0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ One can show that, in general, all ordinals $\alpha$ satisfy the property that $\alpha\notin \alpha$. $\endgroup$ – Hayden Jan 30 '15 at 0:25
  • $\begingroup$ @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. $\endgroup$ – theage Jan 30 '15 at 0:27
  • $\begingroup$ 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. $\endgroup$ – Hayden Jan 30 '15 at 0:29
  • $\begingroup$ 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? $\endgroup$ – Asaf Karagila Jan 30 '15 at 0:55
4
$\begingroup$

All the sets we define and use in the development of ordinary mathematics are automatically well-founded. Essentially, any set you can prove exists can also be proven to be well-founded without using the axiom of foundation.

The role of the axiom of foundation is not to tell us anything new about the sets we're already using, but merely to claim that there are no other sets than the well-founded ones. It puts a limit on which kinds of wild, spurious things may exist in a model of set theory without our asking for them, which is sometimes technically convenient -- but it doesn't say anything we don't already know about the sets we want to exist.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.