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?
An homeomorphism between $\mathbb{R}-\mathbb{Q}$ and $(\mathbb{R}-\mathbb{Q})\cap (0,1)$?
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0You probably mean $g=1/x-1$? – 2012-07-15
1 Answers
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.
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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.
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0@Temitope.A: tomasz is talking about *the* [Baire space](http://en.wikipedia.org/wiki/Baire_space_(set_theory)) $\omega^\omega$. For the sake of confusion there are also [Baire spaces](http://en.wikipedia.org/wiki/Baire_space) in topology: those spaces in which the Baire category theorem holds. – 2012-07-15