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I have a basic and not very clear question. Foundation ensures that there are no infinitely-descending ∈-chains. But given the vastness of a ZFC universe, this seems hard to believe. Shouldn't there be an infinite membership chain somewhere?

Does the absence of such a chain have to do with the finiteness of formulas?

I guess I'm lacking the intuition behind the axiom.

Thanks for your help.

davidp
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  • The natural numbers are pretty vast as well, but this scarcely implies that "somewhere" an infinite descending chain of them exists, $n_1 \gt n_2 \gt \ldots$. – hardmath Sep 26 '16 at 13:29
  • True! But they're all finite, and the whole point of set theory is to investigate the infinite. (I realize that this is a superficial "reason" to expect Foundation to fail. I'm just pointing out that the cases are different.) – davidp Sep 26 '16 at 18:01
  • Of course we can imagine a "universe" in which the Axiom of Foundation fails, so we do need to have such an axiom. See also Axiom of Regularity. That formulas are finite really has no bearing on the absence of infinitely descending membership chains. – hardmath Sep 26 '16 at 18:52

2 Answers2

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No. The absence of such a chain is guaranteed by the Axiom of Foundation (AoF) - that's what this axiom is for after all. In other words, there are models of $\mathsf{ZFC}-\text{AoF}+\neg\text{AoF}$, which behave largely like the usual set theory except that there are infinite $\in$-chains; for example, a Quine atom $x$ with $x\in x$ causes this.

Remains the question why why usually do put AoF into our axiom repository. And the reason is precisely that we are of the opinion that sets with infinite $\in$-chains are unwelcome beasts.

Hagen von Eitzen
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  • Thanks, Hagen! Why do you think that people would find infinitely-descending ∈-chains "unwelcome" in the land of infinity? Again, is the idea related somehow to the thought that one could never prove the existence of such objects in ZFC−AoF because that would require applying, say, Pairing infinitely-many times? – davidp Sep 25 '16 at 20:38
  • @davidp Everything has its use, somewhere. But as the abstraction of the concept of collecting things together, it may seem strange that one is able to unpack infinitely often (matrioshka-like) as that would mean one has never started collecting ... – Hagen von Eitzen Sep 25 '16 at 22:17
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Let's consider for example the von Neumann ordinal numbers..

For finite numbers, it's obvious that there is no infinite descending chain. However what about the first infinite number $\omega$? Well, it contains all finite numbers, and exactly those. Therefore no matter which number you choose, from that point on you can only "descend" a finite number of times. And since you've done only one additional step, the whole chain will be finite as well (because a finite number plus one is still a finite number). That is, the length of your chain is always finite, but you can choose an arbitrary long finite chain.

OK, but $\omega$ is special, so what about $\omega+1$? Well, that differs from $\omega$ only by the extra element $\omega$. So you have two options: Either you select a finite number, then you obviously are in the exact same situation as before. Or you select $\omega$, but for $\omega$ we already know that all chains are finite, and you got just one more. Same argument goes for $\omega+2$ etc. And for $\omega\cdot 2$, we essentially get back to the argument we used for $\omega$.

Indeed, you can make this more general: For any ordinal, no matter which element (that is, which smaller ordinal) you choose first, the length of your chain will always be one more than the length you choose from the chosen ordinal. Therefore if all ordinals below $\alpha$ only admit finitely descending chains, then $\alpha$ does, too, because you need to select one element, and then you've already arrived at a smaller ordinal, so any chain from there is finite, and so the total chain, being one longer, is finite as well.

And indeed, the same argument can be made for the von Neumann hierarchy: Every set has a rank given by an ordinal, and each set only contains sets of lower rank. And thus the very argument I've given above for the ordinals works here, too.

Now what about sets not in the von Neumann hierarchy? Well, it turns out that ZFC doesn't have such sets. All ZFC sets are in the von Neumann hierarchy. And the reason for that is exactly the axiom of foundation.

Basically large sets are large because they have many elements, and therefore also many chains, not because they have long chains. The length of each chain is finite. The number of chains can be arbitrarily large.

celtschk
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  • Thanks, celtschk. I had rehearsed the argument you give here in my head. For the general case, does one argue by induction? Does that assume Foundation? – davidp Sep 25 '16 at 20:56
  • @davidp: Yes, it's transfinite induction; what I wrote is the induction step. To prove it for the von Neumann universe, you don't need the axiom of foundation. What the axiom of foundation effectively tells you is that there are no sets that are not in the von Neumann universe. – celtschk Sep 25 '16 at 21:01