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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?

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I really don't see the connection to [set-theory] here. What sort of structure do you wish to retain from/in $X$? – Asaf Karagila Aug 14 '11 at 8:47
I echo Asaf's comments. For example, for simple cardinality reasons, $\mathbb{R} \times \mathbb{R}$ is in bijection with $\mathbb{R}$... and if you want to think of $\mathbb{R}$ as a concrete set in some axiomatic set theory, you need to be more specific. – Zhen Lin Aug 14 '11 at 8:49
Your edit doesn't help. That's just the definition of cartesian product. Do you want to view $\mathbb R$ as just a set? as a topological space? As a smooth manifold? As a topological group? Since $\mathbb R$ is a particularly nice space having many different structures, there are lots of different categories in which you could ask for a square root. – Aaron Aug 14 '11 at 8:54
Re: last question. Impossible. Since $\mathbb{R}$ is path connected so is $X \times X$, hence $X$ would be path connected, so $X \times X \smallsetminus \{pt\}$ would be path connected, but $\mathbb{R} \smallsetminus \{pt\}$ isn't path connected. See also this MO-thread and this related one. – t.b. Aug 14 '11 at 8:57
@Theo: could you please put your comment as an answer? – Ilya Aug 14 '11 at 9:16
up vote 16 down vote accepted

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 $\mathbb{R} \to \mathbb{R} \times \mathbb{R}$.

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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.

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Impressive argument! Very nice! – Asaf Karagila Aug 14 '11 at 21:56
Do I understand correctly that the spread of a space is the minimal cardinality $\kappa$ such that a discrete subspace must have cardinality $\leq \kappa$? (I never heard that term before) – t.b. Aug 14 '11 at 22:04
@Theo: Essentially: $s(X)=\sup\{|D|:D\subseteq X\text{ is relatively discrete}\}+\omega$. If you take the supremum only over closed, discrete subsets, you get the extent, $e(X)$. The classic reference is I. Juhász, Cardinal Functions in Topology, Math. Centre Tracts 34, Amsterdam, 1971; it’s a bit dated $-$ it doesn’t cover $e(X)$, for instance $-$ but it’s still a valuable reference. – Brian M. Scott Aug 14 '11 at 22:35
@Brian: Thanks for the explanation and the reference. Good to know! The argument for the Sorgenfrey line is very nice. The just for fun part is also nice, but I'm glad that you labeled it just for fun. – t.b. Aug 14 '11 at 22:54
Just noticed that haven't upvoted your answer previously. It's a good occasion to do that now :) – Ilya Mar 9 '15 at 10:50

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.

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Theo: Done; compare with a few months ago, tho. – gary Aug 14 '11 at 17:22

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