# Is the Isbell–Mrówka space $\Psi(\mathcal A)$ with $\lvert\mathcal A\rvert=\omega_1$ starcompact?

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.

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.
• 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? Jan 26 at 10:09
• What are your quantifiers? Do you want it to happen for all $\mathcal{V}$? Or just one finite subfamily of $\mathcal{U}$? Jan 26 at 10:22
• Note: this proof does not depend on $\omega_1$; it will work for every (M)AD family. So $\Psi(\mathcal{A})$ is never starcompact. Jan 26 at 10:24
• We want it to happen for all $\mathcal V$. Jan 26 at 10:39
• 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. Jan 26 at 11:10