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<a_2} and {J\subset (b_1, a_2)}, or {b_2<a_1} 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<a_i\leq b_i<1}. 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<c_2<d_2<1}.)
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<m<n_{l+1}}. 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

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