real analysis - Show that a singleton ${x}$ is negligible

With the following definition of negligible set :

$S \subseteq \mathbb{R}$ is negligible if $$\forall \varepsilon > 0 \hspace{0.2cm} \exists I_{k} : S \subseteq \bigcup\limits_{k \in \mathbb{N}}I_{k} \hspace{0.2cm}, \sum\limits_{k \in \mathbb{N}} |I_{k}| < \varepsilon$$

With $I_{k}$ closed or open intervals of $\mathbb{R}$.

I'd like to prove that a singleton $\{x\}$ is negligible, to be able to say for example that $\mathbb{Q}$ is negligible.

This was my effort :trying with the definition,noticing that $\{x\} \subseteq (x-\frac{1}{k},x+\frac{1}{k}) = |I_{k}|,$

I thought that those could be my interval because $|I_{k}| = \frac{2}{k} \underset{k \to \infty}{\longmapsto} 0$, therefore they satisfy $|I_{k}| < \varepsilon$,

But then i realized i was wrong because i had to sum all the lengths of the intervals,but $\sum\limits_{k \in \mathbb{N}} \frac{1}{k} = +\infty$, is that right ?

If so,any solution or tip to solve the problem would be appreciated.

Answer

Why summing them? For each $\varepsilon>0$, take $k\in\mathbb N$ such that $\frac2k<\varepsilon$ and take only the interval $\left(x-\frac1k,x+\frac1k\right)$. That's all.