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This question asks about properties of functions which are "piecewise" polynomials. I would like to ask a specific question about the meaning of "piecewise" there.

Specifically, consider "partitions" of an open interval of the form $(c,d)=\overline{ \bigcup_{n=1 }^\infty (a_n, b_n)}$, i.e. into a countable infinity of pairwise disjoint open intervals which gives a set dense in $(c,d)$. As I elaborated on a comment of Aaron Tikuisis, this permits ugly constructions like $$(-1,1)=(-1,0)\cup\bigcup_{k=1}^\infty \left(\frac{1}{k+1},\frac{1}{k}\right)$$ or even $$ (0,1)=\bigcup_{k=1}^\infty \bigcup_{j=1}^\infty \left(\frac{1}{k+1}+\frac{1}{k(k+1)}\frac{1}{j+1},\frac{1}{k+1}+\frac{1}{k(k+1)}\frac{1}{j}\right),$$ for which a countable infinity of subintervals (those with $j=1$) have no next neighbour to the right.

My question is, how far can this process be pushed? Can one give a decomposition $(c,d)=\overline{ \bigcup_{n=1 }^\infty (a_n, b_n)}$ where none of the $(a_n,b_n)$ have a next neighbour to the right? or even to both sides? If this is impossible, then can one meaningfully characterize how many of the subintervals can have this property? Has this appeared in the literature?

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up vote 5 down vote accepted

Unless I'm misunderstanding your defintion of having an 'next' interval to the left or right, what about a Cantor set like construction: divide (0,1) into three equal intervals. Let $(\frac{1}{3},\frac{2}{3})$ be included in your decomposition, then divide $(0,\frac{1}{3})$ and $(\frac{2}{3},1)$ into three equal intervals, keeping the middle one of each and further subdividing the left and right pieces. Repeating this process ad nauseum gives a decomposition of (0,1) such that no interval has a nearest neighbor on either side but whose closure is all of (0,1).

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Yes, I think that does it. – Emilio Pisanty Apr 15 '12 at 15:28

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