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.

  • $\begingroup$ I'm in the mood for nitpicking. The highlighted sentence is incorrect if $X$ is empty....+1 $\endgroup$ – DanielWainfleet Jul 8 '19 at 5:15
  • $\begingroup$ 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 $A=\emptyset.$ $\endgroup$ – DanielWainfleet Jul 8 '19 at 5:23
  • $\begingroup$ @DanielWainfleet Yes, inf will do too. It's quite symmetrical.. It's easily shown from Alexander's subbase lemma in either case. $\endgroup$ – Henno Brandsma Jul 8 '19 at 15:45

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