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 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 to and requiring the family of the closed intervals to be non-trivial, or changing it to . 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 ) 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) and , is said to lie between and if either and , or and . If such happens to be a singleton , we say lies between and .
Proof:[One word about the proof: drawing a picture would be a good idea! ]
Suppose is a family of disjoint closed intervals in such that . Let with . We take , and take the smallest such that . (So . )
We then take and take the smallest such that . In the third step, we take and take the smallest such that . (i.e. is the “first” interval lying between and . ) In general, we have
We now take a sequence , we can furthermore assume that converges to some number . (Note that and .)
Claim 1: If for any , then .
To see this, fix any such , then there exists such that . By definition, does not lie between and , whereas lies between and for all . Therefore cannot converges to some point in .
Claim 2: also does not belong to for any .
To see this, fix any . By construction, lies between and for all , whereas does not lie between and . From this it is easy to see that does not converge to any point in .
We conclude that , a contradiction.