Another question on hereditarily lindelöf space Let $X^2$ be hereditarily lindelöf ($X$ is regualr.). How to show the diagonal of $X$ is a $G_\delta$-subset in $X^2$? Thanks ahead:)
 A: Some generalities:
Let $X$ be hereditarily Lindelöf and regular. Let $O \subset X$ be open. Then for every point $x \in O$ find (by regularity) some open subset $V_x$ such that $x \in V_x \subset \overline{V_x} \subset O$. Some countable subset of the $V_x$, say $V_x$ with $x \in C$, $C$ countable, also cover $O$, as $O$ is Lindelöf, so we have that
$$O = \cup_{x \in C} V_x \subset \cup_{x \in C} \overline{V_x} \subset O,$$ so
we have that $O$ is the countable union of closed sets (an $F_\sigma$). Dually we have that every closed subset of $X$ is a $G_\delta$ (taking complements and using de Morgan).  Furthermore, $X$ being Lindelöf and regular implies that it is normal, as shown for example here. This shows that $X$ is perfectly normal.
So for a regular Lindelöf space it is equivalent to be hereditarily Lindelöf or to be perfectly normal. (If an open subset is an $F_\sigma$ in a Lindelöf space it is $\sigma$-Lindelöf and thus Lindelöf too, and this suffices for being hereditarily Lindelöf.)
Now $X$ is closed (Hausdorffness!) in $X^2$ and so if the latter space is hereditarily Lindelöf, $X$ is a $G_\delta$ in its square.
