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Please give me a hint on the following question.

Given $x \in [0, 1]$ show that there exist $a_1, a_2, a_3,...\in S_m$ such that for $s_n=\sum_{k=1}^n \frac{a_k}{m^k}$ we have $x=\lim_{n\rightarrow \infty} s_n$.

I'm not good at proving existence questions.

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    What is $S_m$ ? – 2017-02-17
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    @Fred It's some set that depends on $m\geq 2$ and $m$ is an integer. – 2017-02-17
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    What do you mean it's "some set"? Can we pick $S_m$? – 2017-02-17
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    @lulu Yes, you can. I think the question asks to prove the existence of base-m of a real number. – 2017-02-17
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    As a guess (but it's just a guess) you are just asking about the base $m$ "decimal" expansion of $x$. – 2017-02-17
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    But in that case, you can not choose $S_m$...it has to be $\{0,\cdots,m-1\}$. Oh, and we need $m>1$. – 2017-02-17
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    Hint: use the Greedy Algorithm. Start by subtracting the greatest multiple of $\frac 1m$ less than $x$. Then, of course, we have $x-\frac {a_1}m<\frac 1m$. Now subtract the greatest multiple of $\frac 1{m^2}$ you can from that. We get $x-\frac {a_1}m-\frac {a_2}{m^2}<\frac 1{m^2}$. Repeat. – 2017-02-17

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