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I already asked a similar question about why replacement is true in the hierarchy of sets. After a while of thinking about it and having the discussion there, I finally understand it. Now the next axiom with which I have trouble understanding is the axiom of regularity. It says that if we have a nonempty set $A$ then there is an element $x\in A$ such that $x$ and $A$ are disjoint. The usual explanation of this is that we choose for $x$ an element of $A$ which is created on a stage "as low as possible". Then all elements of $x$ are created before $x$ and all elements of $A$ are created after $x$ or at the same stage as $x$. Thus $x\cap A=\emptyset$.

But how can one assure that there is always such a "minimal stage" for a nonempty set $A$? Why isn't it possible that in the transfinite hierarchy, we can find a set $A$ such that for each member $x$ of $A$, there is another member $y$ of $A$ which is created before $x$?

I found this mathoverflow answer of Andreas Blass to a very similar question:

Begin with some non-set entities called atoms ("some" could be "none" if you want a world consisting exclusively of sets), then form all sets of these, then all sets whose elements are atoms or sets of atoms, etc. This "etc." means to build more and more levels of sets, where a set at any level has elements only from earlier levels (and the atoms constitute the lowest level). This iterative construction can be continued transfinitely, through arbitrarily long well-ordered sequences of levels.

This so-called cumulative hierarchy is what I (and most set theorists) mean when we talk about sets. A set is anything that is formed at some level of this hierarchy. This meaning of "set" has replaced older meanings.

The axiom of regularity is clearly true with this understanding of what a set is. It expresses the idea that the stages of the cumulative hierarchy come in a well-ordered sequence. (Without well-ordering, the instructions for each level, namely "form all sets whose elements are at earlier levels," would not be an inductive definition but a circularity.)

I am always depressed when somebody says that something is "clearly true" and I have trouble understanding it. But on the other hand, mathoverflow is intended for professional mathematicians, and thus Andreas Blass may have written this to be understand by other professional mathematicians (but not by beginners like me).

In this thread, I basically ask for a more detailed explanation that can be easier understood. What have inductive definitions to do with minimal stages?

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  • Do you know what a well-ordered set is? – Eric Wofsey Oct 22 '16 at 22:55
  • @EricWofsey: I think so. The axiom of regularity essentially says that the stages of the hierarchy are well-ordered. Now I ask: why? –  Oct 22 '16 at 22:56
  • No, the stages of the hierarchy are well-ordered by definition. The hierarchy is defined by transfinite recursion on the ordinals, which are well-ordered by definition. – Eric Wofsey Oct 22 '16 at 22:59
  • @EricWofsey: I guess you are mixing up two different things: the "intuitive hierarchy of sets" (that we use as the intended model of set theory) and the "von Neumann hierarchy $V_{\alpha}$". The first is used to justify the axioms, and the latter presupposes that we are already working in a set-theoretic system in which we can speak about models of ZFC (and thus formalize, that is define, the idea of the hierarchy for a given model of ZFC). –  Oct 22 '16 at 23:06
  • @EricWofsey: My point is that the latter is the "meta" perspective where we speak about models of set theory, and the first is just an informal description of the ontology that we have even before writing down the axioms. Though the ideas of a hierarchy are certainly very related. –  Oct 22 '16 at 23:06
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    Because we want it to be :) Our informal conception of the hierarchy has cumulative stages, and has no infinite descending chains of stages — that is, the stages are wellordered. – BrianO Oct 22 '16 at 23:13
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    Even in your informal ontology, the steps being well-ordered should be part of your definition of what you mean by a "hierarchy of sets". Otherwise, it is totally possible to imagine a hierarchy where the steps are not well-ordered, and regularity fails. – Eric Wofsey Oct 22 '16 at 23:15
  • @BrianO: It seems like this is just a reformulation of what I ask. You say that the hierarchy of sets "has no infinite descending chains of stages". I ask: WHY? –  Oct 22 '16 at 23:16
  • @EricWofsey, BrianO: I have the feeling that it's hard for you guys to see the question from my perspective as a beginner. For you, everything is obvious. But for me not. –  Oct 22 '16 at 23:17
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    Well, do you want it to have infinite descending stages? No? I didn't think so. – BrianO Oct 22 '16 at 23:22
  • @BrianO: I am not sure what you mean. Since when does somebody care what I want? For me, the description of the hierarchy makes sense, and now I am wondering why there isn't an infinite descending sequence of stages in this hierarchy. –  Oct 22 '16 at 23:25
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    I'd say it's because the intuition (or desire) is that if something gets created (enters the hierarchy), it does so at some particular stage, before/below which it doesn't exist because its constituents don't all exist yet. There is a first, least stage at which an entity comes into existence, not an infinite regression of stages. This basically means (or: can be modeled as meaning) that the stages are wellordered. – BrianO Oct 23 '16 at 02:07

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The cumulative hierarchy is supposed to proceed by a construction in stages or steps. At each stage, you admit the sets whose members all belong to stages that you already know about.

So you start with nothing at stage $0,$ you get the empty set at stage $1,$ then you move onto stages $2, 3, \dots.$ After all these stages, one introduces a next stage, which is stage $\omega.$ After that, of course, you keep on going, introducing stages $\omega+1, \omega+2,$ etc., and eventually a stage for each ordinal number.

The key point here is that, if this is to be something you can imagine as a construction, then the stages must be well-ordered.

One might ask: Why not introduce, instead of, say, stage $\omega,$ an infinite descending chain of stages after the stages indexed by natural numbers? The answer is that you wouldn't have any way of knowing what to do when. You could specify some non-well-ordered ordering in advance and then possibly rig up a corresponding funny "cumulative" hierarchy that you could show afterwards just happened to satisfy ZF-Foundation, but you wouldn't really be constructing things in stages any more.

Once you know that the construction ordering of stages is a well-ordering, the axiom of foundation (or regularity) follows immediately.

(Just to be clear, when I said at the beginning that this was a construction, I wasn't suggesting that the entire process is actually constructive — each individual stage is produced in a non-constructive way, and also the indexing ends up being over well-orderings that can't be created constructively.)

Mitchell Spector
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  • "The key point here is that, if this is to be something you can imagine as a construction, then the stages must be well-ordered." And my key request in this thread is an explanation of that, and not just saying that it is true. –  Oct 22 '16 at 23:23
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    Look at the situation after the natural-number stages are complete. How do you know what to do next? To be a construction, you have to do something next. We call it stage $\omega.$ If you don't think something has to come next after the the natural-number stages, what do you propose to do at that point in the process? (This is the crux of the matter. Once you're convinced of what has to happen at stage $\omega,$ keep on going in the same fashion; at any point there must always something that comes next if this is to be a well-defined construction; so the whole thing is well-ordered.) – Mitchell Spector Oct 22 '16 at 23:33
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I think the basic thing you're missing is an understanding of what transfinite recursion actually is. In a construction by transfinite recursion, your steps are indexed by a well-ordered set $S$. Intuitively, at any point in the middle of this construction, you have finished some subset $T\subset S$ of the steps. To determine what to do next, you take the least element $\alpha$ of $S\setminus T$ (using the fact that $S$ is well-ordered), and then define the $\alpha$th step in terms of all the steps you've done so far.

Let's see how this goes wrong intuitively if $S$ is not well-ordered. For instance, suppose the elements of $S$ are $$0,1,2,3\dots,\dots\infty-2,\infty-1,\infty,\infty+1,\infty+2\dots$$ where we have an element called "$\infty+n$" for any $n\in\mathbb{Z}$. Note that this is not well-ordered, since $\{\infty+n:n\in\mathbb{Z}\}$ has no least element.

Now imagine doing a transfinite recursion indexed by $S$. First you do step $0$, then step $1$, then step $2$, and so on through all the natural numbers. Having finished step $n$ for every natural number $n$, what is your next step? There's no least element of what's left of $S$ to index your next step! If you choose to do step $\infty+n$ next for some integer $n$, then you've skipped infinitely many steps, namely $\infty+m$ for every $m<n$.

So in order for the intuition of "always having a next step of the recursion" to hold, you need your steps to be well-ordered. Otherwise, at some point you will have completed part of your recursion but won't be able to choose the next step to do. In this way, the stages of the hierarchy being well-ordered is inherent in the concept that the hierarchy is built up recursively.

Eric Wofsey
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