How to understand “Union of balls centered at rational numbers is way less than $\mathbb{R}$

Question

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.

Answer 1

You are looking for something rather like the illustration of the Smith–Volterra–Cantor set from Wikipedia (the set shows as white and its complement as black).

The definition is different but it too is a nowhere dense set with positive measure made by removing intervals around a countable set.

The black blocks represent the removed intervals but these soon get less than a pixel wide making the apparently grey lines which soon fade to almost white: none of the vertical lines in the picture are in fact pure white even if they appear so at first sight.

Answer 2

(This is really just a long comment, so I'm making it CW.)

You should understand this roughly the same way that you understand the resolution of the following counterintuitive fact:

Between two distinct real numbers there is a rational number, but $|\Bbb Q|<|\Bbb R|$.

Order theoretic density (which translates to topological density here as well) is not as connected to cardinality as we first expect. Similarly here, density is not as connected to measure theoretic size as we first expect.

Think of the complement of the union $\bigcup B_{r_n}(q_n)$. This is a closed set of irrational numbers which is nowhere dense, but has infinite measure (or if you limit yourself to $[0,1]$ then $1-\varepsilon$). Baffling isn't it?

The resolution, as in the above case, is that you have to get used to this facts, and then they don't seem so strange after all. They seem quite natural all of a sudden.

Answer 3

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: $0.1010101010101\cdots$).