# arclength as filtered limit

The length (http://planetmath.org/Rectifiable) of a rectifiable curve may be phrased as a filtered limit. To do this, we will define a filter of partitions of an interval $[a,b]$. Let ${\bf P}$ be the set of all ordered tuplets of distinct elements of $[a,b]$ whose entries are increasing:

 ${\bf P}=\{(t_{1},\ldots t_{n})\mid(a\leq t_{1}0)\}$

We shall refer to elements of ${\bf P}$ as partitions of the interval $[a,b]$. We shall say that $(t_{1},\ldots,t_{n})$ is a refinement of a partition $(s_{1},\ldots,s_{m})$ if $\{t_{1},\ldots,t_{n}\}\supset\{s_{1},\ldots,s_{m}\}$. Let ${\bf F}\subset\mathcal{P}({\bf P})$ be the set of all subsets of ${\bf P}$ such that, if a certain partition belongs to ${\bf F}$ then so do all refinements of that partition.

Let us see that ${\bf F}$ is a filter basis. Suppose that $A$ and $B$ are elements of ${\bf F}$. If a partition belongs to both $A$ and $B$ then every one of its refinements will also belong to both $A$ and $B$, hence $A\cap B\in{\bf F}$.

Next, note that, if a partition of $B$ is a refinement of a partition of $A$ then, by the triangle inequality, the length of $\Pi(B)$ is greater than the length of $\Pi(A)$. By definition, for every $\epsilon>0$, we can pick a partition $A$ such that the length of $\Pi(A)$ differs from the length of the curve by at most $\epsilon$. Since the length of $\Pi(B)$ for any partition $B$ refining $A$ lies between the length of $\Pi(A)$ and the length of the curve, we see that the length of $\Pi(B)$ will also differ by at most $\epsilon$, so the length of the curve is the limit of the length of polygonal lines according to the filter generated by ${\bf F}$.

Title arclength as filtered limit ArclengthAsFilteredLimit 2013-03-22 15:49:34 2013-03-22 15:49:34 rspuzio (6075) rspuzio (6075) 13 rspuzio (6075) Result msc 51N05