If a topological space X has $\aleph\_1$-calibre[[definition]](https://mathoverflow.net/questions/78414/aleph-1-calibre/78451#78451), then it must be star countable?
What if the cardinality of the topological space X is additionally < = $2^{\aleph\_0}$?



If a topological space X has $\aleph_1$-calibre, then it must be star countable?
The answer is negative.


* 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](http://www.cs.vu.nl/~vanmill/papers/papers2011/al-ju-vm-tk-wi.pdf),
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 $\aleph\_1$](https://mathoverflow.net/questions/78414/aleph-1-calibre/78451#78451). 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](http://en.wikipedia.org/wiki/Delta_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](https://mathoverflow.net/questions/9641/how-far-is-lindelof-from-compactness/9651#9651).
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. 


