Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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 $m(B(r_{1},2^{-1}))=m((r_{1}-2^{-1},r_{1}+2^{-1}))=1$. –  Thomas E. Mar 7 '13 at 12:02
    
@ThomasE. Fixed. –  JSchlather Mar 7 '13 at 12:22

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
(+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
    
@cardinal Can you please explain, I still don't understand how the above show what you claim, without using any measure theory. –  Theo Mar 7 '13 at 4:17
    
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
    
Yes, I believe that works. We're doing the calculations that we've abstracted when we apply the measure theory. But it's also fairly simple and should be understandable by an undergraduate suitably far in a first course in analysis. I'll leave the question open for a little bit longer to see if anyone has any other thoughts. I imagine with a little more work this type of argument can also show the complement is not nowhere dense. –  JSchlather Mar 7 '13 at 13:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.