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The following proof was suggested: suppose [0,1] was the disjoint countable union of closed intervals. Write the intervals as $[a_n,b_n]$. Start by showing the set of endpoints $a_n, b_n$ is closed. At first I thought this was obvious since it seems like the complement is just the union of $(a_n,b_n)$ which is open but then I thought the complement was the infinite intersection of $(0,a_n) \cup (a_n,b_n) \cup (b_n,1)$ which is not necessarily open any more.

This comes from Taylor's proof in Is $[0,1]$ a countable disjoint union of closed sets?

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  • $\begingroup$ Well, if the interval can be of any length, $[0,1]=\cup \lbrace x: x\in[0,1]\rbrace$. The singletons are closed intervals and $x\neq y$ for different singletons, so they are closed disjoint subsets. But we know that the interval is uncountable, so there are uncountable many singletons. $\endgroup$ Dec 10, 2020 at 17:55

3 Answers 3

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Assume that $[0,1]=\bigsqcup_{i=1}^\infty[a_i,b_i]$.

Start from $x_0=b_1$. Assume $b_1<1$. Then $x_0+\epsilon_1\in (0,1)$ lies in some $[a_i,b_i]$, where $\epsilon_1<\frac{1}{2}$. put $x_1=a_i>x_0$. Then $|x_0-x_1|<\frac{1}{2}$. Then $x_1-\epsilon_2\in (0,1)$ lies in some $[a_j,b_j]$, where $\epsilon_2<\frac{1}{4}$, put $x_2=b_j$, then $x_0<x_2<x_1$ and $|x_1-x_2|<\frac{1}{4}$. Continuing this process, we get $$x_0<x_2<\ldots <x_3<x_1\qquad |x_n-x_{n+1}|<\frac{1}{2^n}$$ thus by Leibniz's criteria, $x_n\to x\in (0,1)$ with $$x_0<x_2<\ldots <x<\ldots <x_3<x_1$$ But $x$ cannot lie in any closed interval. For $x\in [a,b]$, if $a<x$, then it is a must that some $a<x_{2i}<x$, but $x_{2i}$ is a right point of some interval, which is contradict to the disjoint property, thus $a= x$. Similarly, $b=x$. The proof is complete.

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    $\begingroup$ $x_2 < x_0$ could happen for a thin $[a_1, b_1]$, some small $\epsilon_1 < \frac{1}{2}$ and some large $\epsilon_2 < \frac{1}{4}$ where $x_2$ ends up on the left of $[a_1, b_1]$ $\endgroup$
    – Esmailian
    Jan 25, 2021 at 21:35
  • $\begingroup$ That was my first thought too. Thinking about it though, you can always define $\epsilon_{n + 1} = min \{ \frac{\epsilon_{n}}{2}, \frac{1}{2^{n + 1}} \} > 0$ such that, $\epsilon_{n + 1} < \epsilon_n$ and $\epsilon_{n + 1} < \frac{1}{2^n}$ so as to preserve the inequalities $x_0 < x_2 , ... < x_3 < x_1$. $\endgroup$ Sep 19, 2021 at 2:43
  • $\begingroup$ @GiordanoRibeiro to avoid this situation, I believe taking $\epsilon_2<\min\{\frac{1}{4},|x_0-x_1|\}$ and so on is an option. Another one $\epsilon_2<\frac{|x_0-x_1|}{2}<1/4$. I do not get $\epsilon_{n+1}$ as you defined. If you think about it, you have (with the choice of $\epsilon_1<1/2$ as Cubic Bear did) $\epsilon_1/2<1/4\implies \epsilon_2=\epsilon_1/2\implies \epsilon_{n+1}=\epsilon_n/2$. But does this fix Esmailian concern though? $\endgroup$
    – ahmd092
    Jul 27, 2022 at 21:38
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Your first thought was correct. By hypothesis each $x\in[0,1]$ belongs to exactly one of the intervals $[a_n,b_n]$. If it isn't an endpoint of any of the intervals, it must actually belong to the open interval $(a_n,b_n)$. Thus, the complement of the set of endpoints is indeed $\bigcup_{n\in\Bbb N}(a_n,b_n)$.

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  • $\begingroup$ Thanks Brian! So we have that the set of endpoints is closed. It is countable by construction. It seems we want to get a contradiction by showing it must also be uncountable. Since it's closed, if we can show it's perfect then we are done since it is a perfect subset of a complete space and thus uncountable. However, I am worried that some points are isolated: for example, isn't one of the $a_n=0$ and $b_n=1$? Surely these can be isolated? For the other endpoints it seems more likely that they are not isolated since $a_i\neq 0$ must be "pretty darn close" to some $b_j$ for it to cover $[0,1]$. $\endgroup$
    – user224531
    Mar 18, 2015 at 16:49
  • $\begingroup$ why is you so smart sir? $\endgroup$
    – user139708
    Mar 19, 2015 at 0:41
  • $\begingroup$ @user224531: Taylor Martin was a bit careless in writing up his proof. He was actually working in $(0,1)$, not $[0,1]$, though he did point out why it makes no difference. Yes, $0$ is one of the left endpoints, and $1$ is one of the right endpoints; we can number the intervals so that $0=a_0$ and $1=b_1$. Then the intervals $[a_k,b_k]$ for $k>1$ must be a partition of the open interval $(b_0,a_1)$, and his argument shows that $\{a_k:k>1\}\cup\{b_k:k>1\}$ has no isolated points. $\endgroup$ Mar 19, 2015 at 3:01
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A full proof: Assume that there exists a set $A$ of disjoint closed intervals that cover [0,1] and let $B$ be the set of their endpoints excluding 0,1. Since $\bigcup_{\lambda \in A} \lambda =[0,1] \Rightarrow$ $A$ is countable $\Rightarrow$ $B$ is countable. We’ll prove on the other hand that $B$ is a perfect set, and therefore must be uncountable, thus achieving the desired contradiction. B is closed: Let $\ell$ be a limit point of $B$. $B \subset [0,1] \Rightarrow 0\leq \ell \leq 1$. $\ell \neq 1$: Let $\lambda_{1} \in A$ be the interval that contains the point 1 $\Rightarrow \lambda_{1} = [a,1] , a<1$. Since the intervals are disjoint $B\cap (a,1) = \emptyset$. By the reasoning $\ell \neq 0$. Assume that $\ell \notin B \Rightarrow \exists \lambda \in A$ s.t. $ \lambda = [a,b], 0<a,b<1 ,\ell \in (a,b) \Rightarrow \exists e \in B $ s.t. $ |e-\ell|< \min\{\ell-a,b-\ell\} \Rightarrow e \in (a,b), e\neq a,b \Rightarrow$ the interval that one of its endpoints is $e$ intersects $\lambda \Rightarrow$ contradicition $\ell \in B $. Every point in $B$ is a limit point of $B$: Let $\ell \in B \Rightarrow \exists \lambda \in A$ s.t. $ \lambda = [a,\ell], 0\leq a$ or $\lambda = [\ell,b],b\leq 1$. In the first case let $\epsilon< (1-\ell) \Rightarrow [\ell,\ell + \epsilon) \in [0,1]$. Let $x\in [\ell,\ell + \epsilon)$ and let $\lambda_{x} = [a',b']$ be the interval in $A$ that contains $x$. If $a'\leq a \Rightarrow [a',a] \subseteq \lambda \cap \lambda_{x} \Rightarrow$ contradiction $\Rightarrow$ $a<a'<x \Rightarrow |a'-\ell|<\epsilon$. By similar reasoning one can show in the second case that it’s a limit point as well. We can conclude that $B$ is perfect $\Rightarrow$ contradiction $\blacksquare$.

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