Definition talk:Subdivision (Real Analysis)
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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 (Real Analysis) 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 (Real Analysis) 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 (Real Analysis) 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)