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 (?)
How is the closed ordinal space compact hausdorff
compactnessgeneral-topologyorder-topologywell-orders
Best Answer
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 $\beta$.