Does the set of all sets containing one set exist?

I was reading a pdf on set theory that explains that because of the axiom of regularity, there is no set that contains all sets, $\{x: x =x\}$, because it would have to contain itself.

I was wondering if instead a set containing all sets containing one set, $\{x:\exists y\ x=\{y\}\}$, could exist. How could you prove/disprove it's existence?

• Hint: Regularity tells us there can't be a set $x$ such that $x\in x$. But it also tells us that there can't be sets $x, y$ such that $x\in y\in x$. – GME Sep 19 '15 at 19:05
• – Noah Schweber Sep 19 '15 at 19:10
• You can also consider the set $B=\{x\in A:\forall y\in x,x\notin y\}$ (where $A$ is the set of singletons). Prove that this is equal to $\{\{X\}:\{X\}\notin X\}$. Then ask if $\{B\}\in B$. (This does not appeal to regularity.) – Akiva Weinberger Sep 20 '15 at 5:42

No. For any set of sets $\mathcal A, \bigcup \mathcal A = \{x : \exists A \in \mathcal A \text{ with } x \in A\}$ exists (assuming the normal axioms of set theory). But for your supposed set, its union would be the universal set, so your condition does not define a set.