Existence of a function from ordinal to limit ordinal of the same cardinality

Question

Let $\alpha$ be a limit ordinal, such that $\big | \alpha \big | = \omega$. Prove that there is a strictly increasing function $f: \omega \to \alpha$ such that for all $\zeta < \alpha$ there is $n \in \omega$ with $\zeta < f(n)$.

$\textbf{My Attempt:}$ I am trying to construct maybe a recursive function. Since both sets are well-ordered and countable there exists some enumeration on the elements, $\{0,1,2, \dots\}$ and $\{\zeta_0,\zeta_1, \dots\}$

$f(0)=\alpha_{\in_{\zeta_0}}$, the initial segment up to $\zeta_0$

$f(1)=\alpha{\in_{\zeta_1}}$, the initial segment up to $\zeta_1$.

$\vdots$

$f(n+1)=\alpha{\in_{\zeta_n}}$, the initial segment up to $\zeta_n$

Then $f(0) < f(1)$ because $\alpha_{\in_{\zeta_0}} \in \alpha_{\in_{\zeta_1}}$ so the function is strictly increasing.

And for every $\zeta_i \in \alpha$, there exists $n \in \omega$ such that $f(n)=\alpha_{\in_{\zeta_{(i+1)}}}$ and so $\zeta \le f(n)$.

I am not sure if this is correct. Any advice?

Answer 1

HINT: Write the ordinal $\alpha$ as $\{\gamma_n\mid n\in\omega\}$. Now by recursion pick larger and larger ordinals by going over this enumeration, and show that the sequence that you have chosen is the intended function.

Answer 2

To complement Asaf's correct approach, I should point out your approach is incorrect. In particular, it's not necessarily the case that $\zeta_0 < \zeta_1$, so the initial segment up to $\zeta_0$ may not be smaller than the initial segment up to $\zeta_1$.

Think about it like this: suppose $\alpha > \omega$. Then $\omega$ has got to appear as an output of $f$, and only finitely many things can have appeared before it. Therefore, whatever $f$ you come up with to answer this question cannot be surjective (except where $\alpha = \omega$).