Let be $\{\alpha_n\}_{n\in\mathbb{N}}$ a numerable succession of ordinals such that $\alpha_n<\omega_1$: since the length of the succession is $\omega<\omega_1$ is it true that $\mathscr{sup}\{\alpha_n\}_{n\in\omega}:=\bigcup_{n<\omega}\alpha_n<\omega_1$?
I'm sure that it result $\bigcup_{n<\omega}\alpha_n\le\omega_1$, but I'm not sure that $\bigcup_{n<\omega}\alpha_n<\omega_1$: anyway I think it is true since $\omega_1$ is a limit ordinal such that $|\omega|:=\aleph_0<\aleph_1\ :=|\omega_1|$ and so $\{\alpha_n\}_{n\in\omega}$ can't be a bijection in $\omega_1$. Could someone help me, please?
It is true, because $\omega_1$ is uncountable. The union of the countably many countable ordinals $\alpha_n$ is a countable ordinal (it's well-ordered (countable) set, so an ordinal, the exact proof of this will depend on the definition of ordinal one uses) and so is smaller than $\omega_1$ which by definition is the smallest ordinal that is uncountable, so equality with $\omega_1$ is ruled out. As ordinals are linearly ordered $\sup_n \alpha_n < \omega_1$ must hold.