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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.

duodaa
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    Dear @user35739, your question is probably better suited at http://math.stackexchange.com/, see the FAQ of this site. – András Bátkai Jun 26 '13 at 07:48
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    @Andras: would you explain why? – Włodzimierz Holsztyński Jun 26 '13 at 09:41
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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 :-). – Włodzimierz Holsztyński Jun 26 '13 at 09:46
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    The question seems to be well within the scope of MO, compare http://mathoverflow.net/questions/26001/are-the-rationals-homeomorphic-to-any-power-of-the-rationals 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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    I wonder if this question is related to the continuum hypothesis. What about sets $S$ which have an intermediate cardinality? – Włodzimierz Holsztyński Jun 26 '13 at 12:15

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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 $\mathbb{R}$?

Ashutosh
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