Is there any polynomial $f(x,y)\in{\mathbb Q}[x,y]{}\ $ such that $f:\mathbb{Q}\times\mathbb{Q} \rightarrow\mathbb{Q}$ is a bijection?

Edit[ The following is wrong ~ see comments] I don't think so. Suppose $f$ to be surjective. Let $x\mapsto 0$ and $y\mapsto 1$. Now consider two distinct paths $\gamma,\eta:[0,1]\to\mathbb Q\times\mathbb Q$ from $x$ to $y$. Since $f$ is continuous it maps these paths surjectively onto $[0,1]$ (more exactly $[0,1]\subset f\gamma([0,1])$ and $[0,1]\subset f\eta([0,1])$). Thus, $f$ cannot be injective. 


Let us consider a simpler case first. Let $Q_1=Q\cap [0,1]$. Now let us assume $f:Q_1\times Q_1\to Q_1$ is a uniformly continuous bijection. Then, according to Rudin (I believe it is an exercise), there is unique continuous extension, $g$, such that: 1) $f=g$ on $Q_1\times Q_1$. 2) $g$ is uniformly continuous on $\overline{Q_1\times Q_1}=[0,1]^2$. 3) Namely, $g(x_0)=\lim_{x\to x_0} f(x)$. Since the image of a connected, compact set is connected and compact, then $Im(g)=[0,1]$. But, this is impossible because if we consider three distinct rational points $a,b,c$ in $[0,1]^2$, then $g$ restricted to the connected set $[0,1]^2${$a,b,c$} is still continuous, but the image will not be connected since g is bijective on $Q_1\times Q_1$. I think the above case now follows. That is if we consider $f:Q\times Q\to Q$ and $f$ is a polynomial, then $f$ restricted to $Q_1\times Q_1$ would could be a uniformly continuous bijection. However, we won't know what the image is. But this doesn't matter, since when we extend to $g$ the continuous image of a connected compact set is connected and compact. But the only connected compact subsets of $R$ are bounded closed intervals. So, the image would be some $[a,b]$ and would get another contradiction. 


protected by Andres Caicedo Dec 5 '13 at 14:23
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