Nonempty Grothendieck Universe contains Von Neumann Natural Numbers

Theorem

Let $\mathbb U$ be a non-empty Grothendieck universe.

Let $\N$ denote the set of von Neumann natural numbers.

Then $\N$ is a subset of $\mathbb U$.

Proof

We prove the claim by induction.

Basis for the Induction

By Empty Set is Element of Nonempty Grothendieck Universe $\O \in \mathbb U$.

$\Box$

Induction hypothesis

For some fixed $n \in \N$, we have $n \in \mathbb U$.

Induction Step

We have to show, that $n + 1 \in \mathbb U$.

\(\ds n\)

\(\in\)

\(\ds \mathbb U\)

by hypothesis

\(\ds \leadsto \ \ \)

\(\ds \powerset n\)

\(\in\)

\(\ds \mathbb U\)

Grothendieck Universe: Axiom $(3)$

\(\ds \leadsto \ \ \)

\(\ds \set n \cup n\)

\(\in\)

\(\ds \mathbb U\)

Grothendieck Universe is Closed under Subset and $\set n \cup n \subseteq \powerset n$

\(\ds n + 1\)

\(=\)

\(\ds \set n \cup n\)

Definition of Von Neumann Construction of Natural Numbers

\(\ds \leadsto \ \ \)

\(\ds n + 1\)

\(\in\)

\(\ds \mathbb U\)

$\blacksquare$