For any ordinal, does it's cardinality equal its least upper bound

Question

Can you make the claim that for any ordinal, its cardinality equals it's least upper bound.

This is motivated by:

$\bigcup\omega+1=\omega$ and $|\omega+1|=\omega$

where $\bigcup\omega+1$ is also the $\text{sup}(\omega+1)$

Thanks

Answer 1

What is $\bigcup (\omega^2) ?$

Answer 2

If $+$ denotes ordinal arithmetic, then $\cup\omega + 1$ is already not $\omega$ but its successor (since $\cup\omega = \omega$).

If $+$ denotes something else, please tell us what you mean by it