Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Does there exist a continuous bijection $\mathbb{Q}\to \mathbb{Q}\times \mathbb{Q}$?

I am not able to find out how to proceed.

share|improve this question
    
related to mathoverflow.net/questions/21003/… –  Ramanujan Aug 26 '13 at 8:24

2 Answers 2

up vote 10 down vote accepted

Every countable metric space without isolated points is homeomorphic to the rational numbers. In particular $\Bbb{Q\times Q}$.

This is a theorem of Sierpinski, and you can find the details here: http://at.yorku.ca/p/a/c/a/25.htm

share|improve this answer

Not only do such a maps exist, it is possible to construct explicit instances. Here is a sketch of how this can be done.

Let $\pi$ be an irrational real. For every $n \in \mathbb{Z}$ the set $\mathbb{Q} \cap (n+ \pi, n+\pi+1)$ is both open in $\mathbb{Q}$ and homeomorphic to $\mathbb{Q}$. A homeomorphism can be constructed by taking two rational sequences, one increasing and one decreasing to $\pi$ and constructing a piecewise linear, increasing function. Thus we obtain a homeomorpism $f: \mathbb{Q} \to \mathbb{Z} \times \mathbb{Q}$.

It is also possible to obtain an explicit bijection $g: \mathbb{Z} \to \mathbb{Q}$, which is always continuous, since $\mathbb{Z}$ is discrete. If we then put $h_1(x) = g(f_1(x))$ and $h_2(x) = f_2(x)$ then $h: \mathbb{Q} \to \mathbb{Q} \times \mathbb{Q}$ is the desired continous bijection.

share|improve this answer

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.