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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?

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You mean well founded, right? –  Asaf Karagila Jan 19 '13 at 0:07
    
No, I don't. A well-founded relation has the additional property that $A\cap R^{-1}\{x\}$ is a set for any set $x\in A$. I just checked my reference (Takeuti and Zaring, Axiomatic Set Theory) and the also call $R$ a foundational relation, but well-founded has another meaning. –  Mario Carneiro Jan 19 '13 at 0:09
    
No, that's a set-like relation. –  Asaf Karagila Jan 19 '13 at 0:10
    
I'm just quoting definitions. All I can say is that they call this a foundational relation, and a well-founded relation means as I said above. Call my definition whatever you like, as long as you keep the quoted definition (that's exactly why I gave the definition in the first place). –  Mario Carneiro Jan 19 '13 at 0:13
    
Yes. I realized that already. :-) It's an interesting question, but I'm too tired to tackle it right now. By morning it'll probably be answered, but if not I'll give it more thought. (+1 anyway...) –  Asaf Karagila Jan 19 '13 at 0:17

1 Answer 1

up vote 2 down vote accepted

$\newcommand{\rank}{\operatorname{rank}}$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.

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Very nice! This is exactly what I was looking for. Just as a side note, is the Axiom of Infinity required for this proof? –  Mario Carneiro Jan 19 '13 at 18:35
    
(To answer my own comment) If infinity is denied, then all sets are finite, so if $xRy\Leftrightarrow y\in x$ (reverse inclusion), then on any finite subset of $\omega$, there is an $R$-minimal element (the largest number), but $\omega$ itself has no $R$-minimal element. –  Mario Carneiro Jan 19 '13 at 18:39
    
@Mario: It seems to be necessary at least for this argument: it seems that we need that in order to apply replacement to bet $s$. And your example finishes the answer. –  Brian M. Scott Jan 19 '13 at 18:43

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