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[Edit: After digesting the comments, doing more research & more thinking, I have reformulated my question]

On the iterative conception of sets, as it is usually expressed, there is no "completed" $V$, but rather an unending series of stages which are built up iteratively in a process which can never be completed. The obvious objection is that an abstract platonic object is not something which can be "built", nor can it appear in "stages" if it is timeless. -Source

I was previously under the impression that $V=\bigcup\limits_\alpha V_\alpha$, however this comment states that this "should really be thought of as a theorem". Another comment gives the formulation $V=\{x:x=x\}$. The latter formulation seems to incapsulate what I was orginally trying to ask about (a type of "completed" "timeless" universe which is not constructed "from below").

My primary question: does there exist a top-down formulation of set-theory?

My secondary question: would it be accurate to say that the cumulative hierarchy is a (inner?) model of $V$ (eg. $\bigcup\limits_\alpha V_\alpha \in V$)?

user784623
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  • It's not very clear to me what you're asking, since $V=L$ is the axiom of constructibility, simply $V\neq L$ is an alternative axiom (consistent with ZFC) that contradicts constructibility. – Alessandro Codenotti Jun 05 '20 at 09:08
  • @AlessandroCodenotti That is very clear & concise. I added a reference request tag if you have any suggestions. Also, there remains the questions about $0^#$ & $\Delta_3^1$. – user784623 Jun 05 '20 at 09:16
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    Your secondary questions have answers that are quite technical, and to some extent were covered before on this site. You can search the site for answers, and if you don't understand those, then I suggest opening a book, because to truly understand what is $0^#$, or why $\Delta^1_3$ is important here, there is a lot of reading, and also writing on your part (proving things on your own). Any other way is just cheating yourself. – Asaf Karagila Jun 05 '20 at 09:28
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    Here is a short list of relevant questions from the past decade: https://math.stackexchange.com/questions/802763/ https://math.stackexchange.com/questions/1888063/ https://math.stackexchange.com/questions/263862/ https://math.stackexchange.com/questions/39956/ https://math.stackexchange.com/questions/54099/ https://math.stackexchange.com/questions/3210666/ https://math.stackexchange.com/questions/1899879/ and there are probably more (and I didn't cover $\Delta^1_3$ and projective sets here). – Asaf Karagila Jun 05 '20 at 09:46
  • @AsafKaragila Thanks for the links! Do you have any suggestions for books? – user784623 Jun 05 '20 at 09:48
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    Jech, Kunen, Kanamori, you know, the normal books that are recommended everywhere... – Asaf Karagila Jun 05 '20 at 09:55

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The usual way to think about set theory is that there is some universe $V$ of sets that satisfies some axioms. So $V=\{x:x=x\}$ is the correct definition of $V$ as a class. Assuming a strong enough fragment of ZF (e.g. including power set and some replacement), we can define the cumuluative hierarchy $V_\alpha$ by transfinite recursion. And then, if we also assume the axiom of foundation, we can prove that $V=\bigcup_\alpha V_\alpha.$ To emphasize that this is a theorem and not a definition, I'll note that this equation is really short-hand for the theorem $$ \forall x \exists \alpha\; x\in V_\alpha.$$

If we don't have foundation, then we cannot prove this. In this case, we do not have $V=\bigcup_\alpha V_\alpha,$ but $\bigcup_\alpha V_\alpha$ still makes sense as a class. This is called the class $\mathrm{WF}$ of well-founded sets, and if foundation fails it just means that there are sets that are not in this class, i.e $\mathrm{WF}\subsetneq V.$

It is accurate to think of $\mathrm{WF}$ as an inner model of $V$. It is a model of ZF that is a definable submodel of any model of ZF - foundation. (However it is not accurate to say $\mathrm{WF}\in V,$ as you have put it. $\mathrm{WF}$ is a proper class, so it is not an element of $V.$ Perhaps what you meant was $\mathrm{WF}\subsetneq V,$ as I wrote above... which is only true if foundation fails.)

As to what this means about there being a 'top-down approach to set theory', I'm not completely sure. Based on what I gather you mean, the answer is yes, as I've described, what you heard in the comments is the standard mathematical way of thinking about things.

However if we're taking a more philosophical angle, we might not be satisfied with this... do we really have a sharp conception of or motivation for this universe of sets that we think of ZFC as describing? We might find it useful to think about it as being built in stages, at least as a motivating principle. One difficulty here, is we don't really have a good answer for what the "full" power-set of a set is or what "all" the ordinals are... so the principle with which we are constructing it with are vague. It's not any less vague than just saying "we have some universe of sets", but it's not obvious it adds anything other than intuition. On the other hand, there have been philosophically motivated attempts to develop set theory in a way that puts the cumulative structure more at the foreground and less as a consequence... it's just not the way it's typically developed and taught.

spaceisdarkgreen
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  • This is very insightful! Do you know of any examples from $V \setminus \text{WF}$? (Is this itself a class? If so, what sets are in this class?) – user784623 Jun 09 '20 at 01:50
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    @user784623 I'll emphasize again that in ZF, no such example exists since we have the axiom of foundation. However, if we don't take that axiom, it is consistent that such sets exist. An example would be a Quine atom, which is a set that satisfies the equation $x={x}$ (it is its only element). It is consistent with ZF-foundation that there are any number of Quine atoms lying around. And then of course there have to be a bunch of other sets that contain them, or contain sets that contain them, etc, and none of these will be in WF either. – spaceisdarkgreen Jun 09 '20 at 01:55
  • A (hopefully quick) question: is $\forall x \exists \alpha; x\in V_\alpha.$ equivalent to or somehow related to Grothendieck's universe axiom ("For any set $x$, there exists a universe $U$ such that $x \in U$")? Also, in an attempt to try to further clarify my primary question: by top down approach I mean starting with the universe $V$ & then formulating set theory vs. starting with the empty set & then proceeding (ie. bottom up). – user784623 Jun 20 '20 at 03:58
  • @user784623 No, in order to be a universe in Grothendieck's sense, $V_\alpha$ would need to have certain additional properties... in fact these properties imply that it is a model of ZFC and hence we can't prove the existence of a universe in ZFC. ZFC + universe axiom is equivalent to ZFC + there are unboundedly many inaccessible cardinals, and the $V_\alpha$ that are universes are exactly those where $\alpha$ is an inaccessible cardinal. – spaceisdarkgreen Jun 20 '20 at 05:01