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‎The space ‎$ ‎\omega_1‎‎$‎ with its order ‎topology ‎is ‎countably ‎compact‎ and non-‎compact.‎ ‎

‎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 $X = \operatorname{St}(K,\mathscr{U})$.

$St(K, \mathscr{U})=\cup\{u\in \mathscr{U}: u \cap K \neq \emptyset\}‎$‎‎‎ ‎

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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    $\begingroup$ $\omega_1 = [0,\omega_1)$ us dense in $[0,\omega_1] = \omega_1+1$. But of course not sequentially dense. $\endgroup$ – GEdgar Jul 23 at 11:19
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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.

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  • $\begingroup$ what is the definition of star finite? $\endgroup$ – adin Jul 24 at 7:00
  • $\begingroup$ @adin for every open cover $\mathcal{U}$ of $X$ there is a finite subset $F\subseteq X$ such that $\operatorname{St}(F,\mathcal{U}) = X$. $\endgroup$ – Henno Brandsma Jul 24 at 7:02
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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 $\omega_1$.

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  • $\begingroup$ Is there a references to give me more guide about Preliminary and Lemma? Thanks for the answer. $\endgroup$ – adin Jul 24 at 6:41
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    $\begingroup$ @adin see here or look for Fodor’s lemma in a good set theory book, which is more general. $\endgroup$ – Henno Brandsma Jul 24 at 7:04
  • $\begingroup$ Fodor's Lemma is also called the Pressing-Down Lemma. It needs more set-theoretic background than the special case. $\endgroup$ – DanielWainfleet Jul 24 at 11:09
  • $\begingroup$ @HennoBrandsma. Before I learned Fodor's Lemma, I learned the lemma in my A in a course taught by William Weiss. He said the Post Office has a machine that takes a letter from a numbered slot and moves it to a lower-numbered slot. This worked when there were only countably many letters, but when the mail became uncountable, every time they run the machine it immediately jams because uncountably many letters go into the same slot. $\endgroup$ – DanielWainfleet Jul 24 at 11:19

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