10

Consider the rational numbers, we know there are countably many, in the numbering, put at the $i^\text{th}$ point a ball, which has a radius of size $2^{-i}$. Now, the sum of the measure of the ball is the geometric series and finite.

But simultaneously we know that union of these balls must be $\mathbb{R}$ but that has infinite measure.

Paradox? More specifically doesn't this break sigma sub additivity?

Integreek
  • 8,530
Clemens Bartholdy
  • 14,222
  • 3
    I also found this counterintuitive when I first learned measure theory – Smiley1000 Jun 11 '25 at 17:00
  • 12
    I don’t see why you think this covers all of $\mathbb{R}$. It is of course dense, but that is not the same. – Malady Jun 11 '25 at 17:39
  • 2
    So presumably you can turn your argument into a formal proof? Letting $q_i$ be the enumeration of the rationals, you can prove that for every $t \in \mathbb R$ there exists $i$ such that $|t-q_i| < 2^{-i}$? If you have not yet worked that proof out, I suggest that you (try to) do that in detail. – Lee Mosher Jun 11 '25 at 18:01
  • 5
    It's tempting to think if $U$ is open and $\Bbb Q\subseteq U$ then $U=\Bbb R$. It's easy to find a counterexample, like Noah's: $\Bbb R \setminus{\pi}$. I think it feels true for two reasons: 1) it is true if you change "open" to "closed", so if you're not careful you might inadvertently "pattern-recognise" it. And 2) if you think of "open" as "you stay in the set if you move a little bit" and "dense" as "you can get to the set from every point by moving a little bit", then it's obviously true ! But the quantifiers are in the wrong order for this argument to work. – Izaak van Dongen Jun 11 '25 at 18:10
  • 3
    "But simultaneously we know that union of these balls must be R but that has infinite measure." << Actually, you just proved yourself that the union of these balls *cannot* be R – Stef Jun 12 '25 at 08:51
  • Closely related. May be even a duplicate? – Jyrki Lahtonen Jun 12 '25 at 17:11
  • @JyrkiLahtonen: I don't think that one is a dupicate. It's asking for existence of a different radius sequence than $2^{-i}$, with suitably different properties. But anyway, y'all did find a duplicate. – Lee Mosher Jun 12 '25 at 18:26

5 Answers5

31

Here's a simpler version that's easier to figure out why it breaks: let $q_i$ be the $i$th rational, and let $r_i={\vert q_i-\pi\vert\over 2}$. Now think about the collection of open balls $B_{r_i}(q_i)$, whose union contains each rational but by construction doesn't contain $\pi$.

The issue with your construction is exactly the same, only there's no "culprit real" specified in advance so it's harder to see.

Noah Schweber
  • 260,658
10

But simultaneously we know that union of these balls must be $\mathbb{R}$

That is false.

Smiley1000
  • 4,219
  • For any size ball around a real number, there is a rational number in it, so how can it be that there are reals missed? – Clemens Bartholdy Jun 11 '25 at 16:59
  • 8
    @ClemensBartholdy I think it will benefit you if you try to write your argument up formally. You will see that it doesn't work. – Smiley1000 Jun 11 '25 at 17:01
  • 1
    @ClemensBartholdy The order of the quantifiers is important here. If you have a real number x and a sequence of rationals that approaches x, the radiuses of the covering balls corresponding to these rationals are approaching zero, so the question becomes, who is faster? If the radiuses become small faster than the rationals approach x, then it's possible that x is not covered by the balls. – Stef Jun 12 '25 at 08:58
  • @ClemensBartholdy And you fixed an enumeration (q_i) of the rationals first, with radius 2^-i around rational q_i, but then you pick a real number x and find a sequence of rationals (r_k) that approaches x; now (r_k) is mostly a subsequence of (q_i), which means there is some extracting function f:N->N such that r_k = q_f(k), and f might increase very fast, so the radius around r_k is 2^-f(k) might be much smaller than the distance from r_k to x – Stef Jun 12 '25 at 09:02
6

Using the notation $q_i$ for the enumerated rationals, the ball around $q_i$ is the set of all $x$ such that $$|x - q_i| < 2^{-i} $$

In one of your comments you correctlly wrote: "For any size ball around a real number, there is a rational in it." So let's apply this: take any real numnber $t$ and any size ball around $t$. Using radius $r > 0$ as a measure of size, this ball consists of all $y$ such that $|y - t| < r$. As you say, "there is a rational in it". So for some value of the index $i$ we may conclude that $$|q_i - t| < r $$ From what you wrote, it seems that you now want to use this to conclude further that $t$ is in the ball around $q_i$: $$|t - q_i| < 2^{-i} $$ This conclusion is certainly true if it so happens that $r < 2^{-i}$.

But what if $2^{-i} < r$? If that's the case, then the conclusion you want is unwarranted.

So this paradox is resolved by noticing that it arises by not paying attention to the distinction between two radii: the radius $2^{-i}$ of the ball around the rational number $q_i$, and the radius $r$ of the ball around the real number $t$.

Lee Mosher
  • 135,265
3

Responding to your comment on another answer:

For any size ball around a real number, there is a rational number in it, so how can it be that there are reals missed?

Consider a real number x, and a ball of size ε > 0 centered around x.

There is indeed guaranteed to be a rational number y in that ball. In fact, there are guaranteed to be infinitely many rational numbers in that ball. However, that is not enough to prove that x is covered by your rational-centered balls.

For x to be covered by your set of exponentially small balls on the rational numbers, one more condition must be satisfied: At least one of those rational numbers in the ε-size ball around x must have a ball large enough to cover x. It's entirely possible that every single rational within distance ε has a ball even smaller than ε.

Now, you can choose a larger value of ε to add more rational numbers to the set under consideration for x, but doing that also increases the required size of a ball around one of those additional rationals. There are infinitely many real numbers for which the required ball size ε is larger than the largest available ball in the area, for all possible values of ε.

Douglas
  • 161
0

Enumerate the rational numbers $q_i$. Place an open ball $B_i$ of radius $3^{-i}$ about $q_i$. We will construct an explicit real number $\alpha$ that does not lie in the union of the balls.

We will define a sequence $c_n\in\{0,2\}$ and set $$\alpha=2\sum_{n=1}^{\infty}\frac{c_n}{3^{n}}.$$ This sum clearly converges as it is Cauchy.

Consider the interval $[0,2]$. After removing the ball about $q_1$ we are left with either the interval $[0,\frac23]$, or $[\frac43,2]$. In the former case we set $c_1=0$ and in the latter case we set $c_1=2$. If both are free from the ball, we can default to $c_1=0$.

Either way, the interval $$I_1=\left[2\sum_{n=1}^{1}\frac{c_n}{3^{n}}, 2\sum_{n=1}^{1}\frac{c_n}{3^{n}}+\frac2{3^1}\right]$$ is disjoin from the open ball $B_1$.

Now suppose we have chosen $c_1,c_2,\cdots,c_m$ such that the open balls $B_1,\cdots,B_m$ are disjoint from the interval: $$I_m=\left[2\sum_{n=1}^{m}\frac{c_n}{3^{n}}, 2\sum_{n=1}^{m}\frac{c_n}{3^{n}}+\frac2{3^m}\right]$$

Then $B_{m+1}$ will be disjoint from: $$I_{m+1}=\left[2\sum_{n=1}^{m+1}\frac{c_n}{3^{n}}, 2\sum_{n=1}^{m+1}\frac{c_n}{3^{n}}+\frac2{3^{m+1}}\right]$$ for either $c_{m+1}=0$ or $c_{m+1}=2$. In simple terms, all this is saying is that in the worst case removing $B_{m+1}$ removes the middle third of $I_m$, so we always have the left or right third of $I_m$ disjoint from $B_{m+1}$.

We obtain upper and lower bounds for $\alpha$ by replacing all $c_n, n>m$ with $2$ or $0$ respectively. Thus $\alpha\in I_m$ for all $m$. Thus $\alpha\notin B_m$ for all $m$ as required.

tkf
  • 15,315