A question on spaces with calibre-$\aleph_1$ Suppose that $X$ is the $T_1$ space with $k$-in-countable base and $\aleph_1$ is a caliber of $X$. Must $X$ be second countable? Thanks for any help.

A topological space has calibre $\aleph_1$ if for every uncountable family $\{U_\alpha\mid\alpha\lt\aleph_1\}$ of nonempty open sets $U_\alpha\subset X$, there is an uncountable $\Lambda\subseteq\aleph_1$ such that $\bigcap_{\alpha\in\Lambda}U_\alpha\ne\varnothing$.
A family $\mathcal U$ of subsets of a space $X$ is called $k$-in-countable if every set $A \subset X$ with $|A|=k$ is contained in at most countably many elements of $\mathcal U$.

I think it is true.
What I have tried: Let $\mathcal B$ be a $k$-in-countable base of $X$. We only need to show that $\mathcal B$ is countable. Suppose not. Then there exists an uncountable family of open sets of $X$. Since $\aleph_1$ is a caliber of $X$, there is an uncountable subfamily $\Lambda\subset\aleph_1$ with $\bigcap_{\alpha\in\Lambda}U_\alpha\neq\emptyset$. $F=:\bigcap_{\alpha\in\Lambda}U_\alpha$ must have $k$ points of $X$. Otherwise, let $\mathcal U'=\{U_\alpha\setminus F: \alpha \in \Lambda\}$; $\mathcal{U}'$ is still an uncountable family of open sets of $X$. However, $\bigcap \mathcal U' = \emptyset$. This completes the proof.
Am I right? Thanks for your any help.
 A: The argument doesn’t work, I’m afraid. You have the $k$-in-countable base $\mathscr{B}$. You assume, to get a contradiction, that $\mathscr{B}$ is uncountable and let $\mathscr{U}=\{U_\alpha:\alpha\in\omega_1\}$ be a subset of $\mathscr{B}$ of cardinality $\omega_1$. Since $X$ has calibre $\omega_1$, there is an uncountable $\Lambda\subseteq\omega_1$ such that $\bigcap_{\alpha\in\Lambda}U_\alpha\ne\varnothing$. Let $F=\bigcap_{\alpha\in\Lambda}U_\alpha$. $\mathscr{B}$ is $k$-in-countable, so $|F|<k$: $F$ cannot contain $k$ or more points of $X$. This is what you actually want, however, since it does justify your next step; what you had does not.
For the next step let $U_\alpha'=U_\alpha\setminus F$ for $\alpha\in\Lambda$, and let $\mathscr{U}'=\{U_\alpha':\alpha\in\Lambda\}$. $F$ is finite, so each $\mathscr{U}'$ is a family of open sets, and $\bigcap\mathscr{U}'=\varnothing$. However, this doesn’t contradict the hypothesis that $X$ has calibre $\omega_1$: it’s still possible that there is an uncountable $K\subseteq\Lambda$ such that $\bigcap_{\alpha\in K}U_\alpha'\ne\varnothing$.
The proof can be repaired, however. Let $\mathscr{U}_0$ be your original family $\mathscr{U}$, and let $\mathscr{V}_0$ be an uncountable subfamily with non-empty intersection $F_0$. Then $|F_0|<k$, so in particular $F_0$ is finite, and $\mathscr{U}_1=\{U\setminus F_0:U\in\mathscr{V}_0\}$ is an uncountable family of open sets with empty intersection. It therefore has an uncountable subfamily $\mathscr{V}_1$ with non-empty intersection $F_1$. As before we must have $|F_1|<k$, so $\mathscr{U}_2=\{V\setminus F_1:U\in\mathscr{V}_1\}$ is an uncountable family of open sets with empty intersection. Note that $F_1\cap F_0=\varnothing$.
Clearly we can repeat the construction to produce for each $n\in\omega$ uncountable families $\mathscr{U}_n$ and $\mathscr{V}_n$ of open sets and a subset $F_n$ of $X$ such that:


*

*$\mathscr{V}_n\subseteq\mathscr{U}_n$:  

*$F_n=\bigcap\mathscr{V}_n\ne\varnothing$, so $1\le|F_n|<k$; and  

*$\mathscr{U}_{n+1}=\{V\setminus F_n:V\in\mathscr{V}_n\}$.


It follows that
$\quad(1)$ $\bigcap\mathscr{U}=\varnothing$ for all $n>0$,
$\quad(2)$ the sets $F_n$ are pairwise disjoint, and
$\quad(3)$ if $U\in\mathscr{U}_{n+1}$ for some $n$, then $U\cup\bigcup_{k\le n}F_k\in\mathscr{U}_0\subseteq\mathscr{B}$.
Let $F=\bigcup_{n<k}F_n$ and $\mathscr{W}=\{U\cup F:U\in\mathscr{U}_k\}$; $(3)$ implies that $\mathscr{W}$ is an uncountable subset of $\mathscr{B}$. But $\bigcap\mathscr{W}=F\cup\bigcap\mathscr{U}_k=F$ by $(1)$, and $(2)$ implies that $|F|\ge k$, contradicting the hypothesis that $\mathscr{B}$ is $k$-in-countable. Thus, $\mathscr{B}$ must be countable, and $X$ is second countable.
