Let $A$ be a nonempty subset of $\omega$, the set of natural numbers. If $\bigcup A=A$ then $A=\omega$.
I have proved ‘$\bigcup n^+ = n$ for any $n\in\omega$’ and ‘$\bigcup\omega = \omega$’
I think this question should be proved with the above results..
Lemma: If $A$ is a set of ordinals, then $\bigcup A$ is an ordinal.
The proof appears here.
Suppose that $A\subseteq\omega$ then $\bigcup A$ is an ordinal. If $A=\bigcup A$ then $A$ was an ordinal, that is either $n\in\omega$ or $\omega$ itself.
However you know that for $n\neq 0$, $\bigcup n\neq n$, so you are only left with the option $A=\omega$.
To prove this directly in this case you can argue that if $n\in A$ then $n\subseteq\bigcup A$. From this follows that if $A$ is non-empty and $\bigcup A=A$ for every $k\in\omega$ such there is $n\in A$ and $k<n$ we have $k\in\bigcup A$, and therefore $k\in A$. So if $A$ is unbounded $A=\omega$; if $A$ is bounded then $A=n$ but we know that $\bigcup n\neq n$ which is a contradiction.