# can there be an infinite set S whose elements all contain S? [closed]

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?

• 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. Commented Mar 24 at 11:58
• Seems like you are using the word "contains" rather loosely if people can contain people. Commented Mar 24 at 12:02
• You need to specify what "contains" means and what "being a part of" means. Commented Mar 24 at 15:31
• @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. Commented Mar 24 at 15:49
• In my previous comment: "consistent with ZFC" should be "consistent with ZFC minus Foundation" Commented Mar 24 at 17:09

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 $$m. This is a pointed (at $$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.