Initial ordinals

Question

Let $cf(\lambda)$ be the least index $\xi$ such that a limit ordinal $\lambda$ is cofinal with the initial ordinal $\omega_\xi$. How can one prove that $cf(\omega_\Omega)\neq0$ where $\Omega=\omega_1$?

Answer 1

You will need the Axiom of Choice to prove this, so the proof you gave cannot be correct.

With $\mathsf{AC}$ every countable union of countable sets is countable. In particular, if $f:\omega\to\omega_1$ is increasing, then $\lim_{n\in\omega}f(n)=\bigcup_{n\in\omega}f(n)=\alpha$ is a countable union of countable sets (because each $f(n)<\omega_1$) and thus $\alpha$ is countable, which implies that $f$ is not cofinal in $\omega_1$. You could then combine this with the hint that was given in the comments above.

Without $\mathsf{AC}$ it is consistent that there exists a cofinal countable sequence in $\omega_1$. This means that it is consistent with $\mathsf{ZF}$ that $\omega$ and $\omega_1$ have the same cofinality.