## Countable union of closed intervals

Today Zorn Leung asked a question in analysis. In think this is quite a good exercise in elementary analysis. There can be many versions of this question. One version is:
Question: Can the open interval ${(0,1)}$ be expressed as a countable disjoint union of closed intervals?
(There’s a very related post on Terry Tao’s blog, but I haven’t read it yet it is nice to see the differences and similarities of our proof with his (and also the comments therein). ) There are of course many other versions, e.g. by changing ${(0,1)}$ to ${[0,1]}$ and requiring the family of the closed intervals to be non-trivial, or changing it to ${[0,1)}$. However this is essentially the same question as the one I posted. One can also ask the same question in the higher dimensional case, which I haven’t thought about deeply but as mentioned by Terry Tao’s post (and its comments), the higher dimensional case follow from the one-dimensional case (why?). After discussing with Leonard, Raymond etc, we have come up with the following

Theorem 1 (Zorn-Wongting’s Theorem ) ${(0,1)}$ cannot be expressed as a countable disjoint union of closed intervals.

Let me state some conventions (which are not standard) first. For three disjoint closed intervals (not necessarily nontrivial) ${I_1=[a_1, b_1], I_2=[a_2, b_2]}$ and ${J}$, ${J}$ is said to lie between ${I_1}$ and ${I_2}$ if either ${b_1 and ${J\subset (b_1, a_2)}$, or ${b_2 and ${J\subset (b_2,a_1)}$. If such ${J}$ happens to be a singleton ${\{x\}}$, we say ${x}$ lies between ${I_1}$ and ${I_2}$.

Proof:[One word about the proof: drawing a picture would be a good idea! ]
Suppose ${\{I_i\}_{i=1}^\infty}$ is a family of disjoint closed intervals in ${(0,1)}$ such that $\displaystyle (0,1)=\bigcup_i I_i$. Let ${I_i=[a_i, b_i]}$ with ${0. We take ${(c_0, d_0)=(0,1)}$, and take the smallest ${n_1}$ such that ${E_1:=I_{n_1}\subset (c_0, d_0)}$. (So ${E_1=I_1}$. )

We then take ${(c_1, d_1)=(b_{n_1}, d_0)}$ and take the smallest ${n_2}$ such that ${E_2:=I_{n_2}\subset (c_1, d_1)}$. In the third step, we take ${(c_2, d_2)=(c_1, a_{n_2})}$ and take the smallest ${n_3}$ such that ${E_3:=I_{n_3}\subset (c_2, d_2)}$. (i.e. ${E_3}$ is the “first” interval lying between ${E_1}$ and ${E_2}$. ) In general, we have

$\displaystyle \begin{array}{rcl} (c_{2k+1}, d_{2k+1})=(b_{n_{2k+1}}, d_{2k})&, &(c_{2k}, d_{2k})=(c_{2k-1}, a_{n_{2k}}), \\ n_i=\min\{n:I_{n}\subset (c_{i-1}, d_{i-1})\}&, &E_i=I_{n_i} \end{array}$

We now take a sequence ${x_i\in (c_i, d_i)}$, we can furthermore assume that ${x_i}$ converges to some number ${x\in (0,1)}$. (Note that ${\cdots\subsetneq (c_1, d_1)\subsetneq (c_0,d_0)=(0,1)}$ and ${0.)
Claim 1: If ${I_m\neq E_i}$ for any ${i}$, then ${x\notin I_m}$.
To see this, fix any such ${I_m}$, then there exists ${l}$ such that ${n_l. By definition, ${I_m}$ does not lie between ${E_{l-1}}$ and ${E_l}$ , whereas ${x_n}$ lies between ${E_{l-1}}$ and ${E_l}$ for all ${n\geq l}$. Therefore ${x_n}$ cannot converges to some point in ${I_m}$.
Claim 2: ${x}$ also does not belong to ${E_i}$ for any ${i}$.
To see this, fix any ${E_l}$. By construction, ${x_n}$ lies between ${E_{l+2}}$ and ${E_{l+1}}$ for all ${n\geq l+2}$, whereas ${E_l}$ does not lie between ${E_{l+2}}$ and ${E_{l+1}}$. From this it is easy to see that ${x_n}$ does not converge to any point in ${E_l}$.

We conclude that ${\displaystyle \bigcup _i I_i\neq (0,1)}$, a contradiction. $\Box$