Consider the space $\omega_1$ (the first uncountable ordinal) together with the order topology (a convenient base for which is $\{(\alpha, \beta] \mid \alpha, \beta \in \omega_1\} \cup \{0\}$ ). I wish to show:
Any closed interval of $\omega_1$ is compact under the order topology.
At first I thought it false for the following reason: Take $[1, \omega^2]$ ($\omega$ is the first countable ordinal), which contains countably-many countable ordinals ($\omega \cdot n$ for each $n \in \omega$). If the cover of $[1, \omega^2$] is such that no open set contains more than one countable ordinal, then it cannot be reduced to a finite subcover.
The flaw in my reasoning: There is no countable ordinal that is the immediate predecessor of $\omega^2$. Any open set containing $\omega^2$ is going to contain all but finitely-many ordinals of the form $\omega \cdot n$.
As we go out toward $\omega_1$, the open sets around certain limit ordinals are going to take bigger "bites". Can this idea be made rigorous?
If my thoughts are not going to be particularly fruitful, then I would be happy with another approach. (Hints only, please.)
Best Answer
If you want to use a more general approach: for ordered spaces, compactness = order completeness (every subset has a supremum and an infimum), and for closed intervals of $\omega_1$ (where the induced topology = the order topology) this follows easily: we have a min for every non-empty subset, so certainly an infimum, and being bounded above gives us a minimum for the set of upper bounds hence a supremum as well, which by being a closed interval must be in the closed interval too.