If $A$ is a countable subset of $\omega_1$, then there is an $ \alpha < \omega_1$ such that $A \subseteq \alpha$

Question

Prove : if $A$ is a countable subset of $\omega_1$. Then there exists $ \alpha < \omega_1$ with $ A \subseteq \alpha$.

I don't really know where to start, can anyone give a tip first?

Answer 1

HINT: Recall that $\alpha=\bigcup A=\sup A$. Then $\alpha$ is the countable union of countable ordinals.

Answer 2

Hint: Is it possible that $\bigcup A = \omega_1$?

Answer 3

HINT: For each $\alpha\in A$, $\alpha=\{\xi\in\omega_1:\xi<\alpha\}$ is countable.

Answer 4

Also true: if $A$ is a finite subset of $\omega_0$. Then there exists $ \alpha < \omega_0$ with $ A \subseteq \alpha$.