Some say that length is the $\mathbb R^1$ equivalent of $\mathbb R^2$ area or $\mathbb R^3$ volume but you can substract areas or have intersections of areas. I need to do the same in one dimension but I don't which term to use. Is length really the good term for that?
4 Answers
$\begingroup$The typical abstraction of area is a measure which is also an abstraction of length so it would seem likely that it's correct for your application as well.
$\endgroup$ $\begingroup$What I was looking for was an Interval. I find the term way more intuitive and self descriptive than Length of even Measure. Thanks for the answers, but I'll stick with Interval.
Edit: I was wrong, it's a collection of intervals and not a single interval.
$\endgroup$ 10 $\begingroup$The topological volume $V_n(S)$ of a solid $S$ of dimension $n$ is defined as $V_n(S) = \int_S dV$.
As such, $V_2$ represents the surface (area) and $V_3$ represents the volume (usual 3D definition).
If we go through that reasoning, $V_1(S)$ refers to the length. So, your picture is correct if you are speaking above $V_1(S)$.
$\endgroup$ $\begingroup$What you are looking for are the elementary subsets of reals, which is defined as all the subsets that can be generated from the intervals, union, intersection and complement. If you add closure under countable union and countable intersection as well, then you get the Borel subsets.
If you want to be able to assign lengths to them, you might want countable additivity (length of a countable disjoint union is the sum of their lengths). It turns out that the Borel sets satisfy countable additivity under the Lebesgue measure.
These notions extend to higher dimensions. Note that in Euclidean geometry and many real-world applications the (simpler) Jordan measure suffices, though it does not satisfy countable additivity (since the rationals are not Jordan measurable) but only satisfies finite additivity.
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