Disjoint covering of $A \subset [0, 1]$ null set We had the following homework problem: Let $A \subset [0, 1]$ be a null set. Can we find for any arbitrarily small $\epsilon >0$ at most countable many disjoint open intervals $\{I_i\}_{i\in\mathbb{N}}$, such that $A \subset \cup_{i \in \mathbb{N}}I_i$ and $\sum_{i\in\mathbb{N}} |I_i| < \epsilon$?
My group's answer was yes. Since $A$ is a null set, $\forall \epsilon >0$ there is at most countable many open intervals $\{J_i\}_{i\in\mathbb{N}} $, such that $A \subset \cup_{i \in \mathbb{N}}J_i$ and $\sum_{i\in\mathbb{N}} |J_i| < \epsilon$. Now we use induction over the index of $J_n$ to prove, that $\forall n\in \mathbb{N}$ we can construct a collection of disjoint open intervals $I_i$ of $m \le n$ elements such that $\cup_{i \le n} J_i \subset \cup_{i \le m} I_i$ but $\sum_{i\le m} |I_i| \le \sum_{i\le n} |J_i|$. For $n=1$ let $I_1 = J_1$. Now suppose that for some $n \ge 1$ the hypothesis holds, and consider $J_{n+1}$. Remove all $I_i$ interval from the collection of disjoint open intervals that intersects with $J_{n+1}$, and add the union of $J_{n+1}$ and all the intersecting intervals (which will also be an open interval) to the collection of disjoint open intervals. Together with the induction hypothesis, it's obvious that the total length will be at most $\sum_{i \le n+1} |J_i|$, and it'll cover exactly as much, as $\cup_{i \le n+1} J_i$.
Our TA presented an argument along these lines to disprove the claim: suppose there was such a disjoint cover of $[0,1] \cap \mathbb{Q}$, and let $I_k = (a_k, b_k)$ the covering open intervals. We cannot have a "gap" in the covering between some $b_i$ and $a_j$, otherwise some rational number would not be covered (since the rationals are dense in $\mathbb{R}$), thus there must be an enumeration of the endpoints such that $a_1 < b_1 = a_2 < b_2 = a_3 < ...$, and thus $\sum_{i\in\mathbb{N}} |(a_i, b_i)| = 1$. I wasn't there to ask questions, but I see it problematic to assume, that such an enumeration exists.
Was our argument correct?
 A: Neither argument works:) But there is such cover.

*

*Your group argument doesn't work because you infinitely refine cover, possibly never getting any interval that will make it to the final cover.


*Your TA's argument doesn't work, because intervals don't have to be well-ordered (ie it's possible that you can't choose "next" interval). It's easy to cover rationals with disjoint intervals: enumerate them, and for $n$-th rational add an interval of length at most $\frac{\varepsilon}{2^n}$ with irrational endpoints disjoint from previous intervals (there are only finitely many of them, and our rational number isn't endpoint of any of them, so no problems) that contains it, if neither of previously added intervals cover it yet.


*Why it works for any cover: actually, any cover is already a cover by countably many disjoint open intervals, because any bounded open subset of $\mathbb R$ is union of at most countable many disjoint intervals.
To prove it, note that any open interval contains at least one rational number.
Again, enumerate rational numbers, take your cover of $A$, and for each rational in order, if it's covered at all but not covered by previously added intervals, add the interval (maximal connected component of cover that contains our number) that covers it.
For spaces other than $\mathbb R$ (for $\mathbb R^2$ already) it's not the case: segment of length $1$ in plane has zero measure, but can't be covered with disjoint balls with total area less than $\frac{\pi}{4}$.
A: I think the TA's argument is correct. The TA says put $A = [0,1]\cap \mathbb{Q}$ and suppose for the sake of contradiction that such a disjoint cover exists and then derive a contradiction. This is a typical argument by contradiction and I don't think it is problematic.
Next, consider $J_1=(\frac{1}{8},\frac{3}{8})$ and $J_2 = (\frac{1}{4}, \frac{1}{2})$. According to your group's argument, we put $I_1=J_1$. So what is left to cover $J_1 \cup J_2$ is $[\frac{1}{4}, \frac{1}{2})$, which one single open interval $I_2=(\frac{1}{4}, \frac{1}{2})$ cannot do. This is problematic, especially when $\frac{1}{4}\in A$.
