product spaces of rationals - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T15:38:39Z http://mathoverflow.net/feeds/question/108818 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108818/product-spaces-of-rationals product spaces of rationals nemesiso 2012-10-04T13:43:50Z 2012-10-04T21:02:45Z <p>Let $Q$ follow subspace topology from $R$ Then I think it is true that $Q^n$ and $Q^m$ (with product topology) are not homeomorphic.I also think it will be possible to define "rational" homotopy groups by considering $[0,1]\cap Q$ etc. But I can't find any reference regarding topology of product spaces of rationals in general . So I was wondering if someone can give a sketch of a proof and give some references.</p> http://mathoverflow.net/questions/108818/product-spaces-of-rationals/108821#108821 Answer by Todd Trimble for product spaces of rationals Todd Trimble 2012-10-04T14:20:05Z 2012-10-04T21:02:45Z <p><b>Edit:</b> The following is a second attempt to repair problems in earlier proposed solutions, as pointed out by Gerald Edgar in comments. Hopefully this time I've gotten it right this time. </p> <p>It's classical (by a back-and-forth argument; see for instance A Shorter Model Theory by Hodges) that any two countable dense unbounded linear orders without endpoints are isomorphic. For example the linear order $\mathbb{Q}$ is isomorphic to $L = \mathbb{Z}_{(2), (5)} \cap (0, 1)$, referring here to the integers localized at the primes 2 and 5 (i.e., rationals whose decimal expansions don't have an infinite tail of 9's or 0's -- this is for technical reasons). $L$ and $\mathbb{Q}$ are homeomorphic under their order topologies. </p> <p>Let $f: L \times L \to \mathbb{Q} \cap (0, 1)$ be the map that takes a pair of elements $\alpha, \beta \in L$ and forms a rational number by interleaving their decimal expansions, with the decimal expansion of $\alpha$ appearing in odd places and that of $\beta$ in the even places. Let $I$ be the image of $f$. The map $f: L \times L \to I$ is a homeomorphism, and $I$ is again a countable dense linear order without endpoints, hence homeomorphic to $\mathbb{Q}$ again. Then the evident composite </p> <p>$$\mathbb{Q} \times \mathbb{Q} \cong L \times L \stackrel{f}{\to} I \cong \mathbb{Q}$$ </p> <p>is a homeomorphism $\mathbb{Q} \times \mathbb{Q} \cong \mathbb{Q}$. </p>