Can't prove that if $\alpha$ is an ordinal and $A\subseteq\omega_\alpha$ such that $\lvert A\rvert<\aleph_{\alpha}$ then $\bigcup A \in \omega_\alpha$

Question

Prove that if $\alpha$ is an ordinal and $A\subseteq\omega_\alpha$ such that $\lvert A\rvert<\aleph_{\alpha}$ then $\bigcup A \in \omega_\alpha$.

I can't prove that, but I can't prove that it's false either, first I proceeded by contradiction, later by contraposition and failed too. Finally tried with transfinite induction and I didn't make it. Any help will be appreciated.

Answer 1

False. For example $\alpha=\omega$ and $A=\{\omega_n: n\in\Bbb N\}. $ Then $|A|=\aleph_0 <\aleph_{\omega}$ and $\cup A=\omega_{\omega}=\omega_{\alpha}.$

Reading topic: Regular cardinals and singular cardinals.