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Can $\mathbb{R}^n$ be written as $X \times X$ for some topological space $X$? This is obviously true if $n$ is even. Take $X = \mathbb{R}^{n/2}$. However, I'm unsure about $n$ odd.

user141380
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    See: http://mathoverflow.net/questions/60375/is-r3-the-square-of-some-topological-space – Aldo Guzmán Sáenz Apr 07 '14 at 19:36
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    That MO thread does answer the general question. – Ayman Hourieh Apr 07 '14 at 19:37
  • @AymanHourieh True. I have edited my comment. – Aldo Guzmán Sáenz Apr 07 '14 at 19:40
  • I believe that this isn't true. Why? Let $(x,y)\in X\times X$. Then, let $(X,Y,Z,...,X^{2n+1})\in\mathbb{R}^{2n+1}$. We would expect that a point $(x,y)=(X,Y,Z,...,X^{2n+1})$. Note the discrepancy here. There are an odd number of elements on the RHS, however, there are two, i.e., even number of elements on the LHS. –  Apr 07 '14 at 20:21

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I believe that this isn't true. Why? Let $(x,y)\in X\times X$. Then, let $(X,Y,Z,...,X^{2n+1})\in\mathbb{R}^{2n+1}$. We would expect that a point $(x,y)=(X,Y,Z,...,X^{2n+1})$. Note the discrepancy here. There are an odd number of elements on the RHS, however, there are two, i.e., even number of elements on the LHS.

The above statement is right. Let me prove it formally. Let $\mathcal{H}:X\times X\to\mathbb{R}^{2n+1}$ be a homeomorphism. Then, there exists a homeomorphism $X\times X\times X\times X\to\mathbb{R}^{4n+2}$. There exists a map $h:X\times X\times X\times X\to X\times X\times X\times X$ such that $h\circ h$ is orientation preserving. You can show, that when $h\circ h$ is acted upon $\mathbb{R}^{4n+2}$, the map is non-orientation preserving. Hence, $X\times X\times X\times X$ is not homeomorphic to $\mathbb{R}^{4n+2}$, and hence, $X\times X$ is not homeomorphic to $\mathbb{R}^{2n+1}$ either. Q.E.D.