Definition talk:Subdivision (Real Analysis)

Perhaps it would be more useful to define a subdivision as an ordered $\left({n+1}\right)$-tuple? Right now, a subdivision is just a finite subset $P \subseteq \left[{a \,. \, . \, b}\right]$ such that $a, b \in P$.

Also, we would only have to say:
 * "Let $P = \left({x_0, x_1, x_2, \ldots, x_n}\right)$ be a subdivision of $\left[{a \, . \, . \, b}\right]$."

as opposed to:
 * "Let $P = \left\{{x_0, x_1, x_2, \ldots, x_n}\right\}$ be a subdivision of $\left[{a \, . \, . \, b}\right]$, where $x_0 < x_1 < x_2 < \cdots < x_n$."

Comments? --abcxyz (talk) 17:58, 4 November 2012 (UTC)


 * No, even if you specify it as an ordered tuple you still have to specify the fact that $x_0 < x_1 < x_2 < \cdots < x_n$. This is analysis, and it pays us not to get into too much technical jargon from abstract algebra - many readers of a page such as this will be at a level of mathematics where they may not have encountered it. --prime mover (talk) 20:49, 4 November 2012 (UTC)


 * ... besides, the invocation of Definition:Subdivision already has the $x_0 < x_1 < x_2 < \cdots < x_n$ in it as part of the definition, so you already only need to raise the fact that it's a subdivision. Anyway, once Riemann integration has been defined you don't need the concept again. --prime mover (talk) 20:50, 4 November 2012 (UTC)


 * Well, I guess you could call it a finite sequence, if that makes more sense.
 * I meant that if we defined a subdivision $\left({x_0, x_1, x_2, \ldots, x_n}\right)$ as necessarily satisfying $x_0 < x_1 < x_2 < \cdots < x_n$, then we don't have to specify that when we let $\left({x_0, x_1, x_2, \ldots, x_n}\right)$ be a subdivision.
 * I thought that just because $P = \left\{{x_0, x_1, x_2, \ldots, x_n}\right\}$ is a subdivision doesn't mean that $x_0 < x_1 < x_2 < \cdots < x_n$ because, for example, $P = \left\{{x_1, x_0, x_2, \ldots, x_n}\right\}$ as well. So we still have to specify that $x_0 < x_1 < x_2 < \cdots < x_n$, right? --abcxyz (talk) 20:59, 4 November 2012 (UTC)


 * yes but even if you do write $\left({x_0, x_1, x_2, \ldots, x_n}\right)$ then yes you are adding the proviso that they have to go in a particular order, but you still have to mention that $x_0 < x_1 < x_2 < \cdots < x_n$. What's the point? And more relevant, do you have a source to back this up? --prime mover (talk) 21:18, 4 November 2012 (UTC)
 * Last year in my 1st year advanced calculus module it was presented to us like this. So PM is spot on regarding the audience. It is also presented in this way in "An introduction to analysis"-Kirkwood. My Analysis book for this term. --Jshflynn (talk) 21:40, 4 November 2012 (UTC)