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?
Best Answer
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.