# Does the set of sets which are elements of every set exist?

In Zermelo-Fraenkel set theory, does the following set exist? $$A = \{ x \mid \forall y (x \in y) \}$$ I can see why a set which is an element of every set cannot exist (it would break the Axiom of Foundation/Regularity) but then would $$A$$ be empty or not exist at all? I am aware of the fact that: $$B = \{ x \mid \forall y (y \in x) \} = \emptyset$$ with the universal set not existing in ZF set theory. Is then $$A$$ empty for the same reason?

• What do you mean by 'a set contained in' – is it 'a set being a subset of' or rather 'a set being an element of'...? – CiaPan Apr 24 '19 at 21:54
• I meant "being an element of", I will edit the post to make that more clear, thanks! – Jacob Arbib Apr 26 '19 at 13:23
• @CiaPan Incidentally, that's made clear in the body of the question (even pre-edit) by the consistent use of $\in$ instead of $\subseteq$. – Noah Schweber Apr 26 '19 at 13:54

Yes, $$A$$ is just the emptyset.

We don't even need to appeal to Foundation to show this: all we need is that the emptyset exists. To be in $$A$$, you would have to be in every set, so in particular you would have to be in the emptyset - but that's clearly impossible.

Similarly, $$B$$ is just the emptyset (at least, in ZFC): to be in $$B$$ is to be a universal set, and (in ZFC) there aren't any of those.

Note that this is a little more finicky than the analysis of $$A$$: there are set theories which do have a universal set, such as NF, and in such theories the class $$B$$ is not empty. In all the set theories I know, however, the class $$B$$ is a set (whether empty or not): in particular, as long as we have $$(i)$$ Extensionality, $$(ii)$$ Emptyset, and $$(iii)$$ Singletons, we're good (if there are no universal sets then $$B$$ is the empty class, which is a set by $$(ii)$$; if there is at least one universal set, then there is exactly one universal set by $$(i)$$ since any two universal sets have the same elements, and so $$B$$ is the class containing just that universal set, which is a set by $$(iii)$$), and these are fairly un-controversial axioms.

(OK fine there are some interesting set theories without Extensionality; but still, in all the natural examples I'm aware of $$B$$ is a set.)

Really, there's a slight abuse going on here: a priori $$A$$ and $$B$$ are just classes. What's really going on is that I have the formulas $$\alpha(x)\equiv \forall y(x\in y)\quad\mbox{and}\quad \beta(x)\equiv\forall y(y\in x)$$ defining the classes $$A$$ and $$B$$ respectively; I prove in ZFC that "each class is empty," that is, that $$\forall x(\neg\alpha(x))\quad\mbox{and}\quad\forall x(\neg\beta(x)).$$

This now lets me prove "There is a set $$U$$ such that for all $$x$$ we have $$x\in U\iff \alpha(x)$$" - namely, take $$U=\emptyset$$ - and similarly for $$\beta$$. This is the "under-the-hood" version of proving that an expression in set-builder notation actually defines a set: we show that there is a set which is co-extensive with the defining formula of the class. In my opinion, this is an example of a situation where set-builder notation being used at the beginning makes things harder to follow: really, we should be asking (in the case of $$A$$) "Is there a set $$U$$ such that for all $$x$$ we have $$x\in U\iff \forall y(x\in y)$$?" which clearly separates formulas/classes and sets.