The interval $[0,1]$ is not the disjoint countable union of closed intervals.

Question

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?

Answer 1

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.

Answer 2

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 $\bigcup_{n\in\Bbb N}(a_n,b_n)$.

Answer 3

Here is a stronger claim: $[0,1]$ can not be written as the union of a countable disjoint collection of closed sets.

Proof. Suppose not. Then we write $[0,1]=\bigsqcup_{n\geq 1}F_n$ with $F_n$ closed (we assume that each $F_n$ is nonempty, for sure). Notice that each $F_n$ is compact as it is a closed subset of $[0,1]$, a compact set. Then there exists $a\in F_1, b\in F_2$ such that $d(F_1, F_2)=d(a,b)$. Without loss of generality, suppose $a<b$. Then $I^*:=[a,b]$ intersects $F_1$ and $F_2$ at (and only at) $a$ and $b$, respectively. Thus, $I_1:=[(a+b)/2, b]$ is disjoint from $F_1$ and $I_1$ intersect at least two members of $\{F_2,F_3,\cdots\}$.

Repeat the procedure above by updating $[0,1]\leftarrow I_1$ and $F_j\leftarrow F_j\cap I_1$ for all $j\geq 2$ and discarding $F_1$. Then we have nested closed intervals $I_1\supseteq I_2\supseteq I_3\cdots$ with $I_j\subseteq [0,1]\backslash F_j$. Thus, by nested interval theorem (or Cantor intersection theorem), $\bigcap_{j\geq 1}I_j$ is not empty. However, for any $x\in \bigcap_{j\geq 1}I_j$, since $x\in I_j$ for all $j\geq 1$, $x\notin F_j$, which is absurd.

Remark. There is a generalization to abstract topological spaces. See this post for a detailed discussion.

Answer 4

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 $\blacksquare$.