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.
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.