In $\mathbb{R}$ with the usual topology is considered the following equivalence relation $$x \mathfrak{R} y \; \text{if and only if}\ x,y\in \mathbb{Q}\ (\text{or}\ x=y)$$ $\mathfrak{S}$ $\mathfrak{R}$ On the other hand, in $\mathbb{R}^2$ with the usual topology, is defined $$(x,y)\mathfrak{S} (x',y')\; \text{if and only if}\ (x,y),(x',y')\in \mathbb{Q}^2 \; ( \text{or}\ (x,y)=(x',y'))$$ Are $\mathbb{R}^2 / \mathfrak{S}$ and $\mathbb{R}/ \mathfrak{R} \times \mathbb{R} / \mathfrak{R}$ homeomorphic? More generally, can $\mathbb{R}^2/\mathfrak{S}$ be homeomorphic to a product space?
I've been trying to picture how the quotient spaces look like. All points in $\mathbb{Q}$ are one point in $\mathbb{R}/ \mathfrak{R}$, so if I take $x\in \mathbb{Q}$, then
$$[x]=\{ y\in \mathbb{R} \mid x\mathfrak{R} y\}=\{y\in \mathbb{R} \mid y\in \mathbb{Q} \}=\mathbb{Q}$$ and if $x \not \in \mathbb{Q}$, then $$[x]=\{x\}$$ and for the other equivalence relation $\mathfrak{S}$ it should work the same way, but I don't know how to prove or disprove whether the two quotient spaces are homeomorphic or not. Thanks in advance!
In $ℝ^2/\mathfrak{S}$ the point rational×rational is dense while all the other points are closed.
In $(ℝ/\mathfrak{R})^2$ the point rational×rational is dense, the points irrational×irrational are closed, and the points rational×irrational and irrational×rational have nowhere-dense closure of size continuum.