Ordinal Proper Subset Membership

Theorem

Let $A$ be an Ordinal and $B$ be any transitive class, and let them be either proper classes or sets.

Then $B \subset A \iff B \in A$.

Proof

$A$ is an Ordinal, so by Every Ordinal is a Transitive Class and Element of Transitive Class, $B \in A \implies B \subset A$.

$B \subset A$. Therefore, $( A \setminus B ) \not = \varnothing$. By the axiom of foundation, $\exists x \in ( A \setminus B ): \forall y \in ( A \setminus B ): y \not \in x$, which can be restated as $\exists x \in ( A \setminus B ): ( ( A \setminus B ) \cap x ) = \varnothing$.

$A$ is an ordinal, and $x \in A$, so by Initial Segment of Ordinal is Ordinal, $x$ is an ordinal. $( ( A \setminus B ) \cap x ) = ( ( A \cap x ) \setminus B )$. Since $x \in A$, $x \subseteq A$, so $( A \cap x ) = x$. By substitution, $( x \setminus B ) = \varnothing$, so $x \subseteq B$. $\Box$

Now suppose $z \in B$. Because $B \subset A$, $z \in A$. $z,x \in A$, so, because Ordinal is Well-Ordered by Epsilon, $z$ and $x$ satisfy trichotomy -- that is $( z \in x \lor x = z \lor x \in z )$. But $x \in z \lor x = z \implies x \in B$, which is a contradiction. Therefore, $z \in x$. Thus, all members $z$ of $B$ are members of $x$, so $B \subseteq x$. $\Box$

Since $x \subseteq B$ and $B \subseteq x$, $x = B$; hence, $B \in A$ because $x \in A$.

Source

• Gaisi Takeuti: Introduction to Axiomatic Set Theory (1971):$7.7