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My friend threw me a challenge. He told me that he proved the follow proposition:

The topological space $[0,1]$ cannot be partitioned into pairwise disjoint nondegenerate closed intervals (except the trivial partition $\{ [0,1] \}$).

I think that it's false because I prove the follow proposition:

It's impossible to cover $[0,1]$ with pairwise disjoint closed intervals with a finite number of singletons (except for $\{ [0,1] \}$).

Proof?

Let be $0<a_1<a_2<1$, I have that $$[0,1]=[0,a_1] \cup [a_2,1] \cup \left]a_1,a_2\right[.$$

Now I have one "hole" to cover with closed interval.

$\left]a_1,a_2\right[=[a_1+r_1,a_2-r_2] \cup \left]a_1,a_1+r_1\right[ \cup \left]a_2-r_2,a_2\right[$ with $r_1,r_2>0$ such that these are disjoint.

Now I have two "holes" to cover $\left]a_1,a_3\right[$ and $\left]a_4,a_2\right[$ where $a_3=a_1+r_1$ and $a_4=a_2-r_2$.

Following in this way I obtain that $$[0,1]= \bigcup_{i=1}^{\infty}[a_i,a_{i+1}] \cup Z$$ where $Z$ is a countable infinite family of separated "holes". It's impossible to use a finite number of singletons to cover a infinite number of separated holes. $\Box$

What do you think about my proof? Who is right?

user642796
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Skills
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  • The second block quote section is also unclear, what do you mean by "closes subset with a finite number of singleton"? – Asaf Karagila Aug 13 '15 at 21:00
  • That it's impossible to find a covering with a finite number of ${x }$. We know that $[0,1]=\bigcup_{x \in [0,1]} {x }$. So Is it possible to use a finite number of singleton $ {x }$ to cover $[0,1]$? – Skills Aug 13 '15 at 21:04
  • That much is clearly false, since a finite union of singletons is finite, whereas $[0,1]$ is infinite. – Asaf Karagila Aug 13 '15 at 21:05
  • @AsafKaragila: I read the question as: "Can I cover $[0,1]$ by closed intervals (or perhaps more interestingly, sets) such that only finitely many are singletons (in the more general setting, finite)?" –  Aug 13 '15 at 21:07
  • yes but i can use other kind of subsets of $[0,1]$. An example i can cover $[0,1]$ with $[0, \frac{1}{2}[ \cup { \frac{1}{2} } \cup ] \frac{1}{2} , 1]$ but $[0, \frac12[ $isn't a closed subset of $[0,1]$. Now do you understand what i mean? – Skills Aug 13 '15 at 21:07
  • You read right. – Skills Aug 13 '15 at 21:11
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    Taking a single interval $[0,1]$ we get a cover of $[0,1]$ with one closed interval. – Wojowu Aug 13 '15 at 21:14
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    I cannot even parse "coverable with an its partition of closed interval " – Hagen von Eitzen Aug 13 '15 at 21:16
  • @Wojowu yes i know this. But i wanted to mean i finite number of sigleton bigger than 0 – Skills Aug 13 '15 at 21:19
  • @HagenvonEitzen i think that he wanted to say that he used closed intervals and just one singleton to cover $[0,1]$ – Skills Aug 13 '15 at 21:21
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    @Skills I still don't understand the claim. For example $[0,1]=\bigcup_{0\le a\le b\le 1}[a,b]$ is a covering of $[0,1]$ with lots of closed intervals, many of which are singletons and many are not. – Hagen von Eitzen Aug 13 '15 at 21:24
  • @Hagen: He wants them all to be disjoint. –  Aug 13 '15 at 21:26
  • @HagenvonEitzen but i need that $A_i \cap A_j = \emptyset$ for each $i,j$ – Skills Aug 13 '15 at 21:26
  • So you are looking for a disjoint cover. If you exclude the closed intervals $[0,a]$ and $[b,1]$ which are necessary to cover $0$ and $1$ respectively, you are essentially asking about a disjoint cover by closed sets of some open interval $(a,b)$ – Mark Bennet Aug 13 '15 at 21:27
  • Sorry i edited again. I forgot disjoint in my text – Skills Aug 13 '15 at 21:28
  • @MarkBennet Yes. – Skills Aug 13 '15 at 21:29
  • Do you mean "closed subsets" or "closed intervals" in the second block quote? (And you must surely mean "partition" rather than "cover".) – Rob Arthan Aug 13 '15 at 21:54
  • @RobArthan i edited with intervals. Yes i think it's the same if we are talking about topological space. – Skills Aug 13 '15 at 22:00
  • I have tried to clean-up this question and perhaps improve the phrasing. Please look over to ensure that the meaning has stayed the same. – user642796 Aug 14 '15 at 05:23

2 Answers2

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Concerning the current edition of your question, I can say that both propositions are right, by The Sierpiński Theorem. I recall that a continuum is a compact connected Hausdorff space and remark that the unit segment $[0,1]$ can contain at most a countable family of pairwise disjoint nondegenerate closed intervals.

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(cited from “General Topology” by Ryszard Engelking. )

Alex Ravsky
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Your friend is right....(1): If it were possible it would need a countably infinite family...(2):Proof by contradiction: Suppose $I(n)=[A(n),,B(n)] $ with $A(n) < B(n)$ for $n \in N$ with $ \cup \{ I(n) \} _{n \in N} = [0,1]$ and $I(m) \cap I(n) = \phi$ for $m \not =n$. We may assume $A(0)=0$ and $B(1)=1$.For brevity,say that $I(j)$ is between $I(i)$ and $I(k)$ to mean that $$ \max I(i) < \min I(j) < \max I(j) < \min I(k)$$ or that $$\max I(k) < \min I(j) < \max I(j) < \max I(i).$$Now define $F(0)=0, F(1)=1$ and for $n \in N$ let $F(n+2)$ be the least $k > F(n+1)$ such that $I(k)$ is between $I(F(n)$ and $I(F(n+1)$. We have $$B(0)=B(F(0)) < B(F(2)) < B(F(4)) < ... < A(F(5)) < A(F(3)) < A(F(1))=A(1)$$ and F is strictly increasing... Now let $x$ be the lub of $ \{ B(F(2n)) \} _{n \in N}$. Then $x \in I(j)$ for some $j$, and $j$ not equal to any $F(n)$. So $F(n+1) < j < F(n+2)$ for some $n > 0$.(Recall $F(0)=0$ and $F(1)=1$.) But then $I(j)$ is between $I(F(n-1))$ and $I(F(n))$ with $F(N+1)) < j < F(n+2)$, contradicting the definition of $F(n+1)$. QED. Also observe that if we consider only countable families we may allow some $I(n)$ to be degenerate, with an obvious modification to my definition of "between".

DanielWainfleet
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