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. 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!
 A: Hint: Suppose you fix $n$ to be some number, like $100$, and consider numbers of the form $\frac{k}{5^{100}}$. This splits up the real line into small intervals of length $1/5^{100}$. Any real number $x$ lies on one of these intervals (or is one of the endpoints), so it is within distance $1/5^{100}$ of the nearest point of the form $\frac{k}{5^{100}}$.
If you need to find a point of $B$ closer to $x$, try a larger value of $n$.
A: 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 $|x - \frac{k}{5^n}| \le \frac{1}{5^n} < \epsilon$.
A: We denote $A:=\{\frac {k}{5^n}|(k,n)\in\Bbb Z\times \Bbb N\}$. Let $x\in \Bbb R$ and $\epsilon>0$. The Archimedian Property gives us a $n\in \Bbb N$, such that $\frac {1}{5^n}<\epsilon\iff 5^n\epsilon>1\iff 5^n(x+\epsilon)-5^nx>1$, so there exists an integer $k$ in $(5^nx,5^n(x+\epsilon))$, because the length of this interval is more than $1$. This means that $5^nx<k<5^n(x+\epsilon)\iff x<\frac {k}{5^n}<x+\epsilon$, so we found an element of $A$ inside the neighborhood $(x-\epsilon,x+\epsilon)\setminus \{x\}$. Thus $A$ is dense in $\Bbb R$. q.e.d.
