Let us consider the set $A=\{ \sum_{k=1}^{\infty} \frac{a_k}{5^k}: a_k=0,1,2,3,4\}$$,$ a subset of $\mathbb{R}$. Then show that the set $A$ contains an open interval.
My attempt: By the given set $A$, it is clear that $0$ belongs to it, if we consider all $a_k$ be $0$. Also if we take all $a_k$ to be $4$, then $\sum_{k=1}^{\infty} \frac{4}{5^k}=1$. That is $1$ belongs to $A$. Also for every element $x$ in $A$, it is clear to see that $0≤ x≤ 1$. Then it is enough to show that $A=[0,1]$. Then we are done. But I am failed to show that $A=[0,1]$. Please help me to solve this.
The set $A$ you have here is exactly the set of base 5 expansions of real numbers in $0 \leq x \leq 1$ (really it just is the set of all $0 \leq x \leq 1$, except each $x$ is just written out in the series form corresponding to its base 5 representation). Since every number in $[0, 1]$ has a base 5 expansion, $A = [0, 1]$.