# If a topological space X has $\aleph_1$-calibre, then it must be star countable?

If a topological space X has $\aleph_1$-calibre[definition], then it must be star countable? What if the cardinality of the topological space X is additionally < = $2^{\aleph_0}$?

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• A space $X$ is star countable if for every open cover $\cal U$, there is a countable subset $Y\subset X$ such that $\bigcup\{U\in {\cal U}\mid U\cap Y\neq\emptyset\}=X$.

• The space $X$ has calibre $\aleph_1$ if for every uncountable list of nonempty open sets $U_\alpha$ for $\alpha\lt\omega_1$, there is an uncountable sublist $I_0\subset \omega_1$, such that $\bigcap_{\alpha\in I_0}U_\alpha\neq\emptyset$.

According to corollary 1.34 of the paper On the extent of star-countable spaces, it is claimed that $\mathbb{R}^\kappa$ is not star countable for sufficiently large cardinals $\kappa$.

Meanwhile, let's argue that $\mathbb{R}^\kappa$ has calibre $\aleph_1$. In fact, it suffices to argue that the class of spaces with calibre $\aleph_1$ is closed under arbitrary products. And I have now revised my answer to give the proof of this more general fact, which Henno says in the comments is well-known (and he evidently runs in quality circles).

Theorem. The product $\Pi_{i\in I}X_i$ of any family of spaces $X_i$ with calibre $\aleph_1$ has calibre $\aleph_1$.

Proof. First, note that finite products of calibre $\aleph_1$ spaces has calibre $\aleph_1$ by the argument given in this MO question on finite products of calibre $\aleph_1$. Suppose that $U_\alpha$ for $\alpha\lt\omega_1$ is an uncountable family of open sets in the product. By shrinking the sets, we may assume without loss of generality that each $U_\alpha$ is a basic open set, having some finite support $I_\alpha\subset I$. If uncountably many $U_\alpha$ have the same finite support $J$, then by since $\prod_{i\in J}X_i$ is a finite product and hence has calibre $\aleph_1$, it follows that there is an uncountable subfamily of these $U_\alpha$ with nonempty intersection, witnessing this instance of calibre $\aleph_1$ for the product. So we are left with the case where there are uncountably many different supports appearing for the supports of the various $U_\alpha$. Thus, we have an uncountable family of finite sets $I_\alpha$. By the $\Delta$-system lemma, there is an uncountable subfamily $I_0\subset \omega_1$, such that the supports of $U_\alpha$ for $\alpha\in I_0$ form a $\Delta$-system with finite root $J_0$, meaning that any two such supports intersect exactly to $J_0$. Since again $\prod_{i\in J_0}X_i$ has calibre $\aleph_1$, it follows that there is an uncountable subfamily $I_1\subset I_0$ such that $\bigcap_{\alpha\in I_1}U_\alpha\upharpoonright J_0$ is not empty. Since it is a $\Delta$-system, it also follows that $\bigcap_{\alpha\in I_1}U_\alpha$ is not empty in the original space, since the remaining parts of the supports do not conflict with each other, and so $\prod_{i\in I}X_i$ has calibre $\aleph_1$, as desired. QED

In particular, $\mathbb{R}^\kappa$ has calibre $\aleph_1$, but is not star countable, thereby answering the first (original) question.

Perhaps $\mathbb{N}^\kappa$ for uncountable $\kappa$ may be a simpler counterexample, since the theorem also shows it to have calibre $\aleph_1$, and we know at least that it is not Lindelöf by the argument of this MO question on Linedelöfness and compactness. Is it star-countable? I'm not currently sure, but if not, then it may be a simpler counterexample.

But perhaps one needs $\kappa$ to be very large to make this conclusion, in which case even $\mathbb{N}^\kappa$ may not answer the second question.

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It seems that the $\Delta$-system argument shows that the $\aleph_1$-calibre spaces are closed under arbitrary products. – Joel David Hamkins Nov 15 '11 at 1:31
What if the cardinality of the topological space X is additionally < = $2^{\aleph_0}$? – Paul Nov 15 '11 at 1:49
In the paper to which I linked, we have $\kappa\geq 2^{c^+}$, in which case $\mathbb{R}^\kappa$ is similarly large. But if the $\mathbb{N}^{\omega_1}$ example works out, it could still be $\leq 2^{\aleph_0}$, since $2^{\aleph_0}=2^{\aleph_1}$ is consistent with ZFC. So I don't know. – Joel David Hamkins Nov 15 '11 at 1:51
@Joel, indeed the productivity of calibre $\aleph_1$ spaces is well-known. – Henno Brandsma Nov 15 '11 at 18:14
Henno, I edited to give the proof. – Joel David Hamkins Nov 16 '11 at 0:29