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Let $X$ be a topological space of infinite cardinality. Is it possible for any $X$ to be homeomorphic to $X\times X$ $?$

For example, $\mathbb R$ is not homeomorphic to $\mathbb R^{2}$, and $S^{1}$ is not homeomorphic to $S^{1} \times S^{1}$ . What other topological spaces might we consider$?$ What properties of a space may ensure or contradict this possibility$?$ From the little topology I have learnt yet, I have not seen this happening.

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    $\begingroup$ $X = Y^{\mathbb{N}}$ $\endgroup$ – Daniel Fischer Aug 4 '15 at 18:46
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    $\begingroup$ Other common nontrivial examples include $\mathbb Q$ and the Cantor set. $\endgroup$ – user98602 Aug 4 '15 at 18:50
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    $\begingroup$ @Travis: There is a general theorem of Sierpinski: a countable space with no isolated points is homeomorphic to $\Bbb Q$. But I believe you can write down an explicit homeomorphism with continued fraction trickery. I believe this is done somewhere on this site. $\endgroup$ – user98602 Aug 4 '15 at 20:57
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    $\begingroup$ @MikeMiller Thanks, that's quite a nice result. A little searching yields this well-sourced answer: math.stackexchange.com/a/56536/155629 . Sierpinski's original paper (in French) is matwbn.icm.edu.pl/ksiazki/fm/fm1/fm113.pdf . $\endgroup$ – Travis Willse Aug 4 '15 at 21:02
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    $\begingroup$ (@MikeMiller FWIW, that post also includes metrizability among the hypotheses of Sierpinski's Theorem.) $\endgroup$ – Travis Willse Aug 4 '15 at 21:04
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At this level of generality you can make $X=X \times X$ happen quite easily. Take a discrete space of any infinite cardinality, for instance. Or topologize $X=A^B$ by whatever means and compare $X \times X = A^{B \sqcup B}$; under various mild assumptions on $B$ those spaces would be homeomorphic.

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Yes, consider $X := \Bbb Z$ endowed with the discrete topology.

For any topological manifold $M$, $\dim (M \times M) = \dim M + \dim M = 2 \dim M$. Since the dimension of a nonempty topological manifold is well-defined, there is no positive-dimensional topological manifold $M$ for which $M \cong M \times M$, which in particular excludes $R$ and $S^1$ as observed. This implies that the example $X = \Bbb Z$ is the only example that is a (second countable) topological manifold.

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  • $\begingroup$ Here, well defined means that homeomorphic topological manifolds have the same dimension? $\endgroup$ – Augustin Aug 4 '15 at 18:56
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    $\begingroup$ @Augustin: The definition of dimension of a topological manifold is the unique integer $n$ such that $M$ is locally homeomorphic to $\Bbb R^n$. That this is well-defined comes down to that $n$ being unique: that is, that $\Bbb R^n \cong \Bbb R^m$ implies $ n=m$. $\endgroup$ – user98602 Aug 4 '15 at 18:59
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Many Banach spaces are linearly homeomorphic to their Cartesian squares. For instance all classical spaces including $c_0$, $\ell_\infty$, $C(K)$ for $K$ compact metric, $L_p(\mu)$ for $p\in [1,\infty]$ etc.

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