Polynomial bijection from QxQ to Q? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T06:08:44Z http://mathoverflow.net/feeds/question/21003 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21003/polynomial-bijection-from-qxq-to-q Polynomial bijection from QxQ to Q? Z.H. 2010-04-11T12:03:13Z 2013-04-29T16:31:35Z <p>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?</p> http://mathoverflow.net/questions/21003/polynomial-bijection-from-qxq-to-q/21004#21004 Answer by Garlef Wegart for Polynomial bijection from QxQ to Q? Garlef Wegart 2010-04-11T12:49:24Z 2010-04-11T13:00:35Z <p>Edit[ The following is wrong ~ see comments]</p> <p>I don't think so.</p> <p>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.</p> http://mathoverflow.net/questions/21003/polynomial-bijection-from-qxq-to-q/32360#32360 Answer by Daniel Miller for Polynomial bijection from QxQ to Q? Daniel Miller 2010-07-18T13:36:49Z 2010-07-18T13:36:49Z <p>I would be inclined to say no, for the following reasons. First, note that the function $f^{-1}$ is a bijection $\mathbb{Q}\mapsto\mathbb{Q}\times\mathbb{Q}$, and as such, is something that resembles a space-filling curve. But in general, space-filling curves are highly complex, "messy" objects, not something one would expect from the inverse of a polynomial in two variables. Furthermore, note that the polynomials $p(x)=f(x,y)$ and $q(y)=f(x,y)$, for fixed $y$, satisfy $p^{-1}(x)\in\mathbb{Q}$ and $q^{-1}(x)\in\mathbb{Q}$ whenever $x\in\mathbb{Q}$, i.e., the equation $p(x)-r=0$ has a rational solution for every $r\in\mathbb{Q}$. However, as far as I know, the only polynomials that satisfy this are linear functions, which could not provide the bijection required.</p> http://mathoverflow.net/questions/21003/polynomial-bijection-from-qxq-to-q/77027#77027 Answer by Robert Bates for Polynomial bijection from QxQ to Q? Robert Bates 2011-10-03T09:36:40Z 2011-10-03T09:36:40Z <p>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: </p> <p>1) $f=g$ on $Q_1\times Q_1$. </p> <p>2) $g$ is uniformly continuous on $\overline{Q_1\times Q_1}=[0,1]^2$. </p> <p>3) Namely, $g(x_0)=\lim_{x\to x_0} f(x)$. </p> <p>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 $Q_1\times Q_1$. </p> <p>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. </p> http://mathoverflow.net/questions/21003/polynomial-bijection-from-qxq-to-q/82638#82638 Answer by stephanos for Polynomial bijection from QxQ to Q? stephanos 2011-12-04T17:52:27Z 2011-12-04T17:52:27Z <p>No, there is none. Suppose that there was such an $f$. Then for every $a\in\mathbb{Q}$ the function $f(a,\cdot):\mathbb{Q}\rightarrow \mathbb{Q}$ would be an injection. So there should infinitely many injections from $\mathbb{Q}$ to $\mathbb{Q}$, that have disjoint images. But there are only two possibilities for their limits when $x\rightarrow +\infty$. Contradiction.</p>