Collection of open intervals covering rationals in (0,1) where sum of lengths is less than 1?

Question

$A$ is the set of rationals in the interval $(0,1)$. I'd like to find a collection of open intervals $\{I_{n}\}$ covering $A$ where $\sum l(I_{n})<1$.

This collection, I believe, must be infinite.

As I understand it the set of rationals themselves in this interval would be neither closed nor open in the reals: every interval around a rational would contain an irrational, and every interval around an irrational would contain a rational.

I am working on a problem where finiteness of the collection of $\{I_{n}\}$ is a requirement for $\sum I_{n}\geq 1$. (I think I have a proof of this that works just fine).

But I don't understand completely why finiteness is required.

First, enumerate the rationals in $(0,1)$. – David Mitra Sep 05 '15 at 18:33 — David Mitra, Sep 05 '15 at 18:33

Moya · Answer 1 · 2015-10-13T15:10:21.007

3

You can do even better, since $A=\mathbb{Q}\cap (0,1)$ has measure $0$, so $\forall \epsilon>0$, we can find intervals $I_n$ such that $A\subset \bigcup I_n$ and $\sum l(I_n)\leq\epsilon$ so that $m(A)\leq \epsilon$. The idea is that $A$ is countable, so $A=\{q_n\}_{n\in\mathbb{N}}$. Now let $$I_n=\left(q_n-\frac{\epsilon}{2^{n+1}},q_n+\frac{\epsilon}{2^{n+1}}\right)$$ It is easy to see that $\sum l(I_n)\leq\epsilon\sum 1/2^{n}=\epsilon$

edited Oct 13 '15 at 15:10

answered Sep 05 '15 at 18:33

Moya

5,248

it was excellent. – R.N Sep 05 '15 at 18:39
Not what was asked, though. – Oct 13 '15 at 02:05
@JessyCat How so? The question asked for a countable collection of open intervals covering $\mathbb{Q}\cap (0,1)$ with measure less than $1$. This proof shows that by taking $\epsilon<1$. The problem of finiteness was secondary. – Moya Oct 13 '15 at 15:09
Wait, why is $\epsilon \sum 1/2^n = \epsilon$? – mavavilj Sep 08 '19 at 11:01

Collection of open intervals covering rationals in (0,1) where sum of lengths is less than 1?

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