Every successor ordinal is compact in its order topology and every limit ordinal is not; and, assuming countable choice, ${\omega }_1$ is countably compact and sequentially compact while ${\omega }_1+1$ is compact

For claim 1 let $\gamma :={\delta }^{+}$, so that $\delta$ is the greatest element of $\gamma$ and $[0,\delta ]=\gamma$; let $\mathcal{U}$ be an open cover of $\gamma$ and put $S:=\{\eta \in \gamma :\text{finitely many members of }\mathcal{U}\text{ cover }[0,\eta ]\}$.

For claim 2 let $\lambda$ be a limit ordinal; the family $\{[0,\beta ]:\beta \in \lambda \}$ consists of open sets by [L2] and covers $\lambda$, since $\xi \in [0,\xi ]$ for every $\xi \in \lambda$.

For claim 3 assume ${\mathrm{A}\mathrm{C}}_{\omega }$ and let $(y_k)$ be a sequence in ${\omega }_1$; its range is at most countable by [L7], so [L6] gives $\sigma :=\sup \{y_k:k\in \mathbb{N}\}\in {\omega }_1$, and the set $\{k\in \mathbb{N}:y_k\le \sigma \}$ is all of $\mathbb{N}$ and in particular infinite.

Let $\eta \in \gamma$ and suppose $[0,\zeta ]$ is covered by finitely many members of $\mathcal{U}$ for every $\zeta <\eta$. Some $U\in \mathcal{U}$ contains $\eta$, and [L2] gives $B\in {\mathcal{B}}_{\gamma }$ with $\eta \in B\subseteq U$. If $B=[0,\beta ]$ then $\eta \le \beta$ and $[0,\eta ]\subseteq [0,\beta ]\subseteq U$, so $\{U\}$ covers $[0,\eta ]$. If $B=(\alpha ,\beta ]$ then $\alpha <\eta \le \beta$, and $[0,\eta ]\subseteq [0,\alpha ]\cup (\alpha ,\eta ]\subseteq [0,\alpha ]\cup U$ by [L3], so a finite cover of $[0,\alpha ]$ with $U$ adjoined covers $[0,\eta ]$. Either way $\eta \in S$.

A finite subfamily of the cover of step 1.2 is empty, and then covers only $\mathrm{\varnothing }\mathrm{\ne }\lambda$, or is $[0,{\beta }_0],\dots ,[0,{\beta }_n]$ with union $[0,\beta ]$ for $\beta$ the greatest of the ${\beta }_j$, which exists by [L3]; and ${\beta }^{+}\in \lambda$ by [L3] while ${\beta }^{+}\mathrm{\notin }[0,\beta ]$. So no finite subfamily covers $\lambda$ and $\lambda$ is not compact, which is claim 2.

By [L5] and [L6] the set $\{\xi \in {\omega }_1:\{k:y_k\le \xi \}\text{ is infinite}\}$ is a nonempty set of ordinals, $\sigma$ belonging to it by step 1.3, so it has a least element $\tau$ by [L3]; then $P:=\{k\in \mathbb{N}:y_k\le \tau \}$ is infinite while $\{k:y_k\le \alpha \}$ is finite for every $\alpha <\tau$.

By [L4] applied to the well-ordered $\gamma$, step 2.1 gives $S=\gamma$; in particular $\delta \in S$, so finitely many members of $\mathcal{U}$ cover $[0,\delta ]=\gamma$. As $\mathcal{U}$ was arbitrary, $\gamma ={\delta }^{+}$ is compact, which is claim 1.

Let $i\mapsto m_i$ be the strictly increasing enumeration of $P$ given by [L8]; then $(y_{m_i})$ is a subsequence of $(y_k)$ by [L9], and every one of its terms satisfies $y_{m_i}\le \tau$.

$y_{m_i}\to \tau$. Let $U$ be open with $\tau \in U$ and take $B\in {\mathcal{B}}_{{\omega }_1}$ with $\tau \in B\subseteq U$ by [L2]. If $B=[0,\beta ]$ then $\tau \le \beta$ and every term satisfies $y_{m_i}\le \tau \le \beta$, so all terms lie in $B$. If $B=(\alpha ,\beta ]$ then $\alpha <\tau \le \beta$, the set $\{k:y_k\le \alpha \}$ is finite by step 2.3, so $\{i:y_{m_i}\le \alpha \}$ is finite, the map $i\mapsto m_i$ being injective; hence $\alpha <y_{m_i}\le \tau \le \beta$ for all large $i$ and the terms lie in $B$ from some index on. So ${\omega }_1$ is sequentially compact.

By [L10] the space ${\omega }_1$ is therefore countably compact; it is not compact by step 2.2, being a limit ordinal by [L5]; and ${\omega }_1+1={\omega }_1^{+}$ is compact by step 3.1 and [L11]. This is claim 3, and with claims 1 and 2 at steps 3.1 and 2.2 the theorem is proved.