Definition. $\bigcup A = \{ x | \exists a \in A \mbox{ such that } x \in a\}$
Problem. If $A \subset \omega$ and $A \neq \emptyset$ Prove, if $\bigcup A=A$ then $A= \omega$
Attempt. Suppose that $A \neq \omega$ then $\omega -A \neq \emptyset$ and $\omega-A \subset \omega$ then exists $n_0 = \mbox{ min } (\omega-A)$ so, $\forall n \in \omega: n=n_{0} \mbox{ or } n_{0} \in n$.
Then, if there is $m \in A$ such that $n_{0} \in m$ so, $n_{0} \in \bigcup A $ then $n_{0} \in A$ because $\bigcup A = A$ then $\forall m \in A: n_{0} \notin n$, so $\forall m \in A: m \in n_{0} \mbox{ or } n_{0}=m$ but if $n_{0} = m \in A$ is a contradiction so $\forall m \in A: m \in n_{0}.$
But I don't see a contradiction, could you help me?
If $A$ is a set of transitive sets, then $\bigcup A$ is transitive. If $\bigcup A=A$, then $A$ is transitive. And it is a transitive set of ordinals, so it is an ordinal.
So either $A\in\omega$, or $A=\omega$. Now show that if $A\in\omega$, then $\bigcup A\neq A$.