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A space $X$ is said to be starcompact if for every open cover $\mathcal U$ of $X$ there exists a finite subset $\mathcal V\subseteq\mathcal U$ such that $\operatorname{St}(\bigcup\mathcal V,\mathcal U)=X$.

I think the Isbell–Mrówka space $\Psi(\mathcal A)$ with $\lvert\mathcal A\rvert=\omega_1$ is not starcompact. But I can't prove it.

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Enumerate $\mathcal{A}$ as $\langle A_\alpha:\alpha\in\omega_1\rangle$. Enumerate $[\omega_1]^{<\omega}\times\omega$ as $\bigl<\langle F_\alpha,n_\alpha\rangle:\alpha\in\omega_1\bigr>$, with $F_\alpha\subseteq\alpha$ always. Define recursively a sequence of neighourhoods: $U_\alpha$ of $A_\alpha$ such that $U_\alpha\cap(\bigcup\{A_\beta:\beta\in F_\alpha\}\cup n_\alpha)=\emptyset$. Then $\mathcal{U}=\{U_\alpha:\alpha\in\omega_1\}\cup\bigl\{\{n\}:n\in\omega\bigr\}$ is an open cover. If $\mathcal{V}$ is a finite subfamily of $\mathcal{U}$ then there is an $\alpha$ such that $\mathcal{V}\subseteq\{U_\beta:\beta\in F_\alpha\}\cup\bigl\{\{i\}:i\in n_\alpha\bigr\}$. Then $U_\alpha\cap\bigcup\mathcal{V}=\emptyset$, so the point $A_\alpha$ is not in $\operatorname{St}(\bigcup\mathcal{V},\mathcal{U})$.

Addendum (2022-01-30): this proof works for every (M)AD family of any cardinality; the conclusion is that $\Psi(\mathcal{A})$ is never starcompact.

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  • $\begingroup$ Does $St(\bigcup\mathcal V,\mathcal U)$ not contain uncountable number of elements of $\Psi(\mathcal A)$? If not, then can we construct $\mathcal U$ in such a way that it will happen? $\endgroup$
    – Nur Alam
    Commented Jan 26, 2022 at 10:09
  • $\begingroup$ What are your quantifiers? Do you want it to happen for all $\mathcal{V}$? Or just one finite subfamily of $\mathcal{U}$? $\endgroup$
    – KP Hart
    Commented Jan 26, 2022 at 10:22
  • $\begingroup$ Note: this proof does not depend on $\omega_1$; it will work for every (M)AD family. So $\Psi(\mathcal{A})$ is never starcompact. $\endgroup$
    – KP Hart
    Commented Jan 26, 2022 at 10:24
  • $\begingroup$ We want it to happen for all $\mathcal V$. $\endgroup$
    – Nur Alam
    Commented Jan 26, 2022 at 10:39
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    $\begingroup$ Not necessarily; it depends on the AD family: if, in our notation, $\{A_n:n\in\omega\}$ is pairwise disjoint and also $A_n\cap A_\alpha=\emptyset$ whenever $n<\omega\le\alpha$ then there will always be $\mathcal{V}$ with a countable star. $\endgroup$
    – KP Hart
    Commented Jan 26, 2022 at 11:10

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