# A classic example in measure theory, hold the measure theory

I answered a question recently drawing on the following construction. Let $\{r_n\}$ be a enumeration of the rationals and set

$$E:=\bigcup_{n=1}^\infty B(r_n,2^{-n}).$$

Then $$m(E) \leq \sum_{n=1}^\infty m\big(B(r_n,2^{-n})\big)=2$$ and in particular $\mathbb R \setminus E$ is uncountable. I realized when answering this question that I didn't know how to prove that $\mathbb R \setminus E$ is uncountable without measure theory. It seems to me a nice example in point set topology, I recall a period of time when I thought that surely any dense open subset of $\mathbb R$ was only missing a countable number of points (I hadn't read about the Cantor set either at that time). So I was wondering if there was a way to approach this that a student in a typical first class in analysis or point-set topology would understand.

-
There has to be more than just topology going on here, because if $2^{-n}$ is replaced by, say, $1/n$ the result could well be countable (or even empty), depending on the enumeration. – Robert Israel Mar 7 '13 at 3:22
@RobertIsrael That's a fair point. Perhaps measure theory or a measure theory flavored approach is the only way. – JSchlather Mar 7 '13 at 3:24
Note that there is a small typo in the calculation of $m(E)$, since we have already for the first term that $m(B(r_{1},2^{-1}))=m((r_{1}-2^{-1},r_{1}+2^{-1}))=1$. – T. Eskin Mar 7 '13 at 12:02
@ThomasE. Fixed. – JSchlather Mar 7 '13 at 12:22

It is not too hard to prove that if $( (a_i,b_i) : i \in I )$ is an open cover of the unit interval $[0,1]$ then $\sum_{i \in I} (b_i - a_i) > 1$. The proof is just a compactness argument: there would be some finite subcover $( (a_j, b_j) : j < k )$ of the interval, and by considering the endpoints we can show that $\sum_{j < k} (b_j - a_j)$ is already greater than $1$.
This is of course half of the proof that $m([0,1]) = 1$, but you could present it without defining measure at all.
(+1) This is nice in that it shows the stronger statement that for any $a \in \mathbb R$, $[a,a+1]\setminus E$ is uncountable. – cardinal Mar 7 '13 at 3:49
Like Theo I'm having a little bit of trouble seeing the reduction to this. Could you expand a little bit on how assuming $\mathbb R \setminus E$ is countable you get to an open cover of $[0,1]$? This certainly shows that $\mathbb R \setminus E$ is non-empty. – JSchlather Mar 7 '13 at 4:44
@JSchlather: Perhaps this is the argument. Suppose that $\sum_{i \in I} (b_i - a_i) < 1 - \epsilon$, where $\epsilon > 0$, and let $F = \bigcup_{i \in I} (a_i,b_i)$. If $[0,1] \setminus F$ were countable, we could cover it with another sequence of open intervals whose lengths sum to $\epsilon/2$. This would give a new, larger family $\{(a_i, b_i) : i \in I'\}$ for which the lengths still sum to less than 1. By construction, that new family is a cover of $[0,1]$, but by the fact in my answer it can't be a cover because the lengths sum to less than 1. This addresses the question if $m(E)<1$. – Carl Mummert Mar 7 '13 at 12:38