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Suppose $S\subset\mathbb{R}$ is dense without interior point, and for every open interval $I,J\subset\mathbb{R}$, $I\cap S$ is homeomorphic to $J\cap S$.

Is $S\times S$ homeomorphic to $S$?

By Luzin scheme, if $S$ is the set of rationals or irationals , I can see this statement is true.

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Dear @user35739, your question is probably better suited at math.stackexchange.com, see the FAQ of this site. –  András Bátkai Jun 26 '13 at 7:48
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@Andras: would you explain why? –  Wlodzimierz Holsztynski Jun 26 '13 at 9:41
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@user35739: I would first of all wonder about a full characterization of the spaces $S$ with the property of having all (relative) open intervals homeomorphic one to another. There is no need a priori to mention anything about $S$ being dense or having empty interior (it's that elegant :-). –  Wlodzimierz Holsztynski Jun 26 '13 at 9:46
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The question seems to be well within the scope of MO, compare mathoverflow.net/questions/26001/… with an answer containing a reference to several proofs of Sierpinsky's theorem, that provides a partial answer to the OP question. –  Misha Jun 26 '13 at 11:08
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Also, for every $G_{\delta}$-subset $S$ of ${\mathbb R}$ which is dense and has empty interior, one has $S\cong S\times S$. The reason is that such $S$ is a Polish space and, hence, Alexandrov-Uryson theorem implies that both $S$ and $S\times S$ are homeomorphic to ${\mathbb R}\setminus {\mathbb Q}$. –  Misha Jun 26 '13 at 11:21
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up vote 5 down vote accepted

In the following paper, van Engelen constructs a strongly homogeneous, zero dimensional, Borel, dense subspace $Y$ of $2^{\mathbb{N}}$ such that while $Y$ does not admit a topological group structure, $Y^2$ does. I haven't read the proof but maybe you can use it to construct a counterexample in the context of $\mathbb{R}$?

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