_{\alpha+1}\}$, where the von Neumann hierarchy is defined by $V_0=0$, $V_{\alpha+1}=\wp(V_\alpha)$, and $V_\eta=\bigcup_{\xi<\eta}V_\xi$ if $\eta$ is a limit ordinal.
Suppose that $R$ is a foundational relation on $A$, and $\varnothing\ne X\subseteq A$. The idea is to show that if $X$ has no $R$-minimal element, there is a set $s\subseteq X$ that has no $R$-minimal element, contradicting the hypothesis that $R$ is foundational. We start by using the rank function to form a non-empty subset of $X$: let $$X_0=\left\{x\in X:\forall y\in X\big(\rank(x)\le\rank(y)\big)\right\}\;.$$ Note that $X_0\subseteq V_\alpha$ for some $\alpha$, so $X_0$ is a set. Now we want to expand $X_0$ to a set $s$ with the property that if $x\in s$ is not $R$-minimal in $X$, then $x$ is not $R$-minimal in $s$, either. We do this recursively: given a set $X_n$ for $n\in\omega$, we’d like to form $X_{n+1}$ by adding enough elements of $X$ to ensure that if $x\in X_n$ is not $R$-minimal in $X$, there is some $y\in X_{n+1}$ such that $yRx$. At the same time we want to be sure that $X_{n+1}$ is a set, so we use the minimal-rank trick again: let
$$X_{n+1}=X_n\cup\left\{x\in X:\exists y\in X_n\Big(xRy\land\forall z\in X\big(zRy\to\rank(x)\le\rank(z)\big)\Big)\right\}\;;$$
$X_{n+1}$ adds to $X_n$ the minimal-rank representatives of $\{x\in X:\exists y\in X_n(xRy)\}$, and since $X_{n+1}\subseteq X_n\cup V_\alpha$ for some $\alpha$, $X_{n+1}$ is a set.
By the replacement schema we can now form the set $s=\bigcup_{n\in\omega}X_n$. Clearly $0\ne s\subseteq X\subseteq A$. Let $x\in s$; $x\in X_n$ for some $n\in\omega$. If $x$ is not $R$-minimal in $X$, let $y\in X$ be of minimal rank such that $yRx$; then by construction $y\in X_{n+1}\subseteq s$.
Finally, suppose that $X$ has no $R$-minimal element. Then we’ve just shown that for each $x\in s$ there is a $y\in s$ such that $yRx$, i.e., that $s$ has no $R$-minimal element, contradiction the hypothesis that $R$ is foundational.