Originally Posted by

**zzzhhh** If ordinal number $\displaystyle \alpha=\beta+1$, then we call $\displaystyle \alpha$ a successor ordinal. An ordinal is a limit ordinal if it is not a successor. My question is: how to prove that for any countable limit ordinal $\displaystyle \alpha=\{\beta|\beta<\alpha\}, |\alpha|=\aleph_0$, there is a sequence in $\displaystyle \alpha$ converging to $\displaystyle \alpha$? Here we define a sequence in $\displaystyle X$ to be a function on the set of all natural numbers $\displaystyle \mathbb N$ and having values in $\displaystyle X$, and a sequence of ordinals $\displaystyle (\alpha_n)$ converging to ordinal $\displaystyle \alpha$ if for any $\displaystyle \beta<\alpha$, there is a natural number $\displaystyle N$ such that $\displaystyle \beta<\alpha_n\leq\alpha$ for all $\displaystyle n\geq{N}$.

For example, sequence $\displaystyle \alpha_n=\omega+n\rightarrow \omega\cdot2; \beta_n=\omega^n\rightarrow \omega^\omega$, .... But these are specific sequences, I need a proof in general situation.

Thanks!