Density of a set in $\mathbb{R}/\mathbb{Z}$

Question

I want to prove that the set $$\bigg\{\frac{z}{2^{n}-1}+\mathbb{Z} \ \bigg| \ n\in\mathbb{N}, \ z\in\mathbb{Z}\bigg\}$$ is dense in $\mathbb{R}/\mathbb{Z}$ (equipped with the quotient topology). Intuively I see why it is true.

My attempt:

For $n\in\mathbb{N}$ define the set $$A_{n}:=\bigg\{\frac{z}{2^{n}-1} \ \bigg| \ z\in\mathbb{Z}, \ 0\leq z\leq 2^{n}-2\bigg\}\subset[0,1[.$$ Because $[0,1[\to\mathbb{R}/\mathbb{Z}, \ x\mapsto x+\mathbb{Z}$ is a continuous bijection, I think that it suffices to prove that $\cup_{n}A_{n}$ is dense in $[0,1[$. (Maybe by using the pigeonhole principle?)

My questions:

Q1) Does it indeed suffice to prove that $\cup_{n}A_{n}$ is dense in $[0,1[$? And what I struggle most with: Why does it suffice?

Q2) If the answer to Q1 is 'yes', how do I prove that $\cup_{n}A_{n}$ is dense in $[0,1[$?

Any suggestions are greatly appreciated.

Answer 1

$Q1):$ Yes, it's true that density in $[0,1)$ is sufficient. Why? Well, let $x\in \mathbb{R}/\mathbb{Z}$ and pick some $\hat{x}\in p^{-1}(x)\cap [0,1),$ where $p$ is the quotient map. Then, by density (and second countability), there exists a sequence $(a_k)_{k\in\mathbb{N}}\subseteq \cup_n A_n$ such that $a_k\to \hat{x}$. Applying continuity, $p(a_k)\to x$.

$Q2):$ Well, let $x\in [0,1)$ and note that $0\leq x-\frac{\lfloor2^n x\rfloor}{2^n}\leq \frac{1}{2^n}$. Since we have now shown that the sequence $\frac{\lfloor2^n x\rfloor}{2^n}$ converges to $x$, we get density.

Answer 2

Let $S =\{z(2^n-1): z\in \Bbb Z\land n\in \Bbb N\}\setminus \Bbb Z.$

(1). $S$ is dense in $\Bbb R.$

Proof (1-a): It suffices that $S\cap (a,b)\ne \emptyset$ whenever $a,b\in \Bbb R$ with $a<b.$ Take $n_0\in \Bbb N$ with $n_0>1$ and $n_0$ large enough that $(2^{n_0}-1)^{-1}<(b-a)/2.$ Then take $z_0$ to be the largest $z\in \Bbb Z$ such that $z(2^{n_0}-1)^{-1}\le a.$ We have $$z_0(2^{n_0}-1)^{-1}\le a<(z_0+1)(2^{n_0}-1)^{-1}<(z_0+2)(2^{n_0}-1)^{-1}=$$ $$=z_0(2^{n_0}-1)^{-1}+2(2^{n_0}-1)^{-1}\le$$ $$\le a+2(2^{n_0}-1)^{-1}<$$ $$<a +(b-a)=b.$$ So both members of $\{(z_0+1)(2^{n_0}-1)^{-1}, (z_0+2)(2^{n_0}-1)^{-1}\}$ belong to $(a,b)$ and they cannot both be integers because their difference $(2^{n_0}-1)^{-1} $ is in $(0,1),$ so at least one of them is in $S\cap (a,b).$

Proof (1-b): With $n_0$ and $z_0$ as in (1-a) we have $(z_0+1)(2^{n_0}-1)^{-1}\in (S\cup \Bbb Z)\cap (a,b).$ So $S\cup \Bbb Z$ is dense in $\Bbb R.$ So $\Bbb R$ $=\overline {S\cup \Bbb Z}=$ $\overline S \cup \overline {\Bbb Z}=\overline S \cup Z.$ So we have $\overline S \supset \Bbb R \setminus \Bbb Z.$ Therefore $\overline S=\overline {\overline S}\supset \overline {\Bbb R\setminus \Bbb Z}=\Bbb R.$

(2). We may define $\Bbb R /\Bbb Z=(\Bbb R \setminus \Bbb Z)\cup \{p\}$ with $p\not \in \Bbb R$ and define the quotient map $f:\Bbb R\to \Bbb R / \Bbb Z$ by $f(x)=x $ if $x\in \Bbb R \setminus \Bbb Z$ and $f(n)=p$ if $n\in \Bbb Z.$ The topology on $\Bbb R /\Bbb Z$ is defined to be the strongest (i.e. $\subset$-maximum) topology that makes $f$ continuous. This requires that $U\subset \Bbb R/\Bbb Z$ is open iff $f^{-1}U$ is open in $\Bbb R.$ So if $U$ is any non-empty open subset of $\Bbb R /\Bbb Z$ then $f^{-1}U$ is a non-empty open subset of $\Bbb R,$ so there exists $x\in S\cap f^{-1}U,$ so $x= f(x)\in f[S]\cap U=S\cap U.$ So $S\cap U\ne \emptyset.$

Note: In the last sentence above, we have $x=f(x)$ because $x\in S\implies x\not \in \Bbb Z.$ And $f[S]=S$ because $S\subset \Bbb R \setminus \Bbb Z.$

Remarks: (i). The set in your Q is not a subset of $\Bbb R /\Bbb Z$ because it has $\Bbb Z$ as a subset but $\Bbb Z$ is not a subset of $\Bbb R/\Bbb Z....$ (ii). The methods of (1-a) or (1-b) can show that if $A$ is any infinite subset of $\Bbb N$ then $T=\{z/a: z\in \Bbb Z \land a\in A\}$ is dense in $\Bbb R$ and hence $T\setminus \Bbb Z$ is dense in $\Bbb R. $