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)


 * Sorry if it wasn't clear, but I meant that $x_0 < x_1 < x_2 < \cdots < x_n$ would be part of the definition, so if, in some theorem/proof/(other) definition, when we say "let $P = \left({x_0, x_1, x_2, \ldots, x_n}\right)$ be a subdivision", we don't have to mention that $x_0 < x_1 < x_2 < \cdots < x_n$ in the theorem/proof/definition, because it would already be incorporated into the definition of a subdivision.
 * I did not get this idea from any particular source, but there are course notes such as this one which take the approach I mentioned.
 * A lot of the pages that link to Definition:Subdivision just say "let $P = \left\{{x_0, x_1, x_2, \ldots, x_n}\right\}$ be a subdivision" without stipulating that $x_0 < x_1 < x_2 < \cdots < x_n$, and then proceed to define or prove things which only make sense given that the inequality holds. So either we replace $\left\{{ \ }\right\}$ with $\left({ \ }\right)$, or we add the phrase "where $x_0 < x_1 < x_2 < \cdots < x_n$" to all of these pages. My thought was that using $\left({ \ }\right)$ would be both a neater and less verbose way of correcting the pages. Thoughts? --abcxyz (talk) 01:29, 5 November 2012 (UTC)


 * No, I don't understand why it should be necessary. You talk about "correcting" pages as though they are somehow "wrong". --prime mover (talk) 06:12, 5 November 2012 (UTC)


 * The phrase "$x_0 < x_1 < x_2 < \cdots < x_n$" in Definition:Subdivision has no consequence whatsoever. So we will always have to write the condition that $x_0 < x_1 < x_2 < \cdots < x_n$ in all the definitions/proofs that need the $x_i$'s to be so. Currently, this is not done. --abcxyz (talk) 21:12, 9 November 2012 (UTC)


 * How about this. Define it as a finite strictly increasing sequence with the first term as $a$ and the last as $b$. That is certainly in the domain of a decent course on analysis. And of course make a note that loads of other courses present the definition currently up. --Jshflynn (talk) 21:37, 9 November 2012 (UTC)


 * That's basically what I just said. Comments on that? --abcxyz (talk) 21:56, 9 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)