# 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.

general-topologyreal-analysis

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!

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 $$|x - \frac{k}{5^n}| \le \frac{1}{5^n} < \epsilon$$.