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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."

WillG
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  • The result that yada mentioned actually required that $X$ be hereditarily Lindelöf. The example in the first comment there — a strong limit cardinal of countable cofinality in its order topology — is not hereditarily Lindelöf. – Brian M. Scott Sep 17 '20 at 17:15
  • @BrianM.Scott Thanks, I updated my question to read "hereditary Lindelöf." – WillG Sep 17 '20 at 17:54
  • I don't know enough to understand the example. Is it a counterexample to my question? I.e., is it a space that is not hereditary Lindelöf, but does satisfy $\sigma(\mathcal B)=\sigma(\mathcal T)$ for any basis $\mathcal B$? – WillG Sep 17 '20 at 17:58
  • Yes, it’s a counterexample: it’s not hereditarily Lindelöf, since it has the non-Lindelöf subspace $\omega_1$, but every open set is the union of countably many basic open sets for a suitable choice of base. – Brian M. Scott Sep 17 '20 at 18:01
  • My conjecture: Hereditarily Lindelöf $\Leftrightarrow$ Every basis has the Borel property. Suppose a space in NOT hereditarily Lindelöf and has one basis that satisfies the Borel property. This is consistent with the conjecture. – WillG Sep 17 '20 at 18:40
  • Ah, you’re right; my apologies. I’m pretty sure that the usual base of intervals does not have the property, though I’ve not actually gone through the argument. – Brian M. Scott Sep 17 '20 at 18:41

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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).

triple_sec
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