Grothendieck Universes: Atoms

Question

I'm very much beginner in set theory; my apologize for mistakes.

Grothendieck universe: $$\forall X(X\in \mathcal{U}\Rightarrow\mathcal{P}(X)\in U\}$$ $$\forall X\forall x(x\in X\in\mathcal{U}\Rightarrow x\in\mathcal{U})$$ $$\forall X\forall Y(X,Y\in\mathcal{U}\Rightarrow\{X,Y\}\in\mathcal{U})$$ And ...: $$\forall \Lambda\forall X_{\lambda\in\Lambda}(\Lambda\in\mathcal{U}:\ldots)$$ (I'll extend this later!)

Doesn't the empty collection constitute a Grothendieck universe?

May a Grothendieck universe contain any atoms apart the empty set? $$X=\{x\}:\quad X\in U\implies x\in U$$ In that case how is the power set to be understood? $$X=\{x\}:\quad x\in U\implies \mathcal{P}(x)\in U$$ Does set theory in general distinguish atoms?

Answer 1

Doesn't the empty collection constitute a Grothendieck universe?

See Grothendieck universe :

There are two simple examples of Grothendieck universes:

The empty set, and

The set of all hereditarily finite sets $V_{\omega}$.

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Does set theory in general distinguish atoms?

For set theory, due to extensionality, all "objects" without elements are equal; thus, there is only one: the empty set.

In order to allow for atoms (or urelements, as in the original Zermelo's axiomatization), we have to use a "two-sorted" language with a primitive predicate $U$ such that $Ux$ is intended to mean “$x$ is an urelement” (therefore $¬Ux$ will mean “$x$ is a set”).

In this way, we have to reformulate the Extensionality Axiom as :

$$¬U(A) ∧ ¬U(B) → [(∀x)(x ∈ A ↔ x ∈ B) → A = B].$$

See :

• George Tourlakis, Lectures in Logic and Set Theory. Volume 2 : Set Theory (2003), page 99.

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Please, note that $X=\{ x \}$ is not an atom, but a singleton : e.g. $\emptyset \ne \{ \emptyset \}$

There is no power set for an atom: an atom has no elements and thus no subset, while the power set of a singleton $X=\{ x \}$ has two elements : $\emptyset$ and $\{ x \}$.