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If a topological space $X$ has $\aleph_1$-calibre and the cardinality of $X$ is $\le 2^{\aleph_0}$, then it must be star countable? A topological space $X$ is said to be star-countable if whenever $\mathscr{U}$ is an open cover of $X$, there is a countable subspace $A$ of $X$ such that $X = \operatorname{St}(A,\mathscr{U})$.

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Surely the soft-question tag is inapplicable here? – Nate Eldredge Nov 22 '11 at 1:55
Isn't it the same question you asked at MO?… – Martin Sleziak Nov 22 '11 at 12:55
@Martin: It’s not the same as the original MO question, but it is the same as the revised version. – Brian M. Scott Nov 22 '11 at 19:40
up vote 2 down vote accepted

It’s consistent that such a space not be star-countable. Specifically, assume that $2^{\omega_1}=2^\omega=\mathfrak{c}$. Let $D=\{0,1\}$, and let $Y=D^\mathfrak{c}$. For $y\in Y\;$ let $\operatorname{supp}(y)=\{\xi<\mathfrak{c}:y(\xi)=1\}$, the support of $y$, and let $X=\{x\in Y:0<|\operatorname{supp}(x)|\le\omega_1\}$; $|X|=\mathfrak{c}^{\omega_1}=(2^{\omega_1})^{\omega_1}=\mathfrak{c}$.

For any finite partial function $\varphi:\mathfrak{c}\to D$ let $B(\varphi)=\{x\in X:x\upharpoonright\operatorname{dom}\varphi=\varphi\}$; the sets $B(\varphi)$ are a base for the topology on $X$. Let $\{U_\xi:\xi<\omega_1\}$ be a family of open sets in $X$. For $\xi<\omega_1$ let $\varphi_\xi$ be a finite partial function from $\mathfrak{c}$ to $D$ such that $B(\varphi_\xi)\subseteq U_\xi$, and let $D_\xi=\operatorname{dom}\varphi_\xi$. By the delta-system lemma there is an uncountable $\Lambda\subseteq\omega_1$ and a finite $D\subseteq\mathfrak{c}$ such that $D_\xi\cap D_\eta =D$ whenever $\xi,\eta\in \Lambda$ and $\xi\ne\eta$. Since $D$ is finite, there is an uncountable $\Lambda_0\subseteq\Lambda$ such that $\varphi_\xi\upharpoonright D=\varphi_\eta\upharpoonright D$ whenever $\xi,\eta\in\Lambda_0$, and hence $\bigcap\limits_{\xi\in\Lambda_0}U_\xi\supseteq\bigcap\limits_{\xi\in\Lambda_0}B(\varphi_\xi)\ne\varnothing$. Thus, $X$ has calibre $\omega_1$.

Finally, for $\xi<\mathfrak{c}$ let $V_\xi=B(\langle\xi,1\rangle)$; clearly $\mathscr{V}=\{V_\xi:\xi<\mathfrak{c}\}$ is an open cover of $X$. Let $C$ be any countable subset of $X$, and let $S=\bigcup\limits_{x\in C}\operatorname{supp}x$; clearly $|S|\le\omega_1<2^{\omega_1}=\mathfrak{c}$, so there is some $\eta\in\mathfrak{c}\setminus S$. Let $x$ be the unique point of $X$ with support $\{\eta\}$. Suppose that $y\in C\cap V_\xi$. Then $y(\xi)=1$, so $\xi\in S$, $x(\xi)=0$, and $x\notin V_\xi$. Thus, $\operatorname{St}(C,\mathscr{V}\;)\ne X$; and since $C$ was an arbitrary countable subset of $X$, $X$ is not star-countable.

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Brian M. Scott, thanks for your answer. Your answes is aways very helpful for me. The example is very good; X is a subspace of a "good" topological space. – Paul Nov 25 '11 at 3:22

Under CH the space is separable (hence, star countable ).

Proof(Ofelia). On the contrary, suppose that X is not separable .Under CH write $X = \{ x_\alpha : \alpha \in \omega_1 \}$ and for each $\alpha$ in $\omega_1$ define $U_\alpha$ = the complement of $cl ( { x_\beta : \beta \le \alpha } )$ . The family of the $U_\alpha$ is a decreasing family of non-empty open sets; since $\aleph_1$ is a caliber of X, the intersection of all $U_\alpha$ must be non-empty (contradiction!)

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