A generalization of "any countable limit ordinal is the union of a sequence of increasing ordinal" Using the fact that every countable ordinal is isomorphic to a closed subset of $\mathbb Q$, I find out that any countable limit ordinal is the union of a sequence of increasing ordinal. Now I'm trying to generalize the result to what follows:

Let $\alpha\in ON$ be a limit ordinal, and $\kappa:=\vert\alpha\vert$ the corresponding cardinal. Then there always exists a order-preserving function $f:\kappa\to ON$, such that $\bigcup\{f(\xi)\mid\xi\in\kappa\}=\alpha$.

But I have no ideal how to prove or disapprove my conjecture. Can anyone help me? Thanks in advance.
 A: This is not quite possible, but we can slightly fix this.
First let me point out some defining characteristic of cardinals. These are well-orders that every proper initial segment has a strictly smaller cardinality. On the other hand, $\omega_2+\omega_1$ has an initial segment equipollent to the entire order.
So now we run into a problem. If there is an order preserving function from $\omega_2$ into $\omega_2+\omega_1$, then at some point it had to enter the tail segment of order type $\omega_1$. But the domain of the function still has an end segment of type $\omega_2$, which means there is no order preserving function like that.
But at the same time it is obvious what such function might be for $\omega_5+\omega_5$, or $\omega_{53}^{\omega_3}+\omega_{53}$, or whatever.
The solution is to understand that every ordinal has a cofinality, which is exactly the smallest ordinal which has a cofinal embedding into the ordinal. Namely, $$\operatorname{cf}(\alpha)=\min\{\delta\mid\exists f\colon\delta\to\alpha\text{ order preserving and unbounded}\}.$$
You can easily show that $\operatorname{cf}(\alpha)\leq\alpha$, and for limit ordinals $\omega\leq\operatorname{cf}(\alpha)$ (I'm not counting $0$ as a limit ordinal). We can also show that $\operatorname{cf}(\alpha)=\operatorname{cf}(\operatorname{cf}(\alpha))$.
And if we are careful enough, we might be able to prove that $\operatorname{cf}(\alpha)$ is always a cardinal.
