Closed, unbounded subset of a cardinal. I missed two lectures in my set theory course, and now I don't understand the homework problems. 
One is this: let $\kappa$ be a regular uncountable cardinal. Show that the following sets are closed and unbounded in $\kappa.$


*

*$\{\alpha<\kappa\,:\,\alpha \text{ is a limit ordinal}\},$

*$\{\alpha<\kappa\,:\,\mathrm{cf}(\alpha)=\omega\}$ for $\kappa=\omega_1,$

*$\{\lambda<\kappa\,:\,\lambda\text{ is a cardinal}\}$ for $\kappa$ inaccessible.


I don't understand what closed or unbounded subsets of $\kappa$ is. Could you explain this to me and give some pointers on solving the problem?
 A: An unbounded subset of $\kappa$ is some $A\subseteq\kappa$ such that for all $\beta<\kappa$, there is some $\alpha\in A$ such that $\beta<\alpha$.
A closed subset of $\kappa$ is some $A\subseteq\kappa$ such that for all $0<\alpha<\kappa$, if $\sup(A\cap\alpha)=\alpha,$ then $\alpha\in A$.

For the first one, do you know the result that all limit ordinals are of the form $\omega\cdot\alpha$ for some non-zero $\alpha$? In particular, you'll want to show that the limit ordinals less than $\kappa$ are those of the form $\omega\cdot\alpha$ for $0<\alpha<\kappa$. Use continuity of ordinal multiplication to show that this set is closed, and the fact that $\kappa$ is a cardinal to show that it's unbounded.
The second one is just a special case of the first one. (Why?)
The third one won't be too tricky. Do you know the recursive definition of the alephs? (Also, don't forget that the natural numbers are cardinals, too.)
A: Let $\kappa$ be a regular uncountable cardinal. $X \subseteq \kappa$ is unbounded if $\sup X = \kappa$. 
A $\alpha$ is a limit point of $X$ if and only if  $\alpha$ is a limit ordinal and $\sup(X \cap \alpha) = \alpha$. 
$X \subseteq \kappa$ is closed if it contains all its limit points less than $\kappa$. 
1 Let $X = \{\alpha < \kappa : \alpha \text{ is a limit ordinal}\}$. Limit points of $X$ must be limit ordinals. $X$ is closed. Now to show that $X$ is unbounded. Let $\alpha < \kappa$. Since $\kappa$ is a cardinal, $|\alpha| < \kappa$. Then $\alpha + \omega$ is a limit ordinal with cardinality less than $\kappa$. So $\alpha + \omega \in X$. $\alpha + \omega > \alpha$. $X$ is unbounded. 
2 Let $X' = \{\alpha < \omega_1 : cf(\alpha) = \omega\}$. At least in Jech, the cofinality is only defined for limit ordinals. Using this convention, $X' \subseteq X$, where $X$ is the set of ordinals form part 1. The cofinality of any limit ordinal is always a cardinal (in fact regular cardinal). The only cardinality less than $\omega_1$ is $\omega$. It has been shown that if $\alpha < \omega_1$ is a limit ordinal, then $cf(\alpha) = \omega$. So $X' = X$. It is closed and unbounded by part 1. 
3 Suppose $\kappa$ is (weakly) inaccessible. Let $X'' = \{\alpha < \kappa : \alpha \text{ is a cardinal}\}$. It is unbounded since by definition of weakly inaccessible, $\kappa$ is a limit cardinal. Suppose $\alpha$ is a limit point of $X''$. $\alpha$ is a cardinal since $\sup$ of a set of cardinals is always a cardinal. 
