I was wondering if anyone knew whether or not in ZFC (or any other set theory) if the object, $$\Bigg\{ \Big\{ \big\{ \{ \cdots \} \big\} \Big\} \Bigg\}$$ can exist. That is, it is the set containing the set containing the set containing...ad infinitum. Is this even a set? Can it make sense to talk about such a thing? I'm simply curious.
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Now that I think of it, it may not be a set since it is both a subset and an element of itself. So it must not exist in the ZFC universe. – change_picture Oct 08 '15 at 16:23
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There exist sets $A,B$ such that both $A\in B$ and $A\subseteq B$, for example take $A=\varnothing,B={\varnothing}$. The reason why your set doesn't exist is explained in Rob's answer. – Wojowu Oct 08 '15 at 16:36
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Existence of self-containing sets doesn't give us Russell's paradox. – Wojowu Oct 08 '15 at 16:45
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Nevermind, you're right. Either way, we can't have sets which are members of themselves, right? – change_picture Oct 08 '15 at 16:55
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1To amplify Wojowu's comment: Russell's paradox is concerned with the class $U$ of sets $A$ such that $A$ is not a member of $A$. Allowing $U$ as a set causes the paradox. The axiom of foundation implies that $U$ is the entire universe. In non-well-founded set theory, it might be a proper subclass of the universe, but it still can't be a set. – Rob Arthan Oct 08 '15 at 16:58
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Okay, thank you for the clarification. – change_picture Oct 08 '15 at 18:50
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The axiom of foundation (or regularity) explicitly bans your set. This axiom is independent of the other axioms of ZFC and, depending on your point of view is either just a technical convenience or a monster-barring axiom. Proponents of non-well-founded set theory like these monsters.
To see why your set gives rise to a violation of the axiom of foundation, note that (as explained on the wikipedia page), the axiom of foundation implies that that the membership relation is well-founded: i.e., there is no infinite sequence of sets $A_1, A_2, \ldots$ such that $A_1 \ni A_2 \ni A_3 \ldots$, but your set gives rise to such a sequence.
Rob Arthan
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But is it trivial that $S={{\cdots}}$ contains itself? After all, if the definition is "sets are equal if they have the same elements", this gives an infinite recursion when trying to determine whether the element of $S$ is the same set as $S$. – JiK Oct 08 '15 at 20:47
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@JiK: Sets containing themselves are only one kind of thing foundation bans. Here, you want to apply the axiom to the transitive closure of $S$; that is, define the sequence $S_0 = S$ and $S_{n+1} \in S_n$, and apply foundation to ${ S_0, S_1, S_2, \ldots }$ to obtain a contradiction. – Oct 08 '15 at 21:15
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$S = {{ \ldots }}$ is an informational notation for a set that might exist. Its defining property would be $S \in S$ iff $S = S$. The potential infinite recursion involved here is just the kind of thing that proponents of non-well-founded set theory revel in. – Rob Arthan Oct 08 '15 at 22:25
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@RobArthan: There is no guarantee that if $S$ has that shape, then $S \in S$ (of course, the converse holds; if $S = { S }$, then it does have that shape). Having isomorphic membership trees isn't a guarantee that two sets are equal, unless they are well-founded (or you use a stronger form of extensionality). – Oct 08 '15 at 23:50
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As Hurkyl has noted, it does not follow from ZFC - regularity that such a set would contain itself. On the other hand, for example, Aczel's anti-foundation axiom would pose that such a set exists and is unique, which (I think, though I might be wrong in my ignorance of nonstandard set theory ;) ) is enough to conclude that indeed $S\in S$ (because, clearly, the sole element of $S$ has the same membership graph), so $S$ is in fact the Quine atom. – tomasz Oct 09 '15 at 01:03
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Rob, you might want to elaborate just a bit on why the axiom of regularity rejects this set: it promises an element of $S$ distinct from $S$. So either $\left{ \left{ \left{ \dots \right} \right} \right} \cap \left{ \left{ \dots \right} \right} = \varnothing$ or $S$ has another element. (Where the two uses of "$\dots$" are the exact same thing.) – Eric Towers Oct 09 '15 at 03:07
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@EricTowers: This was mentioned elsewhere in the thread. This is a standard thing when using axiom of foundation: the set itself may not directly violate it (if $S\neq {S}$), but surely, the transitive closure does: it is linearly ordered by $\in$, and it either has no least element or the least element is a singleton of itself. – tomasz Oct 09 '15 at 10:28
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Well, I didn't find such a mention when I posted my comment and I don't find such a mention now. Maybe I'm being dense. But I believe your answer would be better if it addressed the question (fully) at the level it was asked. – Eric Towers Oct 09 '15 at 16:23
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@EricTowers: i've added what I think is the simplest explanation of what goes wrong. – Rob Arthan Oct 09 '15 at 19:09
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It depends very much on how you handle the informal "..." and the words ad-infinitum. The short answer is that the notation is not clear enough to answer whether it is a set or a class.
If you define your construct to be "a thing which has the property of having 1 element, which is itself," you have constructed a class, not a set, because a set cannot have those properties in most set theories.
On the other hand, if you construct it using a sequence S which starts with $S_0=\emptyset$ and is recursively constructed according to $S_{n+1}=\{S_n\}$, and define the thing you want to be $S_\omega$, then what you are constructing is not all that far from the standard set theory construction of natural numbers ($S_{n+1}=S_n \cup \{S_n\}$)
Cort Ammon
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1The problem is, you can't really do that, not in standard set theory anyway. What would $S_\omega$ be? There is no relevant notion of convergence here, except for the very naive, as far as I can tell. – tomasz Oct 09 '15 at 01:05
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@tomasz I thought about that, which was why I pointed out how incredibly similar the construction of natural numbers is. To my knowledge, it's just a successor function. Is there a reason why $S(a) = {a}$ is not a valid successor function but $S(a) = a \cup {a}$ is? I always figured the union was just so to tie the number to cardinality, but I might have missed something. – Cort Ammon Oct 09 '15 at 01:10
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1@CortAmmon there is a very specific reason why the two are different. The second one allows an easy limit approach while the first doesn't. What I mean is if your successor function is $S(a)=a\cup{a}$ then you get inclusion giving you a linear ordering (without transitive closure $b < a$ iff $b\in a$) and moreover you get an obvious limit candidate for which the inclusion still gives a linear ordering namely the union of all preceding sets. Note that you still need a special axiom for $\omega$ to exist but it behaves much better. – DRF Oct 09 '15 at 05:03
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Actually to clarify my preceding comment.. You could have $S(a)={a}$ as a successor function it just doesn't behave nearly as nicely as $S(a)=a\cup{a}$. Ignoring the fact that you need transitive closure of $\in$ to get a linear order on such (even finite) ordinals, you need an even more special situation for the limits. Let's consider the easiest limit "\omega" what would that even be? Now in ZFC you have a special axiom saying that the usual $\omega$ exists (axiom of infinity)... – DRF Oct 09 '15 at 06:01
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1If you took ZFC without the axiom of foundation (so that you would be at least allowed to make something like ${{...}}$ you couldn't prove it exists. You would need a special axiom to create it (as someone pointed out probably $W={W}$). The trouble is this set now has nothing to do with the finite ordinals you have defined so far. They aren't part of it in any way. So if you did want this $W$ to be your "$\omega$" you would have to specifically extend the order to it. Worse still your successor function doesn't work on it. $S(W)={W}=W$. – DRF Oct 09 '15 at 06:07
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In conclusion you could define your successor function as $S(a)={a}$. For finite ordinals this is actually Zermelo's definition see https://en.wikipedia.org/wiki/Natural_number. But it doesn't behave nearly as nice as Von Neumann's definition and It doesn't scale to or past "$\omega$" well at all. – DRF Oct 09 '15 at 06:11
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@DRF Thank you for that clarification. Your explanation of why von Neumann's definition is what it is is quite helpful! I do, however, wonder if the OPs actual construct requires $W={W}$. His comment suggests that the thing in his head does require it, but I wanted to provide an answer that pointed out that because the notation is so informal, there are many possible ways it could be interpreted, some are sets and some are not. I may have cheated a bit along the way: I assumed the phrase "ad infinitum" meant I could assume I had a countable infinity to work with in my definition =) – Cort Ammon Oct 09 '15 at 07:01
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Well the issue is even if you do have a countable infinity you can't construct that set.:) That's what I meant when I said even if you had ZFC minus the axiom of foundation that set still doesn't have to exist. Assuming that by "that set" you mean any set such that it contains an infinite descending sequence in terms of $\in$. – DRF Oct 09 '15 at 07:12
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1If you check the wikipedia article on axiom of regularity (foundation) they give the reference for the fact that you can't construct this set. It follows from the result by skolem which shows that ZFC with regularity is equiconsistent with ZFC without regularity. – DRF Oct 09 '15 at 07:19
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@DRF Okay, forgive me if I am being pedantic, but this is an interesting corner of set theory to me, so I want to see if I can get it right. When I look at the axiom of regularity, I see that you cannot have an infinite descending sequence, where $S_{n+1} \in S_n$. I have been constructing an infinite ascending sequence where $S_n \in S_{n+1}$. From your comments, I now understand that the difference between those sequences is much larger than I had previously thought. Am I correct in believing that the successor function $S(a) = {a}$ defines a valid sequence of sets with a limit... – Cort Ammon Oct 09 '15 at 07:25
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... but that the limit of that sequence is not the construct the OP is asking about, because mine is an "ascending sequence" and the OP's request is phrased as a "descending sequence?" – Cort Ammon Oct 09 '15 at 07:27
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