Constructing a Sequence. I am asked to prove that given $x \in [0,1)$ there exist integers $0 \leq x_k \leq 9$ such that for all $n \in \mathbb{N}$
$$\sum_{k = 1}^{n} \dfrac{x_k}{10^k} \leq x \leq \sum_{k = 1}^{n} \dfrac{x_k}{10^k} + \dfrac{1}{10^n}$$
The beginning is quite easy:

Let $E = \left\{ k \in \mathbb{Z}, k \geq 0 \vert k < 10x \right\}$.  Since $10 x < 10 $ we can conclude that $E \subseteq \left\{0,1,\dots,9\right\}$.  Define $x_1 = \operatorname{Sup}(E)$.  Since $x_1 \in E$ then $x_1 < 10 x$.  Note that $x_1+1 \not\in E$ and so $10x<x_1+1$.
Thusly,
$$ \dfrac{x_1}{10} \leq x \leq \dfrac{x_1+1}{10}$$
So I have the first entry in the sequence, but don't know how to proceed to find the others.  I would try the same thing again, but if I let
$$E_2 = \left\{ k \in \mathbb{Z}, k \geq 0 \vert k < 10^2x \right\}$$ then $E_2$ has integers less than 100.
Hints are greatly appreciated.

Possible idea: Apply the same idea to $y = x - \frac{x_1}{10}$. This lies in $[0,\frac{1}{10})$, by what you already know. Then use your $E_2$ with this $y$ (instead of the original $x$).
