Here [A Question on a second countable $T\_2$ space](https://math.stackexchange.com/questions/1215409/a-question-on-a-second-countable-t-2-space), Paul asked if every second countable Hausdorff space has a $G\_\delta$-diagonal. In the comments Brian M. Scott answered that, at the time (2015), the answer to that question might not be known. Interestingly enough, here [Does second countable and functionally Hausdorff imply submetrizable?](https://mathoverflow.net/questions/280359/does-second-countable-and-functionally-hausdorff-imply-submetrizable), Taras Banakh showed two years later that, as a consequence of a more general result, every second countable functionally Hausdorff space is submetrizable (in particular, every such space has a $G\_\delta$-diagonal).


My question is whether the answer to Paul's original question is already known. I couldn't find anything definitive about it.



Second countable vs. $G_\delta$-diagonal
There exists a counterexample to this question of Paul.


It suffices to find a second-countable Hausdorff space $X$ that has two properties:


(1) the space $X\times X$ is Baire;


(2) for any nonempty open sets $U,V\subseteq X$ we have $\overline U\cap\overline V\ne\emptyset$.


Such a space $X$ cannot have $G\_\delta$-diagonal. Indeed, assuming that the diagonal $\Delta\_X$ is of type $G\_\delta$ in $X\times X$, we can write $(X\times X)\setminus \Delta\_X$ as the union $\bigcup\_{n\in\omega}F\_n$ of closed subsets $F\_n$ of $X\times X$. Property (2) implies that $X$ has no isolated points and hence the diagonal $\Delta\_X$ is nowhere dense in $X\times X$. Since the space $X\times X$ is Baire, there exists $n\in\omega$ such that $F\_n$ has non-empty interior in $X\times X$. Then there are nonempty open sets $U,V$ in $X$ such that $U\times V\subseteq F\_n$ and hence $\overline U\times\overline V\subseteq F\_n$. Then $(\overline U\times \overline V)\cap\Delta\_X=\emptyset$ and hence $\overline U\cap\overline V=\emptyset$, which contradicts the property (2).


For the space $X$ one can take [the projective space of the countable product of lines](https://mathoverflow.net/q/46986/61536) $\mathbb R^\omega$. So, $X$ is the quotient space of $\mathbb R^\omega\setminus\{0\}^\omega$ by the equivalence relation $x\sim y$ iff $\mathbb Rx=\mathbb Ry$. Observe that the quotient map $q:\mathbb R^\omega\setminus\{0\}^\omega\to X$ is open. This fact can be used to show that $X$ is Hausdorff and its square $X\times X$ is Baire. Moreover, for any nonempty open set $U\subseteq X$ the closure $\overline{U}$ contains the image $q[(\{0\}^n\times \mathbb R^{\omega\setminus n})\setminus\{0\}^\omega]$ for some $n\in\omega$, which implies that the condition (2) is satisfied.


