If $\sigma(\mathcal B)=\sigma(\mathcal T)$ for every basis $\mathcal B$ of topological space $(X,\mathcal T)$, must $T$ be hereditarily Lindelöf? In the comments of this answer, someone claims:

If topological space $(X,\mathcal T)$ is hereditarily Lindelöf, then $\sigma(\mathcal B)=\sigma(\mathcal T)$ for every basis $\mathcal B$.

Can this theorem be strengthened to read "if and only if"?

To clarify after some discussion in the comments, I'm introducing this terminology:
A basis $\mathcal B$ for topology $\mathcal T$ has the "Borel property" if $\sigma(\mathcal B)=\sigma(\mathcal T)$.
A topology $\mathcal T$ has the "strong Borel property" if every basis for the topology has the Borel property.
The conjecture then becomes: "A topological space $(X,\mathcal T)$ is hereditarily Lindelöf if and only if $\mathcal T$ has the strong Borel property."
 A: I think the reverse implication is false. To show this, consider the following counterexample of a topology with the “strong Borel property” that fails to be Lindelöf, let alone hereditarily so.

Let $(X,\geq)$ be the set of countable ordinals and introduce the following notation for the (strict) lower segment corresponding to a generic $x\in X$:
\begin{align*}
L_x\equiv\{y\in X\,|\,y<x\}.
\end{align*}
Recall that the set of countable ordinals has the following properties:

*

*$X$ is uncountable;

*$\geq$ is a total (complete, antisymmetric, and transitive) order on it;

*$L_x$ is countable for every $x\in X$; and

*every non-empty subset of $X$ has a least element according to $\geq$.


Consider the following family of sets:
\begin{align*}
\tau\equiv\{\varnothing,X\}\cup\{L_x\,|\,x\in X\}.
\end{align*}
This is a topology, because any union of lower segments is either a lower segment or the whole set $X$ itself (you can check this by exploiting the order structure of $(X,\geq)$), and the intersection of two lower segments is also a lower segment.

Let $\mathscr B$ be any topological basis for $\tau$. For any $x\in X$, the open set $L_x$ can be expressed as a union of basic open sets, which must be lower segments themselves. Formally, there exists some $W_x\subseteq X$ such that

*

*$L_w\in\mathscr B$ for every $w\in W_x$; and

*$L_x=\bigcup_{w\in W_x}L_w.$
But then every $w\in W_x$ must satisfy $w\leq x$, so that $W_x\subseteq L_x\cup\{x\}$. Consequently, $W_x$ is countable. Therefore, every lower segment is a countable union of basic open sets, which implies that $\tau\subseteq\sigma(\mathscr B)$. (Note that the empty set $\varnothing$ and the whole set $X$ are trivially included in $\sigma(\mathscr B)$). It readily follows that $\sigma(\tau)=\sigma(\mathscr B)$ for any topological basis $\mathscr B$, so the topological space $(X,\tau)$ has the strong Borel property.

Yet, it cannot be Lindelöf, since
\begin{align*}
X=\bigcup_{x\in X}L_x,
\end{align*}
and the open cover $\{L_x\}_{x\in X}$ has no countable subcover (given that a countable union of countable sets is countable).
