$\bigcup \alpha$ where $\alpha$ is a finite ordinal. Given a finite ordinal, is it correct in saying $\bigcup \alpha = \alpha - 1$?
As an illustrative example consider $3 = \{\emptyset , \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$.
I believe $\bigcup 3 = \{\emptyset , \{\emptyset\}\} = 2$.
Is this accurate?

Assume $\alpha=\beta+1=\beta\cup\{\beta\}$ where $\beta$ is an ordinal (note that this is applicable for all nonzero finite ordinals, but also for many infinite ones). Then $x\in\bigcup\alpha$ iff $x\in y$ for some $y\in \alpha$. And this is equivalent to $x\in\beta$ or $x\in y$ for some $y\in \beta$. Since $x\in y\in \beta$ implies $x\in \beta$ as well, we ultimately have 
$$ x\in\bigcup\alpha\iff x\in\beta$$
which means $$ \bigcup\alpha=\beta.$$
