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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.

Ergonvi
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  • Oh, for sure... Thank you! :) – Ergonvi Dec 26 '15 at 19:18
  • Interestingly, in this paper it is said that the conclusion of the problem you state (and indeed something stronger) is "clear". Now I don't know if I should feel bad about not seeing the solution immeadiately :) – Wojowu Dec 26 '15 at 19:30
  • From ${U_\alpha}{\alpha<\omega_1}$ you can pick, for example, the subfamily ${U\alpha}_{\alpha\geq 1}$ which has non empty intersection (in fact, it is $U_1$) so I think it is not a counterexample but thanks anyway :) – Ergonvi Dec 26 '15 at 20:02
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    My example (from a comment now deleted) actually didn't have an uncountable subfamily with nonempty intersection, because uncountable subset of $\omega_1$ is cofinal in it, and every ordinal $\alpha$ is outside $U_\beta$ for $\beta>\alpha$. This wasn't, however, a counterexample, because $[0,\omega_1]$ doesn't have precalibre $\aleph_1$, as witnessed by $U_\alpha=(\alpha,\alpha+2)$. – Wojowu Dec 26 '15 at 20:05
  • Actually, when reading your comment, I realize I might have written in the old comment $(0,\alpha)$. I meant $(\alpha,\omega_1)$. – Wojowu Dec 26 '15 at 20:06

2 Answers2

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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 $\varnothing\ne \bigcap_{\alpha\in B}\overline {V_\alpha}\subset \bigcap_{\alpha\in B}U_\alpha$.

user642796
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DanielWainfleet
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I think I have a possible answer. Let us consider $\{U_\alpha\}_{\alpha<\omega_1}$ an uncountable family of open sets. From this family we can consider an uncountable subfamily $\{U_\alpha\}_{\alpha\in B}$ with the FIP due to $X$ has $\aleph_1$ as a pre-caliber.

Now, given any finite set $F\in [B]^{<\omega}$, let us consider an open set $V_F$ such that $\overline{V_F}\subset \bigcap_{\alpha\in F} U_\alpha$ (this is possible by regularity and by the FIP). Now, since $X$ has pre-caliber $\aleph_1$ we can pick from $\{V_F\}_{F\in [B]^{<\omega}}$ another time an uncontable family $\{V_G\}_{G\in\mathcal{G}}$ of open sets with the FIP such that $\overline{V_G}\subset \bigcap_{\alpha\in G} U_\alpha$ for every $G\in \mathcal{G}$. So now it's enough to consider as a subfamily of our initial family $\{U_\alpha\}_{\alpha<\omega_1}$ the collection $\{U_\alpha\}_{\alpha\in \mathcal{A}}$ where $\mathcal{A}$ is $\bigcup \mathcal{G}$ because $$\emptyset \neq \bigcap_{G\in\mathcal{G}} \overline{V_G}\subset \bigcap_{\alpha\in\mathcal{A}} U_\alpha$$ by compactness.

Ergonvi
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  • So it's clear that $\bigcap_{G\in\mathcal{G}} \overline{V_G}\subset \bigcap_{G\in\mathcal{G}} \bigcap_{\alpha\in G} U_\alpha$. Now it's enough to see that the right side is included in $\bigcap_{\alpha\in \mathcal{A}} U_\alpha$. But I think (but I'm not pretty sure...) that this is true because if not there were an ordinal $\alpha_0$ such that $x\notin U_{\alpha_0}$. So pick a $G_0\in\mathcal{G}$ such that $\alpha_0\in G$. Then $x\notin \bigcap_{\alpha\in G_0} U_\alpha$ and finally $x\notin \bigcap_{G\in\mathcal{G}} \bigcap_{\alpha\in G} U_\alpha$. – Ergonvi Dec 26 '15 at 21:38