The set of limit points of an unbounded set of ordinals is closed unbounded. Let $\kappa$ be a regular, uncountable cardinal. Let $A$ be an unbounded set, i.e. $\operatorname{sup}A=\kappa$. Let $C$ denote the set of limit points $< \kappa$ of $A$, i.e. the non-zero limit ordinals $\alpha < \kappa$ such that $\operatorname{sup}(X \cap \alpha) = \alpha$. How can I show that $C$ is unbounded? I cannot even show that $C$ has any points let alone that it's unbounded.
 (Jech page 92) 
Thanks for any help.
 A: Fix $\xi\in \kappa$, since $A$ is unbounded there is a $\alpha_0\in A$ so that $\xi<\alpha_0$. Now, construct recursively a strictly increasing sequence $\langle \alpha_n: n\in \omega\rangle$. Let $\alpha=\sup\{\alpha_n: n\in \omega\}.$ Since $\kappa$ is regular and uncountable, we have $\alpha<\kappa.$ It is also easy to see that $\sup(A\cap\alpha)=\alpha$.
A: Since $\kappa$ is regular this means that the order type of $A$ is $\kappa$, and for every $\delta<\kappa$ we have that the cofinality of $\delta<\kappa$ as well.
Now suppose that $A$ is bounded in all limit points above $\beta$, without loss of generality $A\cap\beta=\varnothing$. Define a regressive function on limit ordinals:
$$\delta\mapsto\max\{A\cap\delta\}$$
This is indeed well defined, since $A$ is bounded below $\delta$. Since $\mathrm{Lim}$ is a club set, therefore stationary, this function is constant on a stationary subset.
In turn this means that unboundedly many times $\delta\cap A$ is reaching the same maximum, in particular this means that $A$ is bounded below $\kappa$, in contradiction!
