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I've been thinking about the following question:

Is it possible to write $(a,b)$ as a disjoint union of closed intervals?

My first guess was no, but then I figured the question might be hiding something subtle. I tried various things, some which I asked about on this site, and I can't seem to construct such a union. So I decided to go back to my gut feeling and prove that it cannot be done.

Attempt:

Suppose there is such a union. Then each closed interval contains a rational. Given that the sets are disjoint, any rational specifies a unique interval. Therefore the union must be a countable union, and we can list the closed intervals. Choose some listing $I_1,I_2,\dotsc$ and let: $$(a,b)=\bigcup_{k=1}^{\infty} I_k.$$ Now define $$J_n=(a,b)\setminus \bigcup_{k=1}^n I_k.$$ Specifically, consider that $J_2$ contains a middle interval $(a',b')$ where $a<a'<b'<b$.

Construct a sequence where each $u_n$ is arbitrarily chosen in $J_n \displaystyle\bigcap\, (a',b')$. By the Bolzano-Weierstrass theorem, $u_n$ contains a subsequence $v_n$ such that $v_n\to c$. We must have $c \in (a,b)$ so $J_{\infty}\neq \varnothing$ which is a contradiction.

Put differently, no matter how the closed intervals are chosen, we can find a sequence in $(a,b)$ that converges to a limit in $(a,b)$ which is contained in none of the closed intervals.

Is my proof correct?

Community
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user111064
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  • If you haven't done this, you should prove that $J_n$ must contain an open interval. – Pedro Dec 20 '13 at 05:42
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    I haven't gotten very far, but you almost certainly want to write "disjoint union of non-degenerate closed intervals", as you can certainly write it as a union of disjoint degenerate ones. – dfeuer Dec 20 '13 at 05:42
  • I don't see how Bolzano-Weierstraß applies. – dfeuer Dec 20 '13 at 05:45
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    Why must $c$ be in $(a,b)$? – Pedro Dec 20 '13 at 06:18
  • This version of the question is actually closely related to this question (but the duplicate isn't). Indeed, any such collection $X$ of disjoint closed intervals must be countable, and since they're disjoint there's a natural induced ordering on $X$. Since the intervals are closed, there must be some interval between any two distinct intervals, and there is no first or last interval in the ordering. So actually $X$ is essentially $\Bbb Q$, and that's not possible by the first question I linked. – Izaak van Dongen Mar 26 '23 at 10:17

1 Answers1

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The general strategy can work, but you have to be more careful in choosing the sequence $v_n$. As you have it, $v_n$ might be eventually monotonic, and it might converge to an endpoint of one of the closed intervals, which would contradict your implicit claim that $\lim_n v_n\in J_\infty$. To avoid this possibility, try constructing $v_n$ in such a way that you get $$v_0<v_2<v_4<\cdots<v_5<v_3<v_1.$$ Then, carefully prove that you indeed have $\lim_n v_n\in J_\infty$.

Edit: Here's an explicit counterexample to the original proof. It was phrased as a proof by contradiction, which is hard to work with, so I interpret the proof to claim that given a countable sequence of disjoint closed intervals in $[0,1]$, a point is found which is outside every interval.

Let:

  • $I_1=[0.1,0.2]$
  • $I_2=[0.8,0.9]$
  • $I_n=[0.2+2/3^n, 0.2+3/3^n]$ for $n>2$
  • $u_n=0.2+1/3^n$

Then for any subsequence $u$ of $v$, we have $\lim v = \lim u = 0.2 \in I_1$. Note that $J_\infty$ is not empty... the point is that the sequence $u_n$ doesn't help us find a member of $J_\infty$.

Chris Culter
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  • See my comment to the question, it is not totally trivial. – Igor Rivin Dec 20 '13 at 06:50
  • @IgorRivin Agreed, I didn't mean to suggest that it's trivial. I just wanted to assure the OP that the elementary approach can work... with enough effort! – Chris Culter Dec 20 '13 at 07:05
  • Hi, thank you for the answer. I am bit confused as to why we need $v_n$ to go back-and-forth like that. Can we not impose a little condition on $v_n$ (such that it always lies in the middle of the largest open interval of the form $(a',t_n) \subset J_n$ for instance, where $t_k$ is the lower endpoint of $I_k$) so that convergence to $a'$ drags with it the convergence of a subsequence of $t_n$ to $a'$ as well? (which is a contradiction, because then the resulting union is not disjoint). Thanks for your time, I appreciate it a lot. – user111064 Dec 20 '13 at 14:54
  • For the first question, notice that there is a skipped step in the original argument. You define a number $c$ and prove that $c\in(a,b)$, but you really want $c\in J_\infty$, which isn't proven. If you try to prove $c\in J_\infty$, you'll see what's missing in more detail. For the second question, you can't always define $(a',t_n)\subset J_n$, since it's possible that $a'\in I_3$! – Chris Culter Dec 20 '13 at 17:23
  • @ChrisCulter Hm. But I defined $a'$ as the lower-endpoint of the middle interval of $J_2$ (in other words it is the endpoint of $I_1$ or $I_2$, so we can never have $a'\in I_n$ for any $n>2$, since we have $I_i\cap I_j=\varnothing$ for $i\neq j$). With regards to the limit of $v_n:$ the idea in my previous comment was that we can construct a sequence where $v_n\in J_n$. If its limit is not in $J_{\infty}$ then it must converge to the endpoint of two closed intervals, but this is not possible. I think I might be turning in circles. Could you provide a hint? Thanks again. – user111064 Dec 20 '13 at 17:49
  • Oh, I missed that about the definition of $a'$, but it still doesn't do what you want. I'll add a counterexample to my answer showing that it is indeed possible for the limit to be an endpoint... – Chris Culter Dec 20 '13 at 17:53