Intuitive, possibly graphical explanation of why rationals have zero Lebesgue measure I know that rationals, being a countable set, have zero Lebesgue measure. I think one way to prove it is to find an open set containing rationals that has measure less than $\epsilon$ for every $\epsilon >0$ fixed. You can do it by taking the rational points in sequence and taking intervals of length $\epsilon/2^n$. Then the union of these intervals has measure less or equal than $\epsilon$.
However I was wondering: how can I explain this intuitively? If one thinks of a dense subset, such as $\mathbb{Q}$ in $\mathbb{R}$, one thinks of something that is "so close" to the original set that it is undistinguishable, in a certain way. I think the most intuitive explanation would be that when you take those intervals, you are "scaling down" their lengths faster than how a given sequence of rational points approach a non rational one.
But this may sound a bit confusing, tricky, so I was wondering: is there a simple, intuitive, possibly graphical way of explaining to someone with very little background in math why rationals have measure zero?

You could utilize one of the well known ways to count the rational numbers, namely consider the integer lattice $\mathbb Z^2$ and the subset $\{(a,b)\mid a\geq 1\ \wedge\ b\geq 0\}$ as illustrated here:

This corresponds to the positive rationals, namely $(a,b)\mapsto\frac ba$. It is a surjective covering and it is now simple to see how we might cover all those points using circles of a finite total area $\varepsilon$ for any given $\varepsilon >0$. In the image above, I have done this using circles of exponentially decreasing sizes, which corresponds to using the well known sum
$$
2=\sum \frac n{2^n}
$$
as a finite bound which can then be scaled down ad infinitum.
Thus we can project this representation onto $\mathbb R$ and make a similarly effective covering there.

BTW one way to project this onto the number line $\mathbb R^+$ would be to draw a vertical line at $x=1$. Then given any rational number $q$ one could draw the line from the origin $(0,0)$ through $(1,q)$ and project the circle from the first lattice point this line passes through onto the vertical line $x=1$. This projected circle around $(1,q)$ would effectively translate into an interval around $q$ on the positive $y$-axis (corresponding to $\mathbb R^+$).
