I want to formally prove that $\sup\left(\mathbb D \cap \left[0,\frac{1}{3}\right]\right)=\frac{1}{3}$, where $\mathbb D=\left\{\frac{p}{10^n} ~|~ p \in {\mathbb Z}, n\in \mathbb N\right\}$ is the set of decimal numbers.
Clearly, any $x\in \mathbb D \cap \left[0,\frac{1}{3}\right]$ satisfies $x\leq \frac{1}{3}$. Now, it remains to check that $\frac{1}{3}$ is the least upper bound, i.e. that for any $\varepsilon >0 $, there exists an element $x_0$ of $\mathbb D \cap \left[0,\frac{1}{3}\right]$ such that $x_0 + \varepsilon > \frac{1}{3}$.
If $\varepsilon> \frac{1}{3}$, then any $x_0$ from $\mathbb D \cap \left[0,\frac{1}{3}\right]$ is good. But how do I construct this element in the case when $0 \le \varepsilon \le \frac{1}{3}$. Clearly, this element is of form $x_0=\frac{p}{10^n}$ for natural numbers $p$ and $n$... But how can I justify its existence?
Note that, for each natural $n$,$$\frac13-\sum_{k=1}^n\frac3{10^k}=\sum_{k=n+1}^\infty\frac3{10^k}=\frac3{10^{n+1}}\cdot\frac1{1-\frac1{10}}=\frac1{3\times10^n}.$$Besides, $\sum_{k=1}^n\frac3{10^k}\in\Bbb D\cap\left[0,\frac13\right]$. So, for any $\varepsilon>0$, take $n\in\Bbb N$ such that $\frac1{3\times10^n}<\varepsilon$, and then$$\sum_{k=1}^n\frac3{10^k}\in\Bbb D\cap\left]\frac13-\varepsilon,\frac13\right].$$