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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?

Mario Carneiro
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  • You mean well founded, right? – Asaf Karagila Jan 19 '13 at 00:07
  • 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 00:09
  • No, that's a set-like relation. – Asaf Karagila Jan 19 '13 at 00:10
  • 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 00:13
  • 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 00:17

1 Answers1

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

Brian M. Scott
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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