TITLE: $\mathbb R = X^2$ as a Cartesian product
QUESTION [11 upvotes]: I wonder if it is possible to consider $\mathbb R$ as a Cartesian product $X\times X$ for some set $X$. From the point of view of the dimensionality, there are spaces with a Hausdorff dimension $1/2$ (sort of Cantor sets), but I guess there are other problems in this construction  unrelated to the dimension.
Edited: replying on the comment by Asaf. I want that for all $r\in\mathbb R$ there exists a unique representation $r=\langle x,y\rangle$ where $x\in X$ and $y\in X$. Also, what if we want to have $X$ to be a topological space and $h:X\times X\to\mathbb R$ to be a homeomorphism?

REPLY [6 votes]: Just for fun, here’s a completely different argument for the last question, one that avoids any direct appeal to path connectedness or connectedness.
Let $\Delta$ be the diagonal in $X \times X$, and let $D = h[\Delta]$, where $h:X \times X \to \mathbb{R}$ is a homeomorphism. It’s easy to check that $\Delta$ is a closed, nowhere dense, dense-in-itself, non-compact subset of $X \times X$, so $D$ is a closed, nowhere dense, dense-in-itself, non-compact subset of $\mathbb{R}$.
For $n \in \mathbb{Z}$ choose $r_n \in \left(n-\frac14,n+\frac14 \right) \setminus D$, and let $D_n = D \cap (r_{n-1},r_n)$. Clearly each $D_n$ is either empty or a Cantor set. Since $D$ must be unbounded, this shows that $D$ is homeomorphic to $\omega \times 2^\omega$ (where $2^\omega$ is the product of discrete $2$-point spaces, not the ordinal). But then $D \cong D \times D$, so $\mathbb{R} \cong X \times X \cong \Delta \times \Delta \cong D \times D \cong D \cong \omega \times 2^\omega$. This makes $\mathbb{R}$ itself a countable union of Cantor sets, contradicting the fact that $\mathbb{R}$ is a Baire space.
By the way, the Sorgenfrey line $\mathbb{S}$ is also not a square. If $\mathbb{S} \cong X \times X$, then $X$ is homeomorphic to an uncountable $F \subseteq \mathbb{S}$. But then $\mathbb{S} \cong F \times F$, so $\mathbb{S}$ has an uncountable, closed, discrete subset (via the anti-diagonal in $F \times F$), contradicting the fact that $\mathbb{S}$ has countable spread. (It’s also an immediate consequence of Theorem 2.1 of Burke & Moore, Subspaces of the Sorgenfrey Line: If $X_0,X_1,\dots,X_n$ are uncountable subspaces of the Sorgenfrey line $\mathbb{S}$, then $\prod_{i=0}^nX_i$ does not embed in $\mathbb{S}^n$.)
Added 8 March 2015: Here’s yet another argument, a different way of using connectedness. Clearly $X$ must be connected, and $|X|=|\Bbb R|$. But then $\left\{h\big[\{x\}\times X\big]:x\in X\right\}$ is a partition of $\Bbb R$ into uncountably many non-trivial closed, connected sets, which is impossible, since every such set contains an open interval.