Here we extend our language by the symbol $\emptyset$ (sometimes $\varnothing$ or $\{\}$ is used also), denoting the term (set) with the defining property $\emptyset=\{x\mid \bot\}$.

Does our set theory permit the existence of such a set? Yes, it's granted by the axiom (axiom of empty set (Wikipedia))

$\exists y.\,\neg\exists x.\,x\in y$

(there exists a set $y$, such that there does not exists an $x$ which is in it)

which is equivalent to

$\exists y.\,\forall x.\,\neg(x\notin y)$

i.e.

$\exists y.\,\forall x.\,\left(x\in y\implies\bot\right)$

which, using the tautology $\bot\implies P$, we can extended to

$\exists y.\,\forall x.\,\left((x\in y\implies \bot)\land(\bot\implies x\in y)\right)$

or

$\exists y.\,\forall x.\,\left(x\in y\leftrightarrow \bot\right)$

which is abbreviated by

$\exists y.\,y=\{x\mid \bot\}$

One can show that there is at most one such set (From the extensionality axiom of set theory, and if the domain of discourse is non-empty.)

Remark: We could also avoid the symbol $\bot$ and use the predicate $x\neq x$.

Wikipedia: Empty set, Axiom of empty set (Wikipedia)

First infinite von Neumann ordinal