Show that if $A$ is a non-empty subset of $\omega$ such that $\bigcup A = A$, then $A=\omega$

Question

Let $\omega$ be the set of natural numbers, starting from $0 = \emptyset$. Show that if $\bigcup A = A$, $A = \omega$

I have shown that $A$ is transitive; i.e. $x \in B \in A \implies x \in A$. And, since $A$ is non-empty, $\exists n \in \omega \ s.t.\ n \in A$, and since $A$ is transitive, $0 \in A$.

I want to use the induction principle to show that for all $n \in \omega,\ n \in A$, but I got stuck.

Assume $k \in A$. Since for all $x <k,\ x \in k$, so $0,1,...,k \in A= \bigcup A$. $k \in A \Leftrightarrow \exists X \in A ( k \in X)$

How can I complete my proof?

Answer 1

Recall that a natural number is considered as the set of all natural numbers beneath it.

Hint: Since $\exists X \in A (k \in X)$, and $A$ is a subset of $\omega$, $X$ must be another natural number. Can you finish it from here?

Answer 2

If $\bigcup A = A$ then this entail:$$\forall x (\exists y \in A(x \in y) \iff x \in A)$$

Since $A \neq \emptyset$ and $A \subseteq \omega$ then this entails that $A$ is an infinite ordinal, and the only infinite subset of $\omega$ that is an ordinal is $\omega$ itself.