Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
2  
They're homeomorphic. –  Apollo Oct 4 '12 at 13:50
1  
can you explain how? –  nemesiso Oct 4 '12 at 13:55
2  
They're all countable metric spaces with no isolated points. See at.yorku.ca/p/a/c/a/25.htm –  Apollo Oct 4 '12 at 14:12
    
The result mentioned by Apollo above also shows that for any nonempty countable metric $X$, $X \times \mathbb Q$ is homeomorphic to $\mathbb Q$ - This is one of the results I try to keep in mind to remind myself that my intuition regarding topological spaces is not to be trusted! –  Julien Melleray Oct 4 '12 at 15:24
add comment

1 Answer

up vote 7 down vote accepted

Edit: 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.

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.

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

$$\mathbb{Q} \times \mathbb{Q} \cong L \times L \stackrel{f}{\to} I \cong \mathbb{Q}$$

is a homeomorphism $\mathbb{Q} \times \mathbb{Q} \cong \mathbb{Q}$.

share|improve this answer
2  
Am I confused? $f : \mathbb{Q} \times \mathbb{Q} \to \mathbb{Q}(\sqrt{2})$ sending $(r,s)$ to $r+s\sqrt{2}$ is a homeomorphism? Surely you can have $r_n \to \infty$ and $s_n \to -\infty$ in such a way that $r_n+s_n\sqrt{2} \to 0$? –  Gerald Edgar Oct 4 '12 at 14:51
    
Sorry! You are right. I will edit and try to fix this thing. –  Todd Trimble Oct 4 '12 at 15:13
    
I have substantially edited my answer. Apologies again. If there is still an error, then please (nemesiso) unaccept this answer and I will delete it. I really just wanted something semi-constructive. –  Todd Trimble Oct 4 '12 at 15:33
    
Isn't the product topology on QxQ the same as the order topology on QxQ determined by the lexicographic order? This, together with back-and-forth, would do the trick. –  SJR Oct 4 '12 at 15:46
1  
"Forbidding tails of zeros" ... a slight problem. No pair of numbers interleaves to u = 0.40404040404040... despite its being eventually periodic and not having a tail of zeros. –  Gerald Edgar Oct 4 '12 at 19:31
show 7 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.