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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.)

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up vote 2 down vote accepted

OUTLINE: Without loss of generality start with a cover of $(\alpha,\beta]$ by intervals of the form $(\xi,\eta]$. There must be one of the form $(\xi_0,\beta]$. Then there must be one of the form $(\xi_1,\xi_0]$. In this way construct a strictly decreasing sequence $\langle \xi_n:n\in\omega\rangle$. Therefore?

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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.

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First note that $[1,\omega^2]$ is in fact isomorphic to $omega^2+1$ as an ordinal, since the closed interval implies that $\omega^2$ is in the interval, and it is a maximal element.

Secondly, note that if $[\alpha,\beta]$ is a closed interval of ordinals then it is isomorphic to an ordinal via a continuous and open order preserving isomorphism, therefore it is homeomorphic to an ordinal.

By the above claim we have that $[\alpha,\beta]$ is isomorphic to a successor ordinal (since the set is isomorphic to a successor ordinal: $\mathrm{otp}(\beta-\alpha)+1$) and successor ordinals are always compact in the order topology since a cover of $\gamma+1$ can be made into a decreasing chain of ordinals - which is finite.

Claim: Suppose $[\alpha,\beta]$ is a closed interval of ordinals, then it is homeomorphic to a successor ordinal $\gamma$.

Proof: Since as an ordered set $[\alpha,\beta]$ is well-ordered there exists a unique ordinal and a unique isomorphism $f$ between the interval and the ordinal. First it is obvious that $\gamma$ is a successor ordinal, since $\beta$ is the maximal element in $[\alpha,\beta]$ therefore the order type of $\gamma$ has a maximal element, which can only occur at successor ordinals.

Continuity in well-ordered topologies is to say that if $\delta$ is a limit ordinal then $f(\delta)=\sup\{f(\tau)\mid\tau<\delta\}$. We recall that $f$ is such that $f(\alpha)=0; f(\tau+1)=f(\tau)+1;$ and $f(\delta)=\sup\{f(\tau)\mid\tau<\delta\}$ for limit ordinals $\delta$, so indeed the mapping is continuous.

Lastly we need to show that it is indeed open, which is to say that the image of an open interval is open. This is another result of the definition of $f$, since if $(\alpha+\epsilon,\beta')=[\alpha+\epsilon+1,\beta')$ is an open interval in $[\alpha,\beta]$ then it holds that $f''[\alpha+\epsilon+1,\beta')=[f(\epsilon)+1,f(\beta')]$.

Therefore we have that indeed $[\alpha,\beta]$ is homeomorphic to $\gamma+1$ for some ordinal $\gamma$, and therefore compact.$\square$

Note that if $\gamma$ is countable then $\gamma+1$ is a Polish space, that is a separable, completely metrizable metric space. However compactness does not require metrizability in the case of ordinals, and indeed any successor ordinal is a compact space (when equipped with the order topology).

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