Let $X$ be a strongly monotonically monolithic space with calibre-$\aleph_1$. Must $X$ be Lindelof?
I know $e(X)=l(X)$ for a strongly monotonically monolithic space. So to prove that $X$ is lindelof, we only need to prove that $X$ has countable extent. However under CH, the claim is true since it is separable (under CH) and hence it is second countable. If CH is not true, is the claim still true?
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 $\bigcap_{\alpha\in\Lambda}U_\alpha\ne\varnothing$.
Let $A$ be a subset of $X$. A family $\mathcal B$ of subsets of $X$ is called an an external base of $A$ in $X$ if all elements of $\mathcal B$ are open in $X$ and for any $x\in A$ and any open set $U$ in $X$ with $x\in U$, there exists some $B\in \mathcal B$ such that $x\in B \subset U$.
We say that a space $X$ is strongly monotonically monolithic if for each $A \subset X$ we can assign an external base $\mathcal{O}(A)$ to the set $\overline{A}$ satisfying the following conditions:
a) $\left| \mathcal{O}(A) \right| \le \max\{|A|,\omega\}$;
b) if $A \subset B $, then $\mathcal{O}(A) \subset \mathcal{O}(B)$;
c) if $\alpha$ is an ordinal and we have a family $\{A_\beta: \beta < \alpha \}$ of subsets of $X$ such that $\beta < \beta' $ implies $A_{\beta} \subset A_{\beta'}$, then $\mathcal{O}(\bigcup_{\beta < \alpha} A_\beta) = \bigcup_{\beta < \alpha} \mathcal{O}(A_\beta)$.
Thanks for your help!