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?

  • $\begingroup$ 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'...? $\endgroup$
    – CiaPan
    Apr 24, 2019 at 21:54
  • $\begingroup$ I meant "being an element of", I will edit the post to make that more clear, thanks! $\endgroup$ Apr 26, 2019 at 13:23
  • $\begingroup$ @CiaPan Incidentally, that's made clear in the body of the question (even pre-edit) by the consistent use of $\in$ instead of $\subseteq$. $\endgroup$ Apr 26, 2019 at 13:54

1 Answer 1


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.


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