A topological space is said to be quasi-Polish if it is second-countable and completely quasi-metrizable (see for an introduction de Brecht's article: *de Brecht, Matthew*, [**Quasi-Polish spaces**](http://dx.doi.org/10.1016/j.apal.2012.11.001), Ann. Pure Appl. Logic 164, No. 3, 356-381 (2013). [ZBL1270.03086](https://zbmath.org/?q=an:1270.03086).) ).


These spaces have been introduced because, among other things, they allow the generalization of some results of classical descriptive set theory in non-Hausdorff environments. Now my question is:


* An Hausdorff quasi-Polish space needs to be Polish? Do we have a counter-example of an Hausdorff quasi-Polish space which is *not* Polish?


A strictly related [question](https://mathoverflow.net/questions/221405/hausdorff-open-image-of-a-polish-space), which asks basically the same thing, but is a bit old, so maybe there are new results that came up since then. Moreover, as the OP's comment points out, the question contained in Tkachuk's book is assuming all spaced Tychonoff and I'm not.


Thanks!



Hausdorff quasi-Polish spaces
I hope that the following space $P\mathbb Q^\omega$ is second-countable and quasi-Polish but not Polish.


Let $\mathbb Q$ be the field of rational numbners endowed with the discrete topology. Then its countable power $\mathbb Q^\omega$ is a Polish vector space over the field $\mathbb Q$. Let $P\mathbb Q^\omega$ be the projective space of $\mathbb Q$. So, $P\mathbb Q^\omega$ is the quotient space of $\mathbb Q^\omega\_\circ=\mathbb Q^\omega\setminus\{0\}^\omega$ by the equivalence relation $\sim$ defined as $x\sim y$ iff $\mathbb Qx=\mathbb Qy$. Is is easy to see that the quotient map $q:\mathbb Q^\omega\_\circ\to P\mathbb Q^\omega$ is open, which implies that the space $P\mathbb Q^\omega$ is second-countable. It is known that the space $P\mathbb Q^\omega$ is Hausdorff but not Urysohn (moreover, $P\mathbb Q^\omega$ is superconnected in the sense that for any nonempty open sets $U\_1,\dots,U\_n$ in $P\mathbb Q^\omega$ the intersection $\bar U\_1\cap\dots\cap\bar U\_n$ is not empty). Therefore, $P\mathbb Q^\omega$ is not metrizable and hence not Polish.


Now we show that $P\mathbb Q^\omega$ is quasi-Polish. Let $\mathbb Q^{<\omega}\_\circ=\bigcup\_{n\in\omega}(\mathbb Q^n\setminus\{0\}^n)$ be the countable set of all finite nonzero sequences of rational numbers and let $\{s\_n\}\_{n\in\omega}=\mathbb Q^{<\omega}\_\circ$ be an enumeration of the set $\mathbb Q^{<\omega}\_\circ$. For every sequence $s\in \mathbb Q^{<\omega}\_\circ$ let $\ell(s)$ be its length and ${\uparrow}s=\{x\in\mathbb Q^\omega\_\circ:x{\restriction}\_{\ell(s)}=s\}$.
Let $q[{\uparrow}s]$ be the image of ${\uparrow}s$ under the quotient map $q:\mathbb Q^\omega\_\circ\to P\mathbb Q^\omega$. It follows that $\{q[{\uparrow}s]:s\in\mathbb Q^\omega\_\circ\}$ is a countable base of the topology of $P\mathbb Q^\omega$.


For every $n\in\omega$, consider the quasi-pseudometric
$d\_n:P\mathbb Q^\omega\times P\mathbb Q^\omega\to\{0,1\}$ defined by $d\_n{-1}(1)=q[{\uparrow}s\_n]\times (P\mathbb Q^\omega\setminus q[{\uparrow}s\_n])$.
Finally consider the quasi-metric $d:P\mathbb Q^\omega\times P\mathbb Q^\omega\to[0,1]$ defined by $$d=\sum\_{n\in\omega}\tfrac1{2^{n+1}}d\_n.$$
I hope that the quasi-metric $d$ witnesses that $P\mathbb Q^\omega$ is a quasi-Polish space.


