1
$\begingroup$

Suppose $A_1,\dots,A_k$ are connected open subsets of $[0,1]$ such that $[0,1]=\bigcup_{i=1}^k A_i$. By characterization of connected subsets of $\mathbb{R}$ I know that each $A_i$ is an interval.

I want to show:

There exist $0=a_0<a_1<\dots<a_k=1$ real numbers such that $[a_{i-1},a_i]\subseteq A_i$ for each $i\leq k$ (possibly permuting the $A_i$'s)

My argument

We can suppose $A_1=[0,c_1)$ with $c_1\in A_2=(b_2,c_2)$. Thus we have $b_2<c_1<c_2$.

Take $a_1$ such that $b_2<a_1<c_1$, thus we have $[0,a_1]\subseteq A_1$, and $a_1\in A_2$.

Suppose $c_2\in A_3=(b_3,c_3)$, thus we have $b_3<c_2<c_3$.

Take $a_2$ such that $b_3<a_2<c_2$. Thus we have $[a_1,a_2]\subseteq A_2$, and $a_2\in A_3$.

Continue in this way until we have $[a_{k-2},a_{k-1}]\subseteq A_{k-1}$ and $a_{k-1}\in A_k$.

Being $1\in A_k$ we have also $[a_{k-1},1]\subseteq A_k$.

Is my prof correct? (I know I could have used Lebesgue Number Lemma whch is more powerful but since this case is quite special (since the $A_i$'s are all connected) I wanted to try to prove it in a more elementary way).

EDIT Assume also $A_i\not\subseteq\bigcup_{j\ne i} A_j$ for each $i$.

My argument, n°2

We can suppose $A_1=[0,c_1)$. Since the $A_i$'s cover $[0,1]$ and $c_1\notin A_1$, then it must exists one of the $A_i$ containing $c_1$, (with $i>1$). Let's call it $A_2=(b_2,c_2)$.

I want to show this $A_2$ is unique. Suppose we have two candidates: $A_2=(b_2,c_2)$ and $B=(\alpha,\beta)$ where $B$ is one of the other $A_i$'s. This means we have $c_1\in A_2 \cap B$, so $b_2<c_1$ and $\alpha<c_1$. Suppose for example $\alpha< b_2$. So we have $\beta<c_2$ (otherwise we would have $A_2\subseteq B$) and so $B\subseteq A_1\cup A_2$ whis is a contradiction.

Thus we have a unique $A_2=(b_2,c_2)$ with $b_2<c_1<c_2$.

Take $a_1$ such that $b_2<a_1<c_1$, thus we have $[0,a_1]\subseteq A_1$, and $a_1\in A_2$.

Arguing as done with $A_2$, we have that it exists a unique $A_3$ such that $c_2\in A_3=(b_3,c_3)$, thus we have $b_3<c_2<c_3$.

Take $a_2$ such that $b_3<a_2<c_2$. Thus we have $[a_1,a_2]\subseteq A_2$, and $a_2\in A_3$.

Continue in this way until we have $[a_{k-2},a_{k-1}]\subseteq A_{k-1}$ and $a_{k-1}\in A_k$.

Being $1\in A_k$ we have also $[a_{k-1},1]\subseteq A_k$.

$\endgroup$
  • 1
    $\begingroup$ I'm having trouble with the case $A_1=[0,0.7), A_2=(0.2, 0.3)$ and $A_3=(0.6, 1]$. I might be missing something, but that looks impossible to me. There is no room for permuting the $A_i$ since the end points must be covered by $A_1$ and $A_3$, and $A_2\cap A_3=\varnothing$ means there is no possible $a_2$. $\endgroup$ – Arthur Mar 27 at 14:50
  • $\begingroup$ your problem looks like, you are taking a refinement of some partition of $[0,1]$. $\endgroup$ – Sujit Bhattacharyya Mar 27 at 14:52
  • $\begingroup$ @Arthur I think you should write that down as an answer, not just a comment. $\endgroup$ – 5xum Mar 27 at 14:54
  • $\begingroup$ @5xum Usually I would think so too, but often it's either some detail that I've missed, or that the OP has forgotten to include. It is honestly a request for clarification, not an attempt to answer. $\endgroup$ – Arthur Mar 27 at 14:56
  • $\begingroup$ @Arthur, thank-you! :) With the added hypothesis, it the proof correct? $\endgroup$ – Minato Mar 27 at 14:56
0
$\begingroup$

So without the hypothesis, the proof was incorrect. Here's a small list of points.

  • The order of $A_i$ needs to be explained better, it is integral to the problem and cannot be treated loosely.

  • There is $c_1>0$ such that $[0,c_1] \subset A_1$ is ok. It is not clear why there is some $A_2 \neq A_1$, such that $c_1 \in A_2$. The point $c_1$ could be only in $A_1$ and not be in any other $A_i$! This has to be explained. Also, what happens if there is some other $A_3 = (b_3,c_3)$ in which $c_i$ is contained? You have not shown that there could be a potential clash in construction in this case.

  • The same problem arises for $c_2$. Why should some other $A_3$ contain it? What if more than one contained it?

Now, without the hypothesis, let us see what goes wrong with a given cover $A,B,C$. Take the one given by Arthur, namely $A=[0,0.7)$,$B = (0.2,0.3)$,$C = (0.6,1]$.

Whatever $a_1$ is , it is in $A$. Suppose $a_1 \geq 0.3$ then for any $l \geq 1$ the interval $[a_{l},a_{l+1}]$ cannot intersect $B$ obviously, let alone be contained in it.

If $a_1 \leq 0.2$ then for any choice of $a_2$, the interval $[a_1,a_2]$ is neither fully contained inside $B$ nor $C$.

If $a_1 \in B$, then we must have $a_2 \in B$ as well to fulfill $[a_1,a_2] \subset B$ since $a_1 \notin C$. However, $B \cap C = \emptyset$ so $a_2 \notin C$ must happen, hence no $a_3$ can be found.

Hence, no choice of $a_1,a_2,a_3$ exists, meaning the problem is impossible without further conditions on $a,b,c$.

Small check : Why does the Lebesgue number theorem not imply the above statement? Hint : The Lebesgue number theorem is a statement about sets of sufficient diameter contained in each set. The above also insists that the endpoints of those sets must match, which does not have anything to do with the statement of the LNT.


Now, with the additional hypothesis given, let us try to suggest a proof.

Order the $A_i$ by increasing left endpoint. That is, each $A_i = (b_i,c_i)$ (with $A_1 = [0,c_1), A_k = (b_k,1]$), where we work so that $0 =b_1 \leq b_2 \leq b_3 \leq ... \leq b_k$.

Consider $A_i \cap A_{i+1}$ for $0 \leq i \leq k-1$. I claim this set is non-empty. For if it were, then $A_i$ must lie to the left of $A_{i+1}$, these being disjoint intervals with $b_i \leq b_{i+1}$. Therefore, $c_i < b_{i+1}$. Let $d \in (c_i,b_{i+1})$. Note that $d \notin A_i$, $d \notin A_{i+1}$. Suppose that for any other $A_l$ we had $d \in A_l$. Then, clearly, $b_l < d < c_l$. But then, $b_i < d < b_{i+1}$ and the $A_i$ were chosen to have increasing left endpoints, hence we get that $b_l < b_i$. However, note that $(b_i,c_i) \subset (b_i,d) \subset (b_l,c_l)$ so $A_i \subset A_l$, a contradiction. Thus, the intersection is non-empty.

If we pick $a_i \in A_i \cap A_{i+1}$, then do things work out? Also, I have ignored a few things in the above proof like $d =0$, $d=1$ etc. so please sort this out as it is your proof to finish.

$\endgroup$
  • $\begingroup$ Thank-you enormously for your great work. Please le me ask some some "doubt". (1) According to my original proof, since I had $c_1\notin A_1$, then shurely there was some "$A_2$" cointaining $c_1$. (2) You are right in saying that the order of $A_i$'s is to be explained better, since, for example, there could be two candidade as $A_2$, i.e. two sets containing $c_1$. $\endgroup$ – Minato Mar 27 at 16:07
  • $\begingroup$ (3) With the added hypothesis, I think my proof works because there is only one candidate as $A_2$, as well as for the others. So with the added hypothesis, my alghoritm will automatically (re)order the $A_i$'s in the only unique way possible. To see that there is only one candidate as $A_2$, suppose we have two candidates: $A_2=(b_2,c_2)$ and $B=(\alpha,\beta)$ where $B$ is one of the other $A_i$'s. This means we have $c_1\in A_2 \cap B$, so $b_2<c_1$ and $\alpha<c_1$. Suppose for example $\alpha< b_2$. So we have $\beta<c_2$ and so $B\subseteq A_1\cup A_2$ whis is a contradiction. $\endgroup$ – Minato Mar 27 at 16:07
  • $\begingroup$ Am I missing something? Thank you for your patience and generosity :) $\endgroup$ – Minato Mar 27 at 16:14
  • $\begingroup$ As promised, for the sake of completeness I have carefully elaborated on some steps. If you will take a look I will be happy. Thank you! :) $\endgroup$ – Minato Mar 27 at 16:50
  • $\begingroup$ You are welcome! As for your answer, I admit that I had made some mistakes in checking your answer like the one you've pointed out. I am very happy to say that your improvements are marked, and I will check for correctness shortly. $\endgroup$ – астон вілла олоф мэллбэрг Mar 27 at 16:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.