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.

Are $\mathbb{R}-\mathbb{Q}$ and $(\mathbb{R}-\mathbb{Q})\cap (0,1)$ homeomorphic? My claim is they are and I'm trying using this function:$$f:(\mathbb{R}-\mathbb{Q})\cap (0,1) \rightarrow (\mathbb{R}-\mathbb{Q})\cap (0,\infty)\;\;\;\; \;f(x)=\frac{1}{x}-1$$ which is a restriction of $g=1/x-1$. Proven this, then it would be easy to prove it for ($-\infty$,$+\infty$). So I think I now need to show that $f$ is well defined, which is true because $g$ transform rational numbers into rational and irrational into irrational. So $f$ is well defined, it's bijective, but is it continuous in the subspace topology? I believe it is using the same argument I exposed two lines above. Is my claim false, and/or the proof?

share|improve this question
You probably mean $g=1/x-1$? –  tomasz Jul 15 '12 at 13:51
add comment

1 Answer 1

up vote 4 down vote accepted

Yes, they are homeomorphic. They are both homeomorphic to the Baire space $\omega^\omega$ of all sequences of natural numbers, which is a classical result in descriptive set theory.

Your argument seems correct. Continuity follows from the fact that it is a restriction of a rational function (and rational functions are continuous where defined), and rational functions with rational coefficients preserve rationality. As the inverse of $f$ (that is, $1/(y+1)$) is well-defined and clearly continuous and preserves rationality (implying $f$ preserves irrationality), it is enough.


I just noticed that you intended to show homeomorphism with $\mathbf R\setminus \mathbf Q$ and not $\mathbf R_{>0}\setminus \mathbf Q$.

In this case you should extend your argument a little, like so for example: $(0,1)\setminus \mathbf Q$ is easily homeomorphic with $(-1,1)\setminus \mathbf Q$ (by $h(x)=2x-1$), and then $f$ defined in the same way for positive numbers and separately as $-f(-x)$ for negative numbers yield a homeomorphism onto $\mathbf R\setminus\mathbf Q$. Continuity is still not hard to see.

share|improve this answer
A dense-codense $G_\delta$ in a Polish space is not necessarily homeomorphic to Baire space (it is true for subsets of the real line, however). Take $\mathbf{R} \times (\mathbf{R \smallsetminus Q})$ in $\mathbf{R}^2$, for example. –  t.b. Jul 15 '12 at 14:49
@ tomasz for the edit. Yes it's true, but I too stated it in my question. Knowing that rational functions behave that way, refining the answer is immediate. Thanks. @ t.b. @tomasz; I don't still know what are Baire spaces, it was not part of our course, but I'll try reading something about them. –  Temitope.A Jul 15 '12 at 14:55
@t.b: Right, I think the result was about zero-dimensional Polish spaces. –  tomasz Jul 15 '12 at 15:23
Ah, I see: you're having Mazurkiewicz's theorem in mind: A zero-dimensional dense and codense Polish subspace $X$ of a Hausdorff space is homeomorphic to Baire space. In $\mathbf{R}$ zero-dimensionality of a dense-codense subset is automatic because $(a,b) \cap X$ with $a,b \in \mathbf{R} \smallsetminus X$ is a clopen base of $X$. –  t.b. Jul 15 '12 at 15:41
@Temitope.A: tomasz is talking about the Baire space $\omega^\omega$. For the sake of confusion there are also Baire spaces in topology: those spaces in which the Baire category theorem holds. –  t.b. Jul 15 '12 at 15:45
add comment

Your Answer


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.