# How much foundation do we get for free in $\sf ZFC$?

In $\sf ZFC$ we have the axiom of infinity and thus can define the natural numbers $$\mathbb N \equiv \bigcap\{X:\emptyset\in X\land \forall n(n\in X\implies n\cup\{n\}\in X)\}.$$ From this it's not particularly hard (exercises 1.6 and 1.7 in Jech - Set Theory) to prove that, firstly, every $n\in\mathbb N$ obeys foundation and secondly, $\mathbb N$ itself obeys it, all without explicitly using the axiom of foundation. My question is: does this extend to arbitrary finite sets or even other countable sets? If not, are there any other specific important examples for which we can verify foundation?

• One can show that, in general, all ordinals $\alpha$ satisfy the property that $\alpha\notin \alpha$. – Hayden Jan 30 '15 at 0:25
• @Hayden Does that necessarily imply the pure form of foundation with every nonempty subset of $\alpha$ having a $\in$-minimal element? Excuse my ignorance, I'm quite new to set theory. – theage Jan 30 '15 at 0:27
• Quite alright, to be honest I don't know if that is equivalent to foundation, though I would assume not. Still, it's at least a little bit of regularity that happens to hold for all ordinals, so I figured it was worth mentioning. I'll look around to see if I can find something on the (non)equivalence of the statements, though. – Hayden Jan 30 '15 at 0:29
• I really don't understand the question. Are you asking how many sets are well-founded in $\sf ZFC$? Or are you asking about $\sf ZFC-Fnd$? Or are you talking about foundations of mathematics? – Asaf Karagila Jan 30 '15 at 0:55

The von Neumann herirarchy is defined as $$V_0=\emptyset$$ $$V_{\alpha+1}=P(V_{\alpha})$$ $$V_{\lambda}=\cup_{\alpha < \lambda} V_{\alpha}$$
Now all elements of $V$ satisfy foundation. What foundation really means is that all sets belong to $V$. Since in practice all the sets we naturally deal with are already in $V$ foundation is not so essential to the development of set theory or mathematics.