But for that change, this is what tb had in meant, I think. Just a note: this idea of writing things as a countable union of bounded things comes up often.
Hey, guys, do you think I should or should not have rejected this edit suggestion? I am feeling pangs of guilt. It obviously has been done with good intentions, and the OP is a newbie, so it's good to help the OP improve the question...
Oh, I did not have manners in mind. But I was thinking that this suggestion ought to be rejected because iyengar changed the question. But in this case, changing/improving the question is not that bad after all. [Of course, I don't approve or worry about of the "Thank you" bit.]
Anyway, what's done is done. Thanks, Asaf and Matt.
[JM, tb: if you were reading the transcript till now, I'll be glad to know your opinion as well.]
@Srivatsan Don't feel guilty. It was done in good faith, but I don't think that's what the question asked for and the implicit call for references is always there (and not particularly helpful in this case, or should one refer to Underwood Dudley?)
@tb We should undelete all his downvoted questions and request a global recalc to be triggered, that way he could not delete his questions right after he got the first answer or pointed out that it is a dup.
@Gigili I didn't vote and looked at it only now. Breaking up the problem into even and odd functions doesn't seem to work. Except for pointing out that the constant function $-1$ is also a solution, your answer doesn't contribute much to the solution of the problem. That's just my opinion.
@AsafKaragila well, I don't care much for his deleted questions, but I'm completely opposed to deleting this
If you have a null set in $\mathbb{R}$, for given $\varepsilon$, how do you get a countable number of intervals such that the sum of their measures is less equals epsilon?
@Srivatsan Well, I haven't yet figured out how to get disjointness, but I have learnt some new vocabulary : ) So $\mathbb{R}$ is second countable which apparently implies that it's Lindelöf which means that every open cover has a finite subcover. So I can write every open set as a countable union of open sets.
@Srivatsan If it's open then for every point in it I can find an open ball around the point that is in the set. All these balls form an open cover of the set so I can make it into a countable open cover. A ball is an interval in $\mathbb{R}$.
@Srivatsan It doesn't. I can write it as an arbitrary union of intervals because $\mathbb{R}$ is a metric space. Then Lindeloefness let's me make the cover into a countable one.
@Matt Yes. The problem with the merging idea is this:
Sorry, ignore that comment.
Essentially, I was going to say that if we have infinitely many sets, then it's not clear that we can union pairs of them in some order (even conceptually, if not physically) to get a disjoint family of open sets. But as it is, this is quite obvious.
1. Every point in an open subset of the real line has a connected open neighbourhood. 2. Every connected open neighbourhood is contained in a connected component. 3. Connected components are disjoint. 4. The connected open subsets of the real line are the intervals...
@ZhenLin :) I asked you this long back and I certainly remember this proof. Matt and I was trying out to figure out if the alternate approach could be fixed at all.
So we can start off writing $O$ (an open set) as a countable union of open intervals. Then we want to merge two intervals whenever they overlap, right?
Yes. Actually, the trick is to realise that an arbitrarily long chain of pairwise intersecting open intervals has union an open interval (possibly infinite).
Transfinite induction is just structural induction on the ordinals. The ordinals themselves are very simple. There are three constructors: the constant $0$, the successor (unary), and supremum (accepts any set of ordinals as input)
The key theorem in applying transfinite induction is the Hartogs lemma: for any set, there is an ordinal which does not inject into that set.
@AsafKaragila Just that Zhen was trying to teach me transfinite induction and I didn't follow it. You might safely assume that I know no set theory at all...
@Matt Just use the definition of Lebesgue outer measure: $$\mu^\ast(N) = \inf{\left\{\sum_{n=1}^{\infty} |I_n|\,:\,N \subset \bigcup_{n=1}^\infty I_n\right\}}$$ and a null set is the same thing as a set with outer measure zero.
