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 at 21:54
  • $\begingroup$ I meant "being an element of", I will edit the post to make that more clear, thanks! $\endgroup$ – Jacob Arbib Apr 26 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 $\subseteq$. $\endgroup$ – Noah Schweber Apr 26 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.