$\mathbb R = X^2$ as a Cartesian product

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?

There can be no such square root decomposition as in the last part of your question: Suppose $\mathbb{R}$ is homeomorphic to $X \times X$. Then $X \times X$ is path-connected and this implies that $X$ is path connected (by projecting down). But then $X \times X \smallsetminus \{\textrm{pt}\}$ is also path-connected in contradiction to the fact that $\mathbb{R} \smallsetminus\{\textrm{pt}\}$ is not even connected.

This argument appears e.g. in this MO-thread by Richard Dore. A nice follow-up question with a somewhat surprising answer (which I don't understand in detail but I'm inclined to believe Ryan Budney) is this MO-thread.

Concerning the decomposition of $\mathbb{R}$ into the cartesian product of one set with itself, this is easy for cardinality reasons: $\mathbb{R}$ and $\mathbb{R} \times \mathbb{R}$ have the same cardinality, so there is a bijection $\mathbb{R} \to \mathbb{R} \times \mathbb{R}$.

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.

For the sake of completeness, I will give an argument, given originally by R. Fokkink (see also here), to show that $\mathbb R^{2n+1}$ is not a perfect square for any positive natural $n$; the argument is given only for $n=1$, i.e., this argument shows that $\mathbb R^3$ is not homeo to $X^2$ for any topological space $X$, but it generalizes nicely, to the general case of $\mathbb R^{2n+1}$ being a perfect square.

By contradiction, assume that there is an actual topological space X such that there is a homeo. between $\mathbb R^3$ and $X^2$.

We now then use the fact that homeomorphisms are either orientation-preserving or orientation-reversing (since $\mathbb R$ is orientable, it makes sense to talk about orientability of X). Associated with the concept of orientability is the concept of the degree of a map, so that, if a homeo. $h$ is orientation-preserving, we assign to $h$ a degree $1$, and if $h$ changes orientation, then we assign a degree of $-1$ to $h$.

This concept of degree satisfies the nice property that, given two homeomorphisms $f,g$, then $\deg(f\circ g)=(\deg f)\times(\deg g)$, so that, in particular, $\deg(h\circ h)=1$, i.e., a composition of a homeo. with itself is orientable.

Now, assume $h$ is the given homeo. between $\mathbb R^3$ and $X^2$, so that we also have a homeo. , say $h_1$ between $X^4\rightarrow \mathbb R^6$ taking coordinates $(a,b,c,d)\rightarrow (e,f,g,h,i,j).$

We will get a contradiction by using the assumption of the existence of a homeo $h$, and producing an automorphism $\mathbb R^6 \rightarrow \mathbb R^6$ , whose square does not preserve orientation, contradicting the properties of deg given above.

Consider this: the linear automorphism $L:X^4\rightarrow X^4$ with: $(a,b,c,d)\rightarrow (d,c,b,a)$, so that the composition $L\circ L$ preserves orientation (and I guess it must somehow be funny too ), with $L \circ L:(a,b,c,d\rightarrow (c,d,a,b)$.

Now, we can pullback this linear automorphism $L\circ L $ of $X^4$ to an automorphism of $\mathbb R^6$ using the homeomorphism $h_1$ above, and we would end up with the auto $L\circ L':(e,f,g,h,i,j)\rightarrow (i,j,e,f,g,h)$: But this last is a linear map, and its determinant is $-1$, so the assumption of the existence of a homeo. $h: \mathbb R^3\rightarrow X^2$we constructed a self-homeomorphism of $\mathbb R^6$ whose square is not orientation-preserving.