A topological space $X$ is a $Q$-space if every subset of $X$ is of type $G_\delta$.
The smallest cardinality of a metrizable separable space which is not a $Q$-space is denoted by $\mathfrak q_0$ and is a well-studied cardinal characteristic of the continuum.
It is known that $\mathfrak p\le\mathfrak q_0\le\mathfrak b$. The standard proof of the inequality $\mathfrak p\le\mathfrak q_0$ gives a bit more:
Any second-countable $T_1$-space of cardinality $<\mathfrak p$ is a $Q$-space.
This means that $\mathfrak p\le\mathfrak q_1\le\mathfrak q_0$, where $\mathfrak q_1$ is the smallest cardinality of a second-countable $T_1$-space which is not a $Q$-space.
Question 1. Is $\mathfrak q_1=\mathfrak q_0$?
And what about the cardinal $\mathfrak q_2$ defined as the smallest cardinality of a second-countable Hausdorff space which is not a $Q$-space?
Question 2. Is $\mathfrak q_2=\mathfrak q_0$? Or maybe $\mathfrak q_2=\mathfrak q_1$?
Remark. It can be shown that $\mathfrak q_0$ is equal to the smallest cardinality of a second-countable functionally Hausdorff space which is not a $Q$-space. A topological space $X$ is functionally Hausdorff if for any distinct points $x,y\in X$ there exists a continuous function $f:X\to\mathbb R$ with $f(x)\ne f(y)$. It can be shown that every second-countable functionally Hausdorff space is submetrizable and every submetrizable space of cardinality $<\mathfrak q_0$ is a $Q$-space.
