# does there exist a discrete set whose image is dense

I want to know whether my proof is correct or not : Does there exist a descrete set whose image is dense in $S^1$ under the map $e^{2\pi ix}$ from $\mathbb{R}\rightarrow S^1$? my attempt is : We know that there is a 1-1 correspondence between $S^1$ and $\displaystyle\frac{\mathbb R}{\mathbb Z}$ that is $x$ corresponds to $e^{2\pi i x}$ We know that the numbers of the form $m+n\sqrt2$ are dense in $\mathbb R$ (I don't quite know how to prove this!). So these numbers are dense even in the interval $[0,1]$ This shows that the equivalence classes of $n\sqrt2$ are dense in $\displaystyle\frac{\mathbb R}{\mathbb Z}$ Therefore the numbers of the form $e^{2\pi ix}$ where $x=n\sqrt2$ are dense in $S^1$ Now $A\subset R := A=\{\sqrt{2}n: n\in\mathbb Z\}$ is a discrete subset of $\mathbb R$ Therefore, if $f:\mathbb R\rightarrow S^1:= x\mapsto e^{2\pi ix}$, then $f(A)$ is dense in $S^1$

-
yes, your proof is right! – countinghaus Apr 25 '12 at 6:43
But do I need to show that set is dense while I write in the exam paper or interview? – Taxi Driver Apr 25 '12 at 6:45
For the proof of $m+n\sqrt{2}$ being dense in $\mathbb{R}$, you can consult this topic: math.stackexchange.com/questions/73262/… – Thomas E. Apr 25 '12 at 6:45

## 1 Answer

1. Recall the definition of being dense, i.e. $E \subset A$ is dense in $A$ if $\overline E = A$, or in other words, that every point of $A$ is the limit of some convergent sequence from which the elements come from $E$. Again, in other words, $$\forall x \in A,\quad \exists \{x_n \} \subset E \quad s.t. \quad x_n \to x.$$

2. Note that the subset $E \subseteq S^1$ is dense if and only if under the map $\varphi : \mathbb R / \mathbb Z \to S^1$ which sends $x \mapsto e^{2i\pi x}$ (which is a bijection), $\varphi^{-1}(E)$ is dense in $\mathbb R / \mathbb Z$. Also note that a reformulation of this is that a subset $F$ of $\mathbb R / \mathbb Z$ is dense in $\mathbb R / \mathbb Z$ if and only if its image $\varphi(F)$ is dense in $S^1$. What will be useful to understand this is that your map $\varphi$ is an homeomorphism, i.e. continuous with an inverse that is also continuous, so that you can map back and forth limits of convergent sequences.

3. Now that you know this, try to show that a subset $E$ of $\mathbb R / \mathbb Z$ is dense in $\mathbb R / \mathbb Z$ if and only if the set $E + \mathbb Z = \{ x + n \, | \, \overline x \in E \text{ and } x \in [0,1[, n \in \mathbb Z \}$ is dense in $\mathbb R$. (Note that you need to take care of the fact that by defining this set you're "undoing a quotient".)

4. All of this being done, all you need is a dense subset of $\mathbb R$. Your favorite seems to be $\{ m + n \sqrt 2 \}$, but my favorite is definitely $\mathbb Q$. You want a discrete set? Fine... just spread them all over the place. What I mean by that is this : find a sequence that order the elements of $]0,1[ \, \cap \, \mathbb Q$ (for instance, $\frac 12, \frac 13, \frac 23, \frac 14, \frac 34, \dots$ could be one way to do this ; call them $x_1, x_2, x_3, \dots$). Consider the set $E = \{ x_n + n \, | \, n \in \mathbb N \} \cup \{0\}$. This is clearly a discrete set, and when you look at it in $\mathbb R / \mathbb Z$, you obtain $\mathbb Q / \mathbb Z$. Therefore its image $\varphi(E)$ is dense in $S^1$.

Hope that helps,

-
Actually, I've tried to find the proof that the set $\{ m+n \sqrt 2 \}$ is dense and I gave up on looking when I found my example with $\mathbb Q$. I think you could find your proof easy enough if you had such an argument. It is easy to show that the density of your set is equivalent to $$\underset{m,n \in \mathbb Z}{\inf} |m + n \sqrt 2| = 0.$$ If your set is dense, this is clear ; if the infimum holds, then for any $x$ in $\mathbb R$ you can start with $m_1 + n_1 \sqrt 2 > x$, and keep decreasing with such linear combinations (that you can take as small as you want by construction). – Patrick Da Silva Apr 25 '12 at 7:54