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Yes, this is homework. We're asked to show that the following 3 statements for the Axiom of Foundation are equivalent:

$(1)\quad V = \bigcup_{\alpha} V_{\alpha}$
$(2)\quad \in \text{ is well-founded.}$
$(3)\quad \text{If } x \text{ is a non-empty set, then there is an } a \in x \text{ such that } a \cap x = \varnothing$

I was able to rewrite $(3)$ to show that it claims that there's an $\in$-minimal element $a$ in every non-empty set $x$, implying statement $(2)$. I was also able to show that $(2)$ implies $(3)$ by contradiction after invoking an infinite descending $\in$-chain.

I also showed that $(1)$ implies $(3)$, but what I'm having a hard time doing is showing that $(3)$ implies $(1)$ or $(2)$ implies $(1)$.

Any hints?

user123
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  • Have you heard of the Principle of $\in$-induction? This could very much be useful in showing that (2) implies (1). Once you have this, since (3) implies (2), combining these gives you that (3) implies (1). – Hayden Dec 07 '15 at 22:10
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    Wait, if (3) is not the axiom of foundation, what is? – Asaf Karagila Dec 07 '15 at 23:47

1 Answers1

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First, you want to prove that if $m$ is a set and $m\subset\bigcup_\alpha V_\alpha$, then $m\in \bigcup_\alpha V_\alpha$.

Once you have that, you might be tempted to make the following argument. Let $S=V\setminus\bigcup_{\alpha} V_\alpha$. If $S$ is nonempty, then (2) says it has an $\in$-minimal element $m$. But then every element of $m$ is in $\bigcup_\alpha V_\alpha$, so by the previous paragraph, $m\in \bigcup_\alpha V_\alpha$, which is a contradiction.

Unfortunately, this argument doesn't quite work, because $S$ might not be a set (a priori, it is only a class). To get around this, use the following trick. Suppose $x\in V\setminus \bigcup_\alpha V_\alpha$, and let $T$ be the transitive closure of $\{x\}$. Let $S=T\cap (V\setminus \bigcup_\alpha V_\alpha)$ (this is a set by Separation since $T$ is a set). Now apply the argument above to this $S$.

Eric Wofsey
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  • Hah, I made that first argument and hit a dead end when I realized that $S$ could be a class. This trick seems to resolve it. That was helpful! – user123 Dec 07 '15 at 22:51