TITLE: How to understand “Union of balls centered at rational numbers is way less than $\mathbb{R}$
QUESTION [10 upvotes]: A few month ago I had to prove $\lambda(\mathbb{Q}) = 0$ (where $\lambda$ is the one-dimensional Lebesgue measure).
The idea: Let $\varepsilon \gt 0, r_n := \frac{\varepsilon}{2^n}$ and $\mathbb{Q} = \{q_1, q_2, \dots\}$. Then:
$$
\lambda(\mathbb{Q})
\le \lambda\left(\bigcup_{n \in \mathbb{N}} B_{r_n}(q_n)\right)
= \sum_{n \in \mathbb{N}} \frac{\varepsilon}{2^n}
= \varepsilon.
$$
I got this. However, it means that many irrational numbers are not in the set $S := \bigcup_{n \in \mathbb{N}} B_{r_n}(q_n)$ which is somewhat contra intuitive, as $\mathbb{Q}$ is dense in $\mathbb{R}$.
Is there any (graphical) illustration of this statement? At least for a single surjective function $f: \mathbb{N} \rightarrow \mathbb{Q}$ we should be able to find irrational numbers which are not in $S$, shouldn’t we?
I would love to see visualization of this fact. I already have proven that there are uncountable many irrational numbers in $\mathbb{R} \setminus S$, but I cannot imagine this. Any ideas for a good imagination (even non graphical ones) are welcomed.

REPLY [3 votes]: This is not exactly your scenario, but it might help to gain intuition. Inside the interval $[0,1]$, consider the set of rationals of the form $m/2^n$ for $m,n \in \mathbb{N}$. These are the numbers that, expressed in binary, have finite fractional part. This set is dense in $[0,1]$. However, if we place over each one a ball of radius $r=2^{-(n+3)}$ we still don't cover all the interval - for example, we don't cover the rational $2/3$ (in binary: $0.1010101010101\cdots$).