Grothendieck Universe is Closed under Subset

Theorem

Let $\mathbb U$ be a Grothendieck universe.

Let $u \in \mathbb U$.

Let $v \subseteq u$ be a subset of $u$.

Then $v \in \mathbb U$.

Proof

\(\ds v\)

\(\in\)

\(\ds \powerset u\)

Definition of Power Set

\(\ds u\)

\(\in\)

\(\ds \mathbb U\)

by hypothesis

\(\ds \leadsto \ \ \)

\(\ds \powerset u\)

\(\in\)

\(\ds \mathbb U\)

Grothendieck Universe: Axiom $(3)$

\(\ds \leadsto \ \ \)

\(\ds v\)

\(\in\)

\(\ds \mathbb U\)

Grothendieck Universe: Axiom $(1)$

$\blacksquare$