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I know that every $T_2$ hereditarily Lindelöf space has countable pseudocharacter (each point is a countable intersection of open sets). And, from the same proof idea, I know that every $T_3$ hereditarily Lindelöf space is a $G_{\delta}$ space (each closed set is a countable intersection of open sets). I wanted to find an example of a $T_2$ hereditarily Lindelöf space that isn't a $G_{\delta}$ space, but I haven't found any in $\pi$-Base. $T_3$ is what guarantees, by definition, that given an open set $A$, each point of $A$ has a closed neighborhood contained in $A$. Given a closed set $C$, if the space isn't $T_3$, I doubt that it would be possible, in a general way, to build an open cover of the complement of $C$ verifying that each element of the open cover has its closure disjoint from $C$.

Does anyone know an example of a $T_2$ hereditarily Lindelöf space that isn't a $G_{\delta}$ space?

Almanzoris
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1 Answers1

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Let $(X,\tau)$ be a topological space, $Z \subset X$.

It is easy to see that $\mu = \{U \cup (V\setminus Z): U, V \in \tau\}$ is a topology on $X$, in fact the coarsest topology, which is finer than $\tau$ and such that $Z$ is closed. This is a well-known construction to produce non-regular spaces, see Engelking, General topology, 1.5.6.

It is easy to see that, if $\mathcal B$ is base for $\tau$, then $\{U \cup (V\setminus Z): U, V \in \mathcal B\}$ is a base for $\mu$. Hence, if $\tau$ is second countable, then also $\mu$ is second countable.

If $Z$ is $G_\delta$ with respect to $\mu$, then it is $G_\delta$ with respect to $\tau$.
[$Z = \bigcap_{n \in \mathbb N} (U_n \cup (V_n \setminus Z)) $ implies $Z = \bigcap_{n \in \mathbb N} U_n$.]

Now consider this topology for $\mathbb R$ with the usual topology and $Z = \mathbb Q$. It is second countable, in particular hereditarily Lindelöf, of course, it is $T_2$. Moreover, $\mathbb Q$ is closed, but not $G_\delta$.

Remark
This space is S59 in $\pi$-base. @Steven Clontz: Thank you for pointing out!

Ulli
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