This is a question about the intuition of the rational number having measure zero. Let us consider followng proof:

Let $I = [0,1]$ and $Q = \mathbb Q \cap I$ and let $\lambda$ be the Lebesgue measure. We know $Q$ is countable, so let $Q = \{q_1, q_2, q_3, \ldots\}$. Now consider the balls (intervals) $J_k = B_{b^{-k}/2}(q_k)$ where $b \in (0,1/2)$. Then $\lambda (J_k)= b^{-k}$. Now we have $Q \subset \bigcup_k J_k$, therefore

$$\lambda(Q) \leq \lambda(\bigcup J_k) \leq \sum_k \lambda(J_k) = \sum b^{-k} = b\frac{1}{1-b} \leq 2b$$

So since $b$ can be arbitrarily small, we can conclude that $\lambda(Q)=0$.

This is a classic proof that the rationals have Lebesgue measure zero within the reals. Now intuitively we cover each rational number in $Q$ with an interval of positive length, and since $Q$ is dense in $[0,1]$, one might "feel" like the whole interval $[0,1]$ should be covered by $\bigcup J_k$, but at the same time we know that $Q$ can occupy at most $2b$ of $[0,1]$. So intuitively there must be huge gaps if we e.g. choose $b<<1/2$.

This intuition seems to be misguided in the way that it would make sense for finitely many objects, but as soon as we have infinitely many objects it fails.

So if someone were to ask about these "gaps", can we still somehow find a "better" intuitive reasoning that somehow make sense of these gaps, or should we completely abandon this idea and reason that we ususally do not have very good intuition about infinite collections of things?

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    $\begingroup$ I have never done this, but you should be able to make an explicit enumeration of the rationals, visualize this open cover, and "see" the gaps. I want to play with giving Ford Circles the radius $1/(2q^4)$ instead of $1/(2q^2)$ and see if that would do the trick, but I will not have time for another week or so to play with that sadly. $\endgroup$ Sep 24, 2020 at 12:50
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    $\begingroup$ Giving the Ford circles the radius (1/(4q^4) will work! I didn't need to play because the wikipedia entry for Ford circles actually shows the computation of the total area of the Ford circles, which is basically the same computation of the length of the "shadow" of these "small" Ford circles. It would be neat to actually see the gaps in this picture. $\endgroup$ Sep 24, 2020 at 12:58
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    $\begingroup$ This has been dealt with a lot in Mathematics Stack Exchange. See Lebesgue measure paradox AND Apparent paradox covering the real line AND Paradox as to Measure of Countable Dense Subsets? AND Open Dense Set of Real Numbers and probably several others. $\endgroup$ Sep 24, 2020 at 16:16
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    $\begingroup$ What this leads to is Diophantine approximations and matters such as Roth's Theorem, which says that for $\alpha$ irrational and algebraic, $\vert\alpha-\frac{p}{q}\vert<\frac{1}{q^{2+\epsilon}}$ has only finitely many solutions. It gets technical, but even with just this inequality, you start to see how so many irrational numbers can conspire to escape the covering. $\endgroup$
    – user52817
    Sep 25, 2020 at 1:50

1 Answer 1


Let me build on the idea of Steven Gubkin in his comments. One way to visualize this scenario is to use Ford circles. The standard picture is to plot a circle tangent to the $x$-axis at $\frac{p}{q}$ with radius $\frac{1}{2q^2}$ where we always assume $p$ and $q$ are relatively prime. This gives an intriguing family of circles with special tangency relationships. There is a crowd of circles with small radii just above the $x$-axis that can only be visualized in the mind's eye. These come from rational numbers with large denominators.

Think about this as an open covering of ${\bf Q}\cap [0,1]$: push each circle downwards, so that it is centered on the $x$-axis. The diameter of this depressed circle then marks an open interval of radius $\frac{1}{2q^2}$ centered at $\frac{p}{q}$.

Ford Circles

An important visualization tool afforded by this: let $0<\alpha<1$ and draw a vertical line into the Ford circles emanating from the point $(\alpha,0)$ on the $x$-axis. Then every intersection of this vertical line with a Ford circle corresponds to a rational approximation $\frac{p}{q}$ of $\alpha$ satisfying $|\alpha-\frac{p}{q}|<\frac{1}{2q^2}$. Of course it is difficult to actually see most of these intersections because the vast majority of Ford circles have small radius and blend into the $x$-axis.

Dirichlet's approximation theorem states that for all irrational $\alpha$, there are infinitely many rational numbers such that $|\alpha-\frac{p}{q}|<\frac{1}{2q^2}$. In visual terms, the vertical line emanating from $\alpha$ interests infinitely many of the Ford circles. Stated in terms of the open covering of the rationals corresponding to the depressed Ford circles, each irrational $\alpha$ is in infinitely many of the open intervals. It is important to note that the "total length" of these intervals is infinite. There are $\phi(q)$ circles with radius $\frac{1}{2q^2}$ corresponding to rationals $\frac{1}{q}, \frac{2}{q},\cdots$. Heuristically, $$\hbox{total length}=\sum_{\frac{p_i}{q_i}}\frac{1}{q_i^2}=\sum_{q_i}\frac{\phi(q_i)}{q_i ^2}\approx\sum_q \frac{1}{q}.$$

But now for the next step: let us shrink the radii of the Ford circles from $\frac{1}{2q^2}$ to $\frac{1}{2q^{2+\epsilon}}$. Here, for purposes of plotting, we will take $\epsilon=\frac12.$ So we will shrink radii to $\frac{1}{2q^{5/2}}$.

(In his comment, Steven Gubkin refers to $q^4$ instead of $q^\frac{5}{2}$. This of course works, but I found that the circles got too small to be an effective plot.)

Here is the corresponding visualization. Modified Ford Circles

Notice that as soon as $\epsilon>0$, the beautiful tangency relationships between the Ford circles is lost. We still have an open covering of the rational numbers in $[0,1]$ by depressing these modified Ford circles. But one starts to "see," or least to anticipate, that something will change with the vertical lines and rational approximations to irrational numbers $\alpha$. Indeed, we have Roth's Theorem, which tells us that if $\alpha$ is an algebraic irrational number, then $\vert \alpha-\frac{p}{q}\vert<\frac{1}{q^{2+\epsilon}}$ has at most finitely many solutions. Thus for algebraic irrational numbers, we know that the vertical line will intersect at most finitely many modified (shrunken) Ford circles. In terms of the corresponding open covering of the rationals that we get by depressing the modified Ford circles, such $\alpha$ can be in at most finitely many of the open intervals. Also, we can see more by summing the lengths of the intervals:

$$\hbox{total length}=\sum_{\frac{p_i}{q_i}}\frac{1}{q_i^{\frac52}}=\sum_{q_i}\frac{\phi(q_i)}{q_i ^{\frac52}}<\sum_q \frac{q}{q^{\frac52}}=\sum_q \frac{1}{q^{\frac32}}<\infty.$$

In summary, by shrinking the radii of the Ford circles, we maintain an open covering of the rationals in the unit interval, but can control their radii so that the total length of this open cover is less than 1. It follows that there must be uncountably many vertical lines that do not intersect any modified Ford circle. In terms of the corresponding open cover of the rationals in the unit interval, there are uncountably many irrationals that are not in any of the subintervals.

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    $\begingroup$ Thanks for running with my idea! I don't think I would have found the time to do this, and even if I had I think my answer wouldn't have been as nice as this. $\endgroup$ Sep 27, 2020 at 22:08
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    $\begingroup$ I'm very impressed, this is a beautiful and very visual way to approach this! Thank you very much for this amazing write up! And also thanks @StevenGubkin for the idea in the first place! $\endgroup$
    – flawr
    Sep 27, 2020 at 22:58

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