Let $\alpha$ be an ordinal and $B\subseteq \omega_{\alpha}$ such that $\bigcup B=\omega_{\alpha}$ then $|B|=\aleph_{\alpha}$

Question

I can't prove this problem. My attempt was choosing $\alpha=\omega$ and watch the process, but i can't get the proof. This was a true/false question. I will be very thankful if someone could help me. Thanks

Answer 1

This is false in general, and indeed $\alpha=\omega$ is a counter example (let $\alpha=\omega$ and $B=\{\omega_i\}_{i\in\omega}$, clearly $|B|=\omega<\omega_\omega$, but if $\bigcup B=\beta<\omega_\omega$ then $\beta<|\beta|^+<\omega_\omega$, but $|\beta|^+\in B$ so $\beta>|\beta|^+$, contradiction).

Cardinals $\omega_\alpha$ with this properties (note, the property is of $\omega_\alpha$, not of $\alpha$) are called regular, and generally, the only cardinals you can show have this property are the successor cardinals and $\aleph_0$.

Uncountable limit regular cardinals is known as weakly inaccessible, and they are a form of a large cardinal and we cannot prove their existence is consistent within ZFC.