Here is a Lemma from this paper, which maybe helpful for the answering my question. enter image description here

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

Let $\mathscr{U}$ be an open cover of $X$ and $\mathscr{V}\;$ be a finite subset of $\mathscr{U}$ covering $K$; if $\operatorname{St}(\cup\mathscr{V},\mathscr{U})=X$. Then we called $X$ is $1$-starcompact. Clearly, every star compact space implies $1$-starcompact.

My question is this:

If $X$ is a regular star-compact, first countable, has the cardinality of $\aleph_1$, then is $X$ countably compact?

Thanks ahead:)

One direction: If 1-starcompact in this space is closed hereditarily, I think, the question will be solved.

  • $\begingroup$ Could you explain what you mean by $\operatorname{St}(K,\mathcal{U})$? I have never seen this notation before. $\endgroup$ – Jakub Konieczny Mar 16 '13 at 13:07
  • 2
    $\begingroup$ @Feanor: $=\bigcup\{U\in \mathcal{U}: U\cap K\not=\emptyset\}$ $\endgroup$ – Paul Mar 16 '13 at 13:09

$1$-starcompactness is not closed hereditary. This is mentioned immediately after Example $2.3.3$, which shows that the Tikhonov plank is $1$-starcompact: the set $\{\omega_1\}\times\omega$ is a closed, discrete subset of the plank that is clearly not $1$-starcompact.

The result in your question is at least consistently true. This follows from Lemma $2.2.10$ of the paper:

Lemma $\mathbf{2.2.10}$. If $X$ is a regular first countable $1$-starcompact space with $w(X)<\mathfrak d$, then $X$ is countably compact.

Now let $X$ be a regular, first countable, star-compact space of cardinality $\omega_1$. As you point out in the question, $X$ is $1$-starcompact. Moreover, $w(X)\le\chi(X)\cdot|X|=\omega\cdot\omega_1=\omega_1\le\mathfrak d$. If $\omega_1<\mathfrak d=2^\omega$, which is known to be consistent (e.g., it follows from $\mathsf{MA}+\neg\mathsf{CH}$), then $X$ is countably compact by Lemma $2.2.10$.

Thus, if there is a consistent counterexample, it will have to be in a model in which $\mathfrak d=\omega_1$. And indeed Example $\mathbf{2.3.5}$ of the paper turns out to be such a counterexample. The paper contains a proof that this space $X$ is $1$-starcompact, but in fact it is star-compact.

Proof. Let $\mathscr{U}$ be any open cover of $X$. For $k\in\omega$ define $U_k$ and $f\in D$ as in the proof that $X$ is $1$-starcompact. Let $K=L_f\cup\{g\in D:g\le^*f\}$; I claim that $K$ is compact. $K\cap D$ is homeomorphic to $\alpha+1$ for some countable ordinal $\alpha$ and is therefore compact. Suppose that $V$ is an open nbhd of $K\cap D$, and $A=L_f\setminus V$ is infinite. $L_f\cap\big(\{k\}\times\omega\big)$ is finite for each $k\in\omega$, so $A\cap\big(\{k\}\times\omega\big)$ is finite for each $k\in\omega$, and it follows from the proof of Claim $\mathbf 2$ of the paper that $A$ has a limit point $g\in K\cap D$. But then $g\in V$, so $V\cap A\ne\varnothing$, contradicting the choice of $A$. Thus, $A$ is finite, and $K$ is compact.

Let $F=\{k\in\omega:K\cap U_k=\varnothing\}$; then $F$ is finite, and $\operatorname{st}(K,\mathscr{U})\supseteq K\cup\big((\omega\setminus F)\times\omega\big)\cup(\omega\setminus F)$. Then $K\cup F$ is compact, and $S=(\omega\times\omega)\setminus\operatorname{st}(K\cup F,\mathscr{U})$ is finite, so $K\cup F\cup S$ is compact, and $\operatorname{st}(K\cup F\cup S,\mathscr{U})\supseteq K\cup(\omega\times\omega)\cup\omega$. That is, $X\setminus\operatorname{st}(K\cup F\cup S,\mathscr{U})\subseteq D\setminus K$. But $D\setminus K$ is homeomorphic to $\omega_1$ and is therefore star finite (= strongly $1$-starcompact), so there is a finite $E\subseteq D\setminus K$ such that $K\cup F\cup S\cup E$ is compact and $\operatorname{st}(K\cup F\cup S\cup E,\mathscr{U})=X$. $\dashv$


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