A question (more like three) about a topological space of ordinals. I've been struggling with these for a while now, if anyone is willing to offer a hint I'll be more than grateful. 
Given an ordinal $\varepsilon$, consider the topological space $L_{\varepsilon}$ whose points are the ordinals below $\varepsilon$ and whose topology is generated by the open intervals of the form 
$$
(\alpha,\beta)=\{\gamma < \varepsilon \, :\, \alpha < \gamma < \beta \}.
$$
(a.) Determine the ordinals $\varepsilon$ for which $L_{\varepsilon}$ has the property that any two closed sets of cardinality $|\varepsilon|$ are homeomorphic.  
(b.) Determine the ordinals $\varepsilon$ for which $L_{\varepsilon}$ is countably compact (that is, each countable cover of $L_{\varepsilon}$ has a finite subcover). 
(c.) Suppose $\varepsilon$ is a cardinal. Determine the supremum among the cardinals $\lambda < \varepsilon$ such that any intersection of less than $\lambda$ closed sets of cardinality $\varepsilon$ is nonempty. 
 A: HINTS: (I don’t like using $\epsilon$ for an arbitrary ordinal, so I’ve changed the notation slightly. I’m also dealing only with infinite ordinals, since the finite case is completely straightforward.) I would start with (b), which I think is the easiest, and do (a) last.


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*(a) Suppose that $\alpha$ is an ordinal, and let $\kappa=|\alpha|$. Show first that if $\alpha=\kappa$, then any two closed subsets of $L_\alpha$ of cardinality $\kappa$ are homeomorphic. (In particular, they’re all homeomorphic to $L_\alpha$.) Then show that $L_\alpha$ is compact iff $\alpha$ is a successor ordinal. Use this to show that if $\alpha>\kappa$, then $[0,\kappa]$ is a compact subset of $L_\alpha$ of cardinality $\kappa$. If $\alpha>\kappa$ is a limit ordinal, however, $L_\alpha$ is a non-compact subset of $L_\alpha$ of cardinality $\kappa$, so it only remains to consider the case in which $\alpha$ is a successor ordinal. There is a unique successor ordinal $\beta$ such that $\alpha=\kappa+\beta$, where the addition here is ordinal addition. Show that if $\beta<\kappa$, then $L_\alpha$ is homeomorphic to the disjoint union of $L_\kappa$ and $L_\beta$, which in turn is homeomorphic to $L_\kappa$; use the fact that $\beta+\kappa=\kappa$ (again an ordinal sum). Finally, to handle the case $\beta>\kappa$, answer the following question: is $[0,\kappa+\kappa]$ homeomorphic to $[0,\kappa]$ (where $\kappa+\kappa$ is the ordinal sum)?

*(b) If $\operatorname{cf}\alpha=\omega$, find an open cover of $L_\alpha$ with no finite subcover. (Here $\operatorname{cf}\alpha$ is the cofinality of $\alpha$.) What happens if $\operatorname{cf}\alpha=1$? If $\operatorname{cf}\alpha>\omega$?

*(c) Let $\kappa$ be the cardinal. Show that if $\lambda<\operatorname{cf}\kappa$, and $C_\xi$ is a closed subset of $L_\kappa$ of cardinality $\kappa$ for each $\xi<\lambda$, there is a strictly increasing sequence $\langle\alpha_n:n\in\omega\rangle$ in $L_\kappa$ such that $(\alpha_n,\alpha_{n+1})\cap C_\xi\ne\varnothing$ for each $\xi<\lambda$, and use this sequence to find an ordinal in $\bigcap_{\xi<\lambda}C_\xi$. (If you get completely stuck, look at this Wikipedia article.) What happens if $\lambda\ge\operatorname{cf}\kappa$?
