Let $A=\{ \sum_{n=1}^{\infty}\frac{a_i}{5^i} : a_i=0,1,2,3 \text{ or } 4\}$. What is set $A$?
First thing I noticed is that the elements of set $A$ kind of resembles with how we write numbers of $[0,1]$ in decimal expansion as we know that each number in $[0,1]$ can be written as $\sum_{i=1}^{\infty}\frac{a_i}{10^i}$ where $a_i$ can be any of $1,2,3,4,5,6,7,8,9,0$ . So I think $A$ will also be $[0,1]$. I'm not really sure about my answer and if am I correct how can we rigoursly prove that $A=[0,1]$?
You are right. Actually, if $x\in[0,1]$ and if you write $x$ in base $5$, then$$x=0.a_1a_2a_3\ldots;$$besides, $1=0.444444444\ldots$
Note that if $x\in[0,1)$, you can define $a_1=\lfloor5x\rfloor$. Then, $a_2=\lfloor5^2x-5a_1\rfloor$, $a_3=\lfloor5^3x-5^2a_1-5a_2\rfloor$ and so on. Then$$x=\sum_{n=1}^\infty\frac{a_n}{5^n}.$$