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Suppose that $X$ is the space with a point-countable base and having calibre $\aleph_1$. Must $X$ be second countable?

Definition. A topological space $X$ has calibre $\aleph_1$ if for every uncountable family $\{ U_\alpha : \alpha\lt\omega_1\}$ of nonempty open subsets of $X$ there is an uncountable $\Lambda\subset\omega_1$ with $\bigcap_{\alpha\in\Lambda}U_\alpha\neq\emptyset$.

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up vote 3 down vote accepted

If $X$ has $\omega_1$ as a calibre, then any point-countable family of nonempty open subsets of $X$ must be countable. (If $\mathcal{U}$ were point-countable and uncountable, then $\bigcap \mathcal{U}_0 = \varnothing$ for every uncountable $\mathcal{U}_0 \subseteq \mathcal{U}$, contradicting that $X$ has $\aleph_1$ as a calibre.)

Therefore any point-countable base for $X$ must be countable, witnessing the second-countability of $X$.

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