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Is it possible to create a numerable partition of $[a,b]$? Because I think that it isn't possible, because the last point of the partition must be $b$. But I use the function defined in $[0,1]$ that: $$ f(t)=\frac{1}{2^{n-1}} \mbox{ if } t\in \left[\frac{1}{2^n},\frac{1}{2^{n-1}}\right] $$ I can't create a partition in order to have a point of the partition in every set where $f(t)$ changes its value. What do you think? Thanks for the help!

Peter Taylor
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rusca91
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  • What is a "numerable" partition? – hmakholm left over Monica Dec 18 '13 at 21:04
  • With infinity parts. Maybe the right word is countable, isn't it? Sorry! – rusca91 Dec 18 '13 at 21:05
  • How about letting each rational number in the interval be in its own singleton set, and then adding a single set containing all of the irrationals? That's denumerably many sets that partition $[a,b]$. – hmakholm left over Monica Dec 18 '13 at 21:07
  • I get your idea, but it seems like you never "arrive" to 1. I know, it is stupid what I'm saying... – rusca91 Dec 18 '13 at 21:12
  • What do you mean by "arriving"? One of the sets is ${1}$ since $1$ is rational. It's right there. You know that $\mathbb Q$ is countable, right? Otherwise just take $\left{a+\frac{b-a}{n}\right}$ as your singletons for each $n\ge 1$ and let the zeroth set set contain everything that doesn't have that form. – hmakholm left over Monica Dec 18 '13 at 21:13
  • Ok, but i can't right the partition like this: $a=t_0<\ldots<t_N=b$. That's what i mean – rusca91 Dec 18 '13 at 21:15
  • Usually a "partitioning" is just a set of pairwise-disjoint subsets whose unions is the given set. There's nothing about ordering implied by the term. – hmakholm left over Monica Dec 18 '13 at 21:20
  • Right, I'm a bit confused because in the definition of BV functions you have to work with finite partitions but sometimes you need a denumerable partition in order to get the total variation. Or maybe I'm wrong about this, too. Anyway, thanks a lot for the help! – rusca91 Dec 18 '13 at 21:22
  • Oh -- in that context it looks like (from a cursory review of the Wikipedia article) "partition" is used as a sloppy shorthand for "partitioning into intervals". In that case my examples aren't examples. – hmakholm left over Monica Dec 18 '13 at 21:32
  • So my idea is acceptable in that context? – rusca91 Dec 18 '13 at 21:36

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