Can a space $X$ be homeomorphic to its twofold product with itself, $X \times X$?

Question

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.

Other common nontrivial examples include $\mathbb Q$ and the Cantor set. — , Aug 04 '15 at 18:50
@MikeMiller What is the argument for $\Bbb Q$, say? At a glance this seems nonobvious, assuming that $\Bbb Q$ is given the subspace topology induced by its inclusion into $\Bbb R$. — Travis Willse, Aug 04 '15 at 20:55
@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. — , Aug 04 '15 at 20:57
@MikeMiller Thanks, that's quite a nice result. A little searching yields this well-sourced answer: http://math.stackexchange.com/a/56536/155629 . Sierpinski's original paper (in French) is http://matwbn.icm.edu.pl/ksiazki/fm/fm1/fm113.pdf . — Travis Willse, Aug 04 '15 at 21:02
(@MikeMiller FWIW, that post also includes metrizability among the hypotheses of Sierpinski's Theorem.) — Travis Willse, Aug 04 '15 at 21:04
@Travis: Ah, fair point. You need some decent hypothesis since it's obviously false if you allow non-Hausdorff spaces. Thanks for the correction. — , Aug 04 '15 at 21:05
See also: Homeomorphism between Space and Product, Homeomorphism between topological space and product space, — Martin Sleziak, Nov 19 '16 at 17:18

Travis Willse · Answer 1 · 2015-08-04T20:19:27.140

10

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.

edited Aug 04 '15 at 20:19

answered Aug 04 '15 at 18:49

Travis Willse

108,056

Here, well defined means that homeomorphic topological manifolds have the same dimension? – Augustin Aug 04 '15 at 18:56
2

@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$. – Aug 04 '15 at 18:59

score 5 · Accepted Answer · answered Aug 04 '15 at 18:50

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.

score 4 · Answer 3 · answered Aug 04 '15 at 18:52

4

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.

answered Aug 04 '15 at 18:52

Tomasz Kania

16,996

Can a space $X$ be homeomorphic to its twofold product with itself, $X \times X$?

3 Answers3

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