Q:

Denseness of set S in R

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 .

A:

$S$ 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.

A:

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 $\Bbb{R}$.

