@AsafKaragila Yes, namely that if $x = x^\prime$ then $(x,y) = (x^\prime, y^\prime)$ in $R$. That's what I meant when I wrote "...it's a function..." above.
@tb I don't want to up vote if I don't read the question. I don't know if there is anything wrong with it but I'd presume there is not. Except for the fact that there is no one left that's still interested in an answer given that it was asked in April.
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.
I have been asked to prove that the altitudes of triangle intersect eachother at a single point, I just need a start if any of you could give me a little hint...
Dear community, I wanted to ask you all, if there's any way in which all of you can be notified. And, a message to me reads (removed). What does it mean?
@KannappanSampath No, you cannot address the whole community. // It's not about you in particular. The author of the message chose to delete it after posting it, that's all.
The idea behind induction is the same as recursion.
You want to be able and "unfold" the definition.
The only way it can be mathematically valid is if you can unfold in a finite number of steps. The recursion theorem tells us that it is the other way around as well - if you can always get to the bottom in finitely many steps then you can define recursively.
But can you see why Zhen suggested transfinite induction for my problem. This is my problem: "If $A$ is a union of open intervals in $\mathbb R$, then $A$ is the union of disjoint intervals." (I know that the disjoint intervals is essentially the connected components of $A$, but let us forget this for now.)
The idea is this: I want to merge pairs of overlapping intervals and replace them by the union. Of course, this is not a finite process, else ordinary induction would've sufficed.
Er, I meant to say: I know that I can assume $A$ is a union of countably many intervals. From there, I do not know how to go to disjoint intervals.
I am quoting Zhen's hint here: "Actually, the trick is to realise that an arbitrarily long chain of pairwise intersecting open intervals has union an open interval (possibly infinite)."
Let $U_i$ be the list of open intervals, $i\in\mathbb N$.
No... wait. Let me think it through.
Okay. I think I got it.
We go by induction. Let $\mathcal V_0=\{U_0\}$.
Suppose $\mathcal V_n$ was defined, let $\mathcal T_{n+1}$ the collection of $\{V\in\mathcal V_n\mid V\cap U_{n+1}\neq\varnothing\}$, and define $\mathcal V_{n+1} = (\mathcal V_n\setminus T_{n+1})\cup(\bigcup\mathcal V_n\cup U_{n+1})$
Why not? What's the difference between this and the following example: $\mathbb N$ is finite. Proof: By induction on $n$. For every $n$, $\{ 1, 2, \ldots, n \}$ is finite (by induction, for instance). Therefore, $\mathbb{N}$ is finite.
We only define $\mathcal V_n$ using $U_k$ for $k<n$.
So there is no danger that the process will go into an "infinite loop". You just run through the $U_n$'s and then you finish after $\omega$ many steps.
Oh, I see. That's right. So $\mathcal V_n$ "stabilises" after the $n$th step. But our goal is to write the set $A$ into disjoint intervals, right? How is that related to $\mathcal V_n$?
I think this can be done using ordinary induction. I mean: For given intervals $U_i$, $i\in\mathbb N$ you define inductively systems of intervals $\mathcal V_i$ such that: A. The system $\mathcal V_i$ is non-overlapping; B. $\mathcal V_i\subseteq \mathcal V_{i+1}$; C. $\bigcup\mathcal V_k = \bigcup_{i=1}^k U_i$.
@Srivatsan The $\mathcal V_n$'s are collection of disjoint intervals, and the union of the intervals within $\mathcal V_n$ is the same as the union of $U_0\cup\ldots\cup U_n$.
I meant this: Do you agree that the union will be a system of non-overlaping intervals? Do you agree that union is the same as $\bigcup_{i=1}^\infty U_i$? But perhaps it's better not to mix two approaches together - maybe have a look at Asaf's solution first.
@AsafKaragila I see. This one statement -- this sums up what I am uncomfortable about. What really does it mean to say "after we've exhausted all Un's"?
Wherever you are... sounds intriguing and filled with women, drugs and alcohol. Are in your secret layer, doing cocaine and PCP while having sex with three illegal immigrants? Are you drinking all my scotch while you're at it?? If you do, please stop!
@Srivatsan well standard induction + the one more step (But you do not have to look at it as transfinite induction - you can see it as using standard induction to produce some auxiliary system and then using this system to construct what you want.)
@Srivatsan I am not sure about that. At least I did not describe how I construct intervals in n-th step in detail. (So at least this is a difference, perhaps there are more differences.)
Oh I see. I don't want $\mathcal V_n$ to consist of any old intervals. I want them to be open (sorry if I missed it before; all this comes from some general topology discussion with Matt).
But sets from Asaf's construction are increasing in a different sense: If $U\in\mathcal V_n$ then there exists $U'\in\mathcal V_{n+1}$ such that $U\subseteq U'$.
@Asaf you do not define $\mathcal V_\omega$ in your construction? I mean if it is supposed to be transfinite induction, there should be some step for limit ordinal...
The set which we started with is $A=\bigcup_{i=1}^\infty U_i$. For any $x\in A$ there is $n_0$ such that for $n>n_0$ we have $x\in\bigcup \mathcal V_i$, i.e. there is unique $U_x^i\in\mathcal V_i$ such that $x\in U_x^i$. We could define $A_x:=\bigcup_{i=1}^\infty U_x^i$. Then the system $A_x$ is the system of open pairwise disjoint intervals.
Perhaps this could work, unless I've overlooked something again.
