Union of Ordinal is Subset of Itself
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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 $1.20 \ (1)$