An homeomorphism between $\mathbb{R}-\mathbb{Q}$ and $(\mathbb{R}-\mathbb{Q})\cap (0,1)$? 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?
 A: 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.
==edit==
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.
