If $X$ is a subset of $\omega_{\alpha}$ such that $|X| < \aleph_{\alpha}$, then $|\omega_{\alpha} - X| = \aleph_{\alpha}$

Question

If $\alpha=0$, then $\omega_0=\mathbb{N}$ and $\aleph_0=$ countable. So $|\omega_{\alpha} - X| = \aleph_{\alpha}$ becomes $|\mathbb{N}-X|=|\mathbb{N}|$ which is true (the function $f:\mathbb{N} \rightarrow \mathbb{N}-X$ by $f(n)=n+|X|$ is a bijection).

But when it is changed to $\alpha$, I have no idea how to proceed. I also have tried to use transfinite induction on $\alpha$ but I am getting nowhere.

Answer 1

Try to show that $|X \cup Y| = \text{max}(|X|, |Y|)$ when at least one of $X, Y$ is infinite.