a question about $\omega_1‎$ ‎The space ‎$  ‎\omega_1‎‎$‎ with its order ‎topology ‎is ‎countably ‎compact‎ and non-‎compact.‎
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‎A topological space $X$ is said to be star compact if whenever $\mathscr{U}$ is an open cover of $X$, there is a compact subspace $K$ of $X$ such that $X = \operatorname{St}(K,\mathscr{U})$.
$St(K, \mathscr{U})=\cup\{u\in \mathscr{U}: u \cap K \neq \emptyset\}‎$‎‎‎
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Can anyone help me to show:

Why ‎is‎ ‎$ ‎\omega_1‎ $‎ star ‎compact?‎
why ‎is ‎not‎ ‎$ ‎\omega_1‎ $‎ closed in ‎$ ‎\omega_1‎ +‎ ‎1‎ $‎?‎
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 A: As mentioned in this answer, for Hausdorff spaces like $\omega_1$, countable compactness is equivalent to being star finite. So $\omega_1$ is countably compact and so it is star finite and thus trivially star compact as well.
In the space $\omega_1 + 1$, which consists of the set of all ordinals $\le \omega_1$, the point $\omega_1$ is its maximum, and so in the order topology it has basic neighbourhoods of the form $(\alpha, \omega_1]$ with $\alpha < \omega_1$. And any such basic open set intersects $D=\omega_1$ (as a subset of $\omega+1$) (e.g. in $\alpha+1$ or any larger countable ordinal, in fact $|(\alpha, \omega_1] \cap D| = \aleph_1$), so $\omega_1$ is a limit point of $\omega_1$ (and not in $\omega_1$) so it’s a dense, not closed subset. 
A: A direct proof that $\omega_1$ with the $\epsilon$-order topology is star-compact.
$(1).$ Preliminary. Def'n: A function $f:\omega_1\to \omega_1$ is regressive iff $f(x)<x $ whenever $0<x<\omega_1.$
Lemma: If $f:\omega_1\to \omega_1$ is regressive then there exists $y\in \omega_1$ such that $f^{-1}\{y\}$ is uncountable.
Proof of Lemma. For  $0<x <\omega$: Let $f^0(x)=x,$ and for $n\in \omega$ let $f^{n+1}(x)=f(f^n(x)).$ For each $x$ there exists $n\in \omega$ with $n> 0$ such that $f^n(x)=0,$ otherwise $(f^n(x))_{n\in \omega}$ would be a strictly decreasing infinite sequence of ordinals. So $$(*)\quad \omega_1\setminus \{0\}=\cup_{0<n<\omega} G_n$$ where $G_1=f^{-1}\{0\}$ and $G_{n+1}=f^{-1}G_n$ for $1\le n <\omega.$
Now the RHS of $(*)$ is the union of a countable family so $G_n$ is uncountable for some $n.$ Let $n_0$ be the least $n$ such that $G_n$ is uncountable.
If $n_0=1$ then $G_1=f^{-1}\{0\}$ is uncountable. 
If $n_0>1$ then $f$ maps the uncountable set $G_{n_0}$ onto the countable set $G_{n_0-1}$ so $f^{-1}\{y\}$ is uncountable for some $y\in G_{n_0-1}.$
$(2).$ Let $U$ be an open cover of $\omega_1.$ Define $f(0)=0$ and $f(x+1)=x$ for $x\in \omega_1.$ For $0<x=\cup x \in \omega_1$ choose $U(x)\in U$ with $x\in U(x)$ and choose $f(x)<x$ such that the interval $[f(x),x]\subset U(x).$
Then $f$ is regressive so by the Lemma let $y\in \omega_1$ such that $f^{-1}\{y\}$ is uncountable.
And let $K=y+1.$
Let $S=\{x\in \omega_1: 0<x=\cup x\land f(x)=y\}.$  Then $S$ is uncountable because $f^{-1}\{y\}=S\cup \{y+1\}.$ Observe that $U(x)\cap K\ne \emptyset$ for all $x\in S.$
Now if $y<z<\omega_1$ then there exists $x\in S$ with $z<x,$ so $z\in [y,x]\subset U(x)\in U,$ so $z\in St(K,U)$ .
And of course if $z\le y$ then $z\in K$ so $z\in St(K,U)$ because $U$ is a cover of $\omega_1$.
