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I was surprised I couldn't find the proof of this here. The problem is to prove the image of $\{(r,r\sqrt 2)\mid r\in\mathbb R\}$ is dense in the torus where we think of the torus as $I\times I$ with opposite edges identified in the usual way, where $I=[0,1]$.

I'm just looking for an elegant way to prove this.

Gregory Grant
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    Can you prove, for instance, that ${0,\sqrt{2},2\sqrt{2},3\sqrt{2},4\sqrt{2},\ldots}$ is dense mod $1$? Because if you look at any vertical cross section, you get a translation of this. – Milo Brandt Mar 28 '15 at 23:30
  • I am surprised about your statement. You can find it all over the place, I saw this question a couple of times in the past, e.g. here http://math.stackexchange.com/questions/79051/showing-a-subset-of-the-torus-is-dense – Daniel Valenzuela Mar 29 '15 at 00:53
  • @Dan Thanks I found that link. But it does not prove what I asked. If you look carefully at it he only states my question, and then he says he knows how to prove it, his question is something different. – Gregory Grant Mar 29 '15 at 02:25
  • @Dan But since you say you can find it "all over the place" then please post another link that actually proves it, because I still can't find it. Thank you. – Gregory Grant Mar 29 '15 at 02:26
  • @Meelo Thank you, yes the proof in one dimension is easy, you just use the pigeon-hole principle. But how do I lift this to two dimensions? It's not immediately obvious to me. It must be simple, my old brain is just not working like it used to. I'm surprised nobody has come up with anything in two hours. – Gregory Grant Mar 29 '15 at 02:27
  • @GregoryGrant I don't get it. I think he asks precisely about the density and the answer gives you a nice way. If you want another rigorous solution: http://math.stackexchange.com/questions/449756/a-mapping-from-mathbbr1-to-a-dense-subset-of-the-surface-of-torus-in-mat/449815#449815 or rather the intuitive approach (which also tells you how to solve it) http://math.stackexchange.com/questions/1141342/lie-subgroup-example-explanation – Daniel Valenzuela Mar 29 '15 at 04:37
  • @Dan Ok, I did misread the original, there is supposedly a solution in there. But it's not really a proof, it's an idea. Same with all of the links you posted. I'm looking for a rigorous and complete write-up with details, because I'm having trouble providing the details. The proof Meelo gave (below) comes closer, I'm trying to write it up with i's dotted and t's crossed and I'm not quite there yet, but hopefully this morning with a fresh mind I'll get over the hump. It will be good to have a complete (with details) straightforward proof of this, because that'll be a first. – Gregory Grant Mar 29 '15 at 13:24

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Given that you can prove that $S=\{\ldots,-2\sqrt{2},-\sqrt{2},0,\sqrt{2},2\sqrt{2},3\sqrt{2},\ldots\}$ is dense when taken in the reals mod $1$, one can quickly reach the desired conclusion. In particular, take some point $(x,y)$ on the torus, and consider the points in your set with the same $x$ coordinate - which is exactly those of the form $(r,r\sqrt{2})$ where $r-x$ is an integer (i.e. they're equal mod $1$). You can easily check that this is that same as saying the points are those of the form $(x,x\sqrt{2}+z)$ where $z\in S$. However, then $y-x\sqrt{2}$ is arbitrarily close to elements of $S$, implying it is an accumulation point of your set.

Milo Brandt
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  • I follow all the steps of your proof, but I am still having trouble connecting the last dot. Yes $y-x\sqrt2$ is an accumulation point of $S$. I'm having trouble using this to get the 2-d result with zero hand-waving. Would you mind writing down the exacting details. I would really appreciate it. – Gregory Grant Mar 29 '15 at 14:54
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    @GregoryGrant Well, one way to be rigorous would be considering the map $I\rightarrow I\times I$ (with the appropriate identifications) defined by $f(z)=(x+x\sqrt{2}+z)$ - which is continuous - meaning, in particular, that the preimage of any open set $O$ around $(x,y)$ is an open set in $I$. Any open set in $I$ contains an element of $S$ and the image of any element in $S$ is in your given set. Thus $O$ contains an element of your set - so any neighborhood of $(x,y)$ intersects your set. (You might be using different definitions though; replacing "open set" with "open ball" works too) – Milo Brandt Mar 29 '15 at 15:10
  • Thank you, your proofs are correct and understandable, I'm going to give it the thumbs up. I'm going to post how I would like to think about it. Is this proof bulletproof? Thanks again for your help! – Gregory Grant Mar 29 '15 at 15:42
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    Let $\mathbb Z \times \mathbb Z$ act on $\mathbb R \times \mathbb R$ by $(a,b)\cdot(r,s) = (r+a,s+b)$. Let $T$ be the identification space. Let $(\overline p,\overline q)\in T$. Then the distance in $T$ is $d_T(\overline p,\overline q)=\min_{a\in\mathbb Z\times\mathbb Z}||(p+a)-q||$. Let $(x,y)\in I\times I$. Let $\epsilon>0$. Choose $m,n\in\mathbb Z$ s.t. $|(y-x\sqrt2)-(m+n\sqrt2)|<\epsilon$. Let $p=(x,x\sqrt2+n\sqrt2+m)$. Then $||(x,y)-p||=||(0,(y-x\sqrt2)-(n\sqrt2+m)||<\epsilon$. And $\overline p=\overline{(x,(x+n)\sqrt2+m)}=\overline{(x+n,(x+n)\sqrt2)}$ which is in the curve. – Gregory Grant Mar 29 '15 at 15:46
  • Or does it not make sense to think of the distance on $T$? I'm thinking it might not satisfy the triangle inequality. – Gregory Grant Mar 29 '15 at 15:54
  • Ok I take it all back. I'm still not comfortable with this proof. I can't believe this is stumping me up, it's downright embarrassing. Also, in your comment, that's a map from $I$ to $I$, not to $I\times I$, did you mean $(x,x\sqrt2+z)$? – Gregory Grant Mar 29 '15 at 15:59
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    @GregoryGrant Yes, your last comment's correction is right. Your proof looks correct to me - your definition of distance on $T$ is perfectly natural and is definitely a metric (since if we take a path from $p\in \bar p$ to $q\in \bar q$ and one from $q'\in\bar q$ to $r'\in\bar r$, we can translate the whole path from $q'$ to $r'$ so that it starts at $q$ and ends at $r\in \bar r$ - so it corresponds to a path from $\bar p$ to $\bar q$ to $\bar r$ in normal Euclidean space). If you have any particular question about your own proof, I'd be happy to answer, but it looks perfectly nice to me. – Milo Brandt Mar 29 '15 at 16:12
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    Thank you very much I really appreciate you walking me through this!! I think I've written it up cleaner without need to mention $d_T$, but it's good to know that definition is sensible. The key to me was to write down exactly what I need to show as follows:

    For $(x,y)\in I\times I$ we must find $r$ s.t. $||(r,r\sqrt2)-(x+k,y+\ell)||<\epsilon$ for some $k,\ell\in\mathbb Z$. From there it's easy, just set $r=x+n$, $k=n$ and $\ell=m$, where $n\sqrt2+m$ approximates $x\sqrt2-y$. This is essentially exactly your original suggested proof. Whew, I finally got it.

     – Gregory Grant Mar 29 '15 at 16:19
  • May I ask why $n\sqrt{2}+m$ approximates $x \sqrt{2}-y$ at the end? Thanks! – Toasted_Brain Jan 28 '23 at 21:11