# How is the closed ordinal space compact hausdorff?

So the original question in the paper was whether every compact hausdorff space automatically metrizable? I found out that closed ordinal space with order topology is a counter example. Since the nature of $$\mathcal w$$ is uncountable, and every metric space is first countable, thus this cant be metrizable. That makes sense. But how do I go about proving that it is compact hausdorff? I don't think that the finite subcover cover of any open cover definition of compact would help here (?)

Consider the closed ordinal space given by $$X=\{\xi\mid \xi\leq \gamma\}$$ for some limit ordinal $$\gamma$$.

Let $$\mathscr U=\langle U_\eta\mid \eta<\alpha\rangle$$ be an open cover of $$X$$. Then let $$\beta$$ be the largest ordinal such that the interval $$[0,\delta]$$ has a finite subcover in $$\mathscr U$$ for every $$\delta<\beta$$.

If $$\beta\leq\gamma$$, then $$\beta\in U_\xi$$ for some $$\xi<\alpha$$. But then $$U_\xi$$ contains an open interval $$(\beta_1,\beta_2)\ni\beta$$ (or if $$\beta=\gamma$$ an open ray $$(\beta_1,\gamma]$$) with $$\beta_1<\beta$$. However $$[0,\beta_1]$$ has a finite subcover $$\langle U_{\eta_1},\dots,U_{\eta_n}\rangle$$, as $$\beta_1<\beta$$.

But this means that $$\langle U_{\eta_1},\dots,U_{\eta_n},U_\xi\rangle$$ is a finite subcover of $$[0,\beta]$$, contradicting the maximality of $$\beta$$.

Therefore $$\beta\geq\gamma+1$$, which means $$[0,\gamma]=X$$ has a finite subcover.

To show $$X$$ is Hausdorff, take $$\alpha<\beta$$, then the intervals $$[0,\alpha+1)$$ and $$(\alpha,\gamma]$$ separate $$\alpha$$ and $$\beta$$.

Here's yet another proof of compactness.

Let $$x_1 = \gamma$$, the greatest element in the given closed ordinal space $$X$$. Let $$U_1$$ be any element of the open cover that contains $$x_1$$, let $$B_1 \subset U_1$$ be an open interval containing $$x_1$$, and let $$y_1$$ be the least element in $$U_1$$. Since $$B_1$$ is open, $$y_1$$ is not a limit ordinal. If $$y_1$$ is the least element of $$X$$, we stop. Otherwise let $$x_2$$ be the predecessor of $$y_1$$, and so $$x_1 > x_2$$.

Continue by ordinary induction.

This induction must stop after finitely many steps because otherwise we obtain an infinite decreasing sequence $$x_1 > x_2 > ...$$ which is impossible.

But the only way that the induction can stop is if some $$y_n$$ is the least element, in which case $$U_1,...,U_n$$ is a finite subcover.

If $$X$$ is the set of all ordinals $$\le \gamma$$ for some ordinal $$\gamma$$ we can apply a well-known criterion for compactness of ordered topological spaces:

A (non-empty) ordered set $$(X,<)$$ is compact in the order topology iff for all $$A \subseteq X$$, $$\sup(A)$$ exists in $$X$$.

Apply to $$X$$: $$\sup(X)=\gamma$$ surely exists, as does $$\sup(\emptyset)=\min(X)=0$$. For all other $$A$$, $$A$$ has a non-empty set of upper bounds $$U$$ ($$\gamma \in U$$) and so by well-orderness $$\min(U) \in X$$ exists and this equals $$\sup(A)$$ by definition.

I give an alternative proof for your case here, using only the definition of the order topology and a minimality argument to exploit well-orderedness.

• I'm in the mood for nitpicking. The highlighted sentence is incorrect if $X$ is empty....+1 – DanielWainfleet Jul 8 '19 at 5:15
• We also have: A non-empty ordered set $(X,<)$ is compact in the order topology iff for all $A\subseteq X,\; \inf(A)$ exists in $X$. The important case here is $A=\emptyset.$ – DanielWainfleet Jul 8 '19 at 5:23
• @DanielWainfleet Yes, inf will do too. It's quite symmetrical.. It's easily shown from Alexander's subbase lemma in either case. – Henno Brandsma Jul 8 '19 at 15:45