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Just as the title says:

Can there be an infinite set S whose elements all contain S?

If so, how is it called?

An example of a finite set could be the set of all Humans, where each human h knows about all other humans (where 'knowing' means being part of a set).

If there are infinite things in reality, and all things interact with each other, it could also be an example of the former infinite set.

Does this have a name in mathematics?

arod
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  • Related: https://math.stackexchange.com/questions/1046863/how-can-a-set-contain-itself – Adam Rubinson Mar 24 '24 at 11:51
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    You're talking about non-well founded sets. We usually take one of the axioms of set theory to be that such sets can't exist. But sometimes mathematicians don't assume this, and non-well founded sets have been studied. Non-well founded sets don't have to contain infinitely many copies of themselves though. One is enough, and there are ways of making non-well founded sets without any copies of themselves. – Zoe Allen Mar 24 '24 at 11:58
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    Seems like you are using the word "contains" rather loosely if people can contain people. – Thomas Andrews Mar 24 '24 at 12:02
  • An element of $S$ cannot contain more than this element itself. The only case , this is possible is a singleton set like {$x$}. This contains $x$ , which is however not {$x$} , so strictly speaken even this does not do the job. Assuming that sets can contain themselve led to Russels paradox which forced mathematicians to modify ZFC and make it much weaker. – Peter Mar 24 '24 at 12:45
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    You need to specify what "contains" means and what "being a part of" means. – Zuhair Mar 24 '24 at 15:31
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    @Peter Your comment contains a number of misconceptions. (1) I don't understand what you're saying in the first two sentences. Certainly an element of a set $S$ can contain lots of other sets... (2) There is nothing paradoxical about assuming sets can contain themselves (and in fact the axiom AFA mentioned in my answer is consistent with ZFC, assuming ZFC is consistent). What's paradoxical is assuming we can gather up all sets which don't contain themselves into a new set. (3) ZFC is the system that was developed after Russell's paradox, not modified in response to it. – Alex Kruckman Mar 24 '24 at 15:49
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    In my previous comment: "consistent with ZFC" should be "consistent with ZFC minus Foundation" – Alex Kruckman Mar 24 '24 at 17:09

1 Answers1

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In ZFC, the answer is of course no, by the axiom of foundation.

But under AFA (see here) the answer is yes. Consider the directed graph whose vertex set is $\mathbb{N}\cup \{S\}$. There is an edge from $S$ to $n$ and from $n$ to $S$ for each $n$, and there is an edge from $n$ to $m$ if and only if $m<n$. This is a pointed (at $S$) accessible directed graph.

By AFA, there is a set $S$ such that the membership relation on the transitive closure of $S$ is the graph described above. Now one can show that each $n$ corresponds to a distinct set, and they are all members of $S$, so $S$ is infinite. And every element of $S$ contains $S$, as desired.

Alex Kruckman
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