Denseness of set S in R

Question

Consider the set $S= \{\frac{p}{2^n} : p,n \in \mathbb{Z}\}$. Is it dense in $\mathbb{R}$ ?? To me intuitively it seems to be dense but I cannot come up with any analytic proof or disprove .

Answer 1

$S$ is an additive subgroup of the reals. Such subgroups are either dense or discrete. As $S$ is obviously not discrete it is dense.

If you don't know the result that I mentioned about the additive real subgroups, the best is to prove it. The proof will be very similar to prove directly that $S$ is dense and the result more general.

Answer 2

Let $r \in \Bbb{R}$ and $\varepsilon > 0$. We must show there is an $s \in S$ such that $|s-r| < \varepsilon$. Let $n = \lceil 2^{- \log_2(\varepsilon)} \rceil + 1$ and note both $n \in \Bbb{Z}_{\geq 1}$ and $2^{-n} < \varepsilon$. Let $p$ be the nearest integer to $2^n r$, i.e. $|p - 2^n r| < 1$, so that $\left| \dfrac{p}{2^n} - r \right| < 2^{-n} < \varepsilon$. By construction, $\dfrac{p}{2^n} \in S$. Consequently, $S$ is dense in $\Bbb{R}$.

Answer 3

Let $x\in\Bbb R$ be arbitrary. Consider $[x]$ written in base $2$ as some string of $0$s and $1$s. Then $[x] = \displaystyle\sum_{k=0}^N d_i 2^i, d_i\in \{0,1\}$.

Now how to get the rest of $x$? Well clearly $[x]\le x$. So now construct the series $S_N$ with $S_0 = [x]$ and $S_N = S_{N-1}+ f(N)$ where $f(N) = \displaystyle\inf_{y\in S}\{S_{N-1} + y\le x\}$. Then this is a monotone, bounded (by $x$) sequence of real numbers which converges to its least upper-bound. But if $x'< x$ is the lub, clearly you can exceed $x'$ as if $\epsilon = x-x'$ we know for all $n$ there is a $k$ such that ${k\over 2^n}< \epsilon \le {k+1\over 2^n}$. So $x$ is the least upper bound of this series, proving density.

Answer 4

For each $n \in \mathbb N,$ let $S_n = \{p/2^n: p \in \mathbb Z\}.$ Fix $x\in \mathbb R.$ Then for each $n\in \mathbb N$ there is $y_n\in S_n$ such that $|x-y_n|\le 1/2^n.$ The sequence $y_n$ thus converges to $x.$ Since each $y_n \in S,$ $S$ is dense in $\mathbb R.$