The square of X which is $\aleph\_1$-calibre is still $\aleph\_1$-calibre?



$\aleph_1$-calibre 
The answer is yes.


A topological space has *calibre* $\aleph\_1$ if for every uncountable sequence $\langle U\_\alpha\mid\alpha\lt\aleph\_1\rangle$ of nonempty open sets $U\_\alpha\subset X$, there is an uncountable subfamily $\Lambda\subset\aleph\_1$ with $\bigcap\_{\alpha\in\Lambda}U\_\alpha\neq\emptyset$. 


**Theorem.** If a topological space $X$ has calibre $\aleph\_1$, then so does $X\times X$.


Proof. Suppose $X$ has calibre $\aleph\_1$, and consider a family of nonempty open sets $U\_\alpha$ in $X\times X$, for $\alpha\lt\aleph\_1$. The idea will be to reduce first on the first coordinate, and then on the second. Since $U\_\alpha$ is open in the product, we may choose a nonempty open rectangle $U\_\alpha^0\times U\_\alpha^1\subset U\_\alpha$. Consider the sequence $\langle U\_\alpha^0\mid\alpha\lt\aleph\_1\rangle$ of open sets arising from the first coordinate. By the calibre $\aleph\_1$ property of $X$, it follows that there is $\Lambda\_0\subset\aleph\_1$ with some $x\in\bigcap\_{\alpha\in\Lambda\_0}U\_\alpha^0$. Consider now the enumeration $\langle U\_\alpha^1\mid \alpha\in\Lambda\_0\rangle$. By the calibre property again, we may find a smaller uncountable family $\Lambda\subset\Lambda\_0$ with some $y\in\bigcap\_{\alpha\in\Lambda}U\_\alpha^1$. It follows that $(x,y)\in\bigcap\_{\alpha\in\Lambda}U\_\alpha^0\times U\_\alpha^1\subset\bigcap\_{\alpha\in\Lambda}U\_\alpha$, and so $X\times X$ has calibre $\aleph\_1$, as desired. QED


