Union of Ordinal is Subset of Itself

Theorem

Let $\alpha$ be an ordinal.

Then:

$\bigcup \alpha \subseteq \alpha$

where $\bigcup \alpha$ denotes the union of $\alpha$.

Proof

Let $x \in \bigcup \alpha$.

Then:

$\exists \beta \in \alpha: x \in \beta$

By Element of Ordinal is Ordinal, $\beta$ is an ordinal.

Thus:

$x \in \beta$ and $\beta \in \alpha$

Hence by Ordinal Membership is Transitive:

$x \in \alpha$

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Theorem $1.20 \ (1)$