TITLE: Compact Hausdorff Spaces with pre-caliber $\aleph_1$ has caliber $\aleph_1$
QUESTION [7 upvotes]: Let us recall that a topological space has $\aleph_1$ pre-caliber (resp. caliber) if given any family of open sets $\{U_\alpha\}_{\alpha<\omega_1}$ there exists an uncontable set $B\subset\omega_1$ such that the subfamily $\{U_\alpha\}_{\alpha\in B}$ has the finite intersection property (resp. $\bigcap_{\alpha\in B} U_\alpha\neq\emptyset$).
I was trying to prove that a compact Hausdorff space has $\aleph_1$ pre-caliber if and only if it has $\aleph_1$ caliber. In this regard, I would construct a subfamily of closed sets with the FIP inside the family $\{U_\alpha\}_{\alpha\in B}$ and then using the compactness ensure that $\bigcap_{\alpha\in B} U_\alpha\neq\emptyset$. Probably using the normality of the space $X$ we can construct that subfamily but I'm not sure.
Can anybody help me? Thanks in advance.

REPLY [2 votes]: Let $F=\{U_\alpha :\alpha\in \omega_1\}$ be an open family with $\varnothing\not \in F$. For $\alpha\in \omega_1,$ let $V_\alpha$ be open with $\varnothing\ne V_\alpha\subset \overline {V_\alpha}\subset U_\alpha$. Let $B$ be an uncountable subset of $\omega_1$ such that $\{V_\alpha :\alpha\in B\}$ has the F.I.P. Then $\{\overline V_\alpha :\alpha\in B\}$ has the F.I.P., so by compactness we have $\varnothing\ne \bigcap_{\alpha\in B}\overline {V_\alpha}\subset \bigcap_{\alpha\in B}U_\alpha$.