Show that the set of numbers of the form $\frac{k}{5^n}$, where $k$ is an integer and $n$ a positive integer, is dense in the real line.

My solution starts:

A set $B = \{\frac{k}{5^n} : k \in \mathbb{Z}, n \in \mathbb{N}^+\}$ is dense in $\mathbb{R}$ if every point in $\mathbb{R}$ is either in $B$ or a limit point of $B$.

So far, I've divided it into the cases of integers and non integers. If integers, then select $k, n$ such that $k = x \cdot 5^n$ is satisfied; then that integer is in the set $B$.

Then, the main point is to find if something is a limit point of $B$. In the case of non-integers, I'm not sure how to go about it. Suppose we have some non-integer real like $x$. I'm thinking about forming neighborhoods $(x – \frac{k}{5^n}, x + \frac{k}{5^n}$) and trying to show that there is at least one $y_n \neq x \in B$ to show that $x$ is the limit point of $B$. Not sure if that is the right track though – any hints would be appreciated!

## Best Answer

Hint: Thinking about intervals that cover the real line is a good start. I think it is simpler to use the equivalent characterization of a dense subset $B$ of $\Bbb{R}$ that says that $B$ is dense, iff for every $x$ and $\epsilon > 0$, there is $b \in B$ such that $|x -b| <\epsilon$. So, let $x$ and $\epsilon >0$ be given. We can find $n \in \Bbb{N}^+$ such that $\frac{1}{5^n} < \epsilon$. Now consider all the closed intervals $I_k =[\frac{k}{5^n}, \frac{k+1}{5^n}]$ for $k \in \Bbb{Z}$ (so that $I_k$ has length $\frac{1}{5^n} < \epsilon$). Using the Archimedean property of $\Bbb{R}$ (which we have actually used already to find $n$), you can show that $x \in I_k$ for some $k$. But then $|x - \frac{k}{5^n}| \le \frac{1}{5^n} < \epsilon$.