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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?

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You probably mean $g=1/x-1$? –  tomasz Jul 15 '12 at 13:51
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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.

==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.

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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
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